JEE Electrochemistry Questions

We need to find the total volume of oxygen evolved at STP during the entire electrolysis process.

We are given that 300 mg (0.3 g) of Cu is deposited, a current of 600 mA (0.6 A) is applied for 28 minutes, the molar mass of Cu is 63.54 g/mol, and the Faraday constant is 96500 C/mol.

We start by calculating the moles of Cu deposited:

$$\frac{0.3}{63.54} = 4.72 \times 10^{-3}$$ mol

Since each Cu²⁺ requires two electrons for reduction, the reaction is:

$$Cu^{2+} + 2e^- \rightarrow Cu$$

so the moles of electrons used are:

$$2 \times 4.72 \times 10^{-3} = 9.44 \times 10^{-3}$$ mol

At the anode during this phase, water is oxidized according to:

$$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$

so the moles of O₂ produced are:

$$\frac{9.44 \times 10^{-3}}{4} = 2.36 \times 10^{-3}$$ mol

After all Cu²⁺ is exhausted, only water electrolysis continues. The total charge passed is:

$$0.6 \times 28 \times 60 = 1008$$ C

which corresponds to:

$$\frac{1008}{96500} = 0.01044$$ mol of electrons

Using the same oxidation reaction at the anode, the moles of O₂ produced in this phase are:

$$\frac{0.01044}{4} = 2.61 \times 10^{-3}$$ mol

The total moles of O₂ evolved are the sum of both phases:

$$n_{O_2} = 2.36 \times 10^{-3} + 2.61 \times 10^{-3} = 4.97 \times 10^{-3}$$ mol

At STP, one mole of gas occupies 22400 mL, so the volume of O₂ is:

$$V = n \times 22400 = 4.97 \times 10^{-3} \times 22400 = 111.3 \text{ mL}$$

Therefore, rounding to the nearest integer gives a total volume of approximately 111 mL of O₂ at STP.

Electrolysis of 600 mL aqueous NaCl for 5 min changes pH to 12. Find the current.

We use the relation $$pH + pOH = 14$$, so with pH = 12, $$pOH = 14 - 12 = 2$$.

Therefore $$[OH^-] = 10^{-pOH} = 10^{-2} = 0.01$$ M.

With a volume of 600 mL (0.6 L), the moles of $$OH^-$$ produced are $$n_{OH^-} = [OH^-]\times V = 0.01\times0.6 = 0.006$$ mol.

At the cathode the reaction $$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$ shows that 2 moles of electrons produce 2 moles of $$OH^-$$, so the moles of electrons transferred equal the moles of $$OH^-$$, namely 0.006 mol.

The total charge passed is $$Q = n_e \times F = 0.006 \times 96500 = 579$$ C.

Since the electrolysis runs for 5 min (300 s), the current is $$I = \frac{Q}{t} = \frac{579}{300} = 1.93 \approx 2$$ A.

The answer is 2 A.

Given a 0.2 % (w/v) NaOH solution, 0.2 g of NaOH is present in 100 mL of solution.
Hence in $$1000\ \text{mL} = 1\ \text{dm}^3$$, mass of NaOH = $$0.2\times 10 = 2\ \text{g}$$.

Molar mass of NaOH = $$40\ \text{g mol}^{-1}$$.
Therefore molarity $$c$$ is
$$c = \frac{2\ \text{g}}{40\ \text{g mol}^{-1}} = 0.05\ \text{mol dm}^{-3}$$ $$-(1)$$

Resistivity of the solution, $$\rho = 870.0\ \text{m}\Omega\ \text{m}$$
$$1\ \text{m}\Omega = 10^{-3}\ \Omega$$, so
$$\rho = 870.0 \times 10^{-3}\ \Omega\ \text{m} = 0.870\ \Omega\ \text{m}$$ $$-(2)$$

Conductivity (specific conductance) $$\kappa$$ is the reciprocal of resistivity:
$$\kappa = \frac{1}{\rho} = \frac{1}{0.870} = 1.149\ \text{S m}^{-1}$$ $$-(3)$$

To work with molar conductivity we first convert $$\kappa$$ to $$\text{S cm}^{-1}$$:
$$1\ \text{S m}^{-1} = 0.01\ \text{S cm}^{-1}$$ (because $$1\ \text{m}=100\ \text{cm}$$ and conductivity $$\propto 1/{\text{length}}$$).
Hence
$$\kappa = 1.149 \times 0.01 = 0.01149\ \text{S cm}^{-1}$$ $$-(4)$$

Formula for molar conductivity $$\Lambda_m$$ (in $$\text{S cm}^2\ \text{mol}^{-1}$$):
$$\Lambda_m = \frac{1000\,\kappa}{c}$$ $$-(5)$$

Substituting $$\kappa$$ from $$(4)$$ and $$c$$ from $$(1)$$:
$$\Lambda_m = \frac{1000 \times 0.01149}{0.05} = 229.8\ \text{S cm}^2\ \text{mol}^{-1}$$ $$-(6)$$

Convert $$\Lambda_m$$ to the requested unit $$\text{mS dm}^2\ \text{mol}^{-1}$$.
Relation:
$$1\ \text{S cm}^2 = (1000\ \text{mS})(0.01\ \text{dm}^2) = 10\ \text{mS dm}^2$$.
Therefore
$$\Lambda_m = 229.8 \times 10 = 2298\ \text{mS dm}^2\ \text{mol}^{-1}$$ $$-(7)$$

Expressed as $$\Lambda_m = 22.98 \times 10^2\ \text{mS dm}^2\ \text{mol}^{-1}$$.
Nearest integer = $$23$$.

Hence the molar conductivity of the solution is 23 $$\times 10^2$$ mS dm$$^2$$ mol$$^{-1}$$.

The quinhydrone electrode acts as the anode and the silver electrode acts as the cathode.

Anode reaction

$$QH_2 \rightarrow Q+2H^++2e^-$$

$$E^\circ_{Q/QH_2}=+0.7\text{ V}$$

Cathode reaction

$$2Ag^++2e^- \rightarrow 2Ag$$

$$E^\circ_{Ag^+/Ag}=+0.8\text{ V}$$

The standard cell potential is

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$

$$E^\circ_{cell}=0.8-0.7=0.1\text{ V}$$

The overall cell reaction is

$$QH_2+2Ag^+ \rightarrow Q+2H^++2Ag$$

For the quinhydrone electrode, $$[Q]=[QH_2]$$, therefore they cancel out in the reaction quotient. Also, $$[Ag^+]=1\text{ M}$$.

The Nernst equation becomes

$$E_{cell}=E^\circ_{cell}-\frac{0.06}{2}\log[H^+]^2$$

Substituting the given values,

$$0.4=0.1-0.03\log[H^+]^2$$

$$0.4=0.1-0.06\log[H^+]$$

Since

$$pH=-\log[H^+]$$

we get

$$0.4=0.1+0.06(pH)$$

$$0.3=0.06(pH)$$

$$pH=5$$

The solution in the anodic compartment contains $$NH_4X$$, a salt of a weak base and a strong acid.

For such a salt,

$$pH=7-\frac{1}{2}pK_b-\frac{1}{2}\log C$$

Given,

$$pH=5$$

$$C=0.01\text{ M}=10^{-2}\text{ M}$$

Substituting,

$$5=7-\frac{1}{2}pK_b-\frac{1}{2}\log(10^{-2})$$

$$5=7-\frac{1}{2}pK_b+1$$

$$5=8-\frac{1}{2}pK_b$$

$$\frac{1}{2}pK_b=3$$

$$pK_b=6$$

Therefore,

$$\boxed{pK_b=6}$$

Initial data (volume of each solution = 1 L):

• $$[Cu^{2+}]_0 = 1.5\;M \Rightarrow n(Cu^{2+})_0 = 1.5\;mol$$
• $$[Ag^{+}]_0 = 0.2\;M \Rightarrow n(Ag^{+})_0 = 0.2\;mol$$

Step 1 : Effect of electrolysis on the $$Cu^{2+}/Cu$$ half-cell

Reaction at cathode: $$Cu^{2+}+2e^- \rightarrow Cu$$

1 F = 1 mol e⁻, therefore 1 F deposits $$\dfrac{1}{2}=0.5\;mol$$ of $$Cu^{2+}$$.

Remaining $$Cu^{2+}$$ moles: $$1.5-0.5=1.0\;mol$$
New concentration: $$[Cu^{2+}] = \dfrac{1.0}{1\;L}=1.0\;M$$

Step 2 : Effect of electrolysis on the $$Ag^{+}/Ag$$ half-cell

Reaction at cathode: $$Ag^{+}+e^- \rightarrow Ag$$

0.1 F = 0.1 mol e⁻, therefore 0.1 mol of $$Ag^{+}$$ is removed.

Remaining $$Ag^{+}$$ moles: $$0.2-0.1=0.1\;mol$$
New concentration: $$[Ag^{+}] = \dfrac{0.1}{1\;L}=0.1\;M$$

Step 3 : Construction of the galvanic cell

Standard reduction potentials: $$E^\circ_{Ag^{+}/Ag}=0.80\;V,\;E^\circ_{Cu^{2+}/Cu}=0.34\;V$$

Hence Ag⁺/Ag will act as the cathode and Cu/Cu²⁺ as the anode:

Cell: $$Cu(s)\,|\,Cu^{2+}(1.0\,M)\,||\,Ag^{+}(0.1\,M)\,|\,Ag(s)$$

Step 4 : Standard emf of the cell

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.80-0.34=0.46\;V$$

Step 5 : Emf under the given concentrations (Nernst equation)

Overall cell reaction: $$Cu(s)+2Ag^{+}\rightarrow Cu^{2+}+2Ag(s)$$

Number of electrons transferred, $$n=2$$.

Reaction quotient: $$Q=\dfrac{[Cu^{2+}]}{[Ag^{+}]^{2}}=\dfrac{1.0}{(0.1)^{2}}=100$$

Nernst equation (298 K):
$$E_{cell}=E^\circ_{cell}-\dfrac{0.0591}{n}\log Q$$
Using the given value $$\dfrac{2.303RT}{F}=0.06\;V$$:
$$E_{cell}=0.46-\dfrac{0.06}{2}\log 100$$

Since $$\log 100 = 2$$,
$$E_{cell}=0.46-(0.03)(2)=0.46-0.06=0.40\;V$$

Answer: The emf of the cell is $$0.40\;V$$, i.e. $$400\;mV$$.

For weak electrolytes such as $$NH_4OH$$ the degree of dissociation $$\alpha$$ is obtained from
$$\alpha \;=\;\dfrac{\Lambda_m(c)}{\Lambda_m^{\infty}}$$
where $$\Lambda_m(c)$$ is the molar conductance at concentration $$c$$ and $$\Lambda_m^{\infty}$$ is the molar conductance at infinite dilution.

Step 1: Calculate the ionic conductance of the ammonium ion.

By Kohlrausch’s Law of Independent Migration of Ions, for any electrolyte
$$\Lambda_m^{\infty} \;=\; \lambda_{+}^{\infty} \;+\; \lambda_{-}^{\infty}$$

For $$NH_4Cl$$,
$$\Lambda_m^{\infty}(NH_4Cl) \;=\; 185\;S\;cm^{2}\;mol^{-1}$$
and the ionic conductance of $$Cl^-$$ is
$$\lambda_{Cl^-}^{\infty} \;=\; 70\;S\;cm^{2}\;mol^{-1}$$

Hence,
$$\lambda_{NH_4^+}^{\infty} \;=\; \Lambda_m^{\infty}(NH_4Cl)\;-\;\lambda_{Cl^-}^{\infty}$$ $$\lambda_{NH_4^+}^{\infty} \;=\; 185\;-\;70 \;=\; 115\;S\;cm^{2}\;mol^{-1}$$

Step 2: Obtain $$\Lambda_m^{\infty}$$ for $$NH_4OH$$.

For $$NH_4OH$$ the ions present are $$NH_4^+$$ and $$OH^-$$, so
$$\Lambda_m^{\infty}(NH_4OH) \;=\; \lambda_{NH_4^+}^{\infty} \;+\; \lambda_{OH^-}^{\infty}$$

The ionic conductance of $$OH^-$$ is
$$\lambda_{OH^-}^{\infty} \;=\; 170\;S\;cm^{2}\;mol^{-1}$$

Therefore,
$$\Lambda_m^{\infty}(NH_4OH) \;=\; 115 \;+\; 170 \;=\; 285\;S\;cm^{2}\;mol^{-1}$$

Step 3: Compute the degree of dissociation.

Given molar conductance of $$0.02\;M$$ $$NH_4OH$$ solution:
$$\Lambda_m(0.02M) \;=\; 85.5\;S\;cm^{2}\;mol^{-1}$$

Hence,
$$\alpha \;=\;\dfrac{\Lambda_m(0.02M)}{\Lambda_m^{\infty}(NH_4OH)} \;=\;\dfrac{85.5}{285}$$

$$\alpha \;=\; 0.30 \;=\; 3.0 \times 10^{-1}$$

Step 4: Expressing in the required form.

The question states $$\alpha = x \times 10^{-1}$$. Comparison gives $$x \approx 3$$ (nearest integer).

Hence, the required value of $$x$$ is 3.

The given reduction half-cell is
$$Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^- \rightarrow 2Cr^{3+}(aq)+7H_2O(l)$$

Step 1: Write the Nernst equation
For a general reduction reaction, $$E = E^{0}-\dfrac{0.059}{n}\log_{10}Q$$ where
$$n =$$ number of electrons transferred (= 6 here) and
$$Q =$$ reaction quotient.

Step 2: Express the reaction quotient $$Q$$
Only aqueous species appear in $$Q$$, hence
$$Q = \dfrac{[Cr^{3+}]^{2}}{[Cr_2O_7^{2-}][H^+]^{14}}$$

Step 3: Insert the given concentration ratio
We are told $$\dfrac{[Cr^{3+}]^{2}}{[Cr_2O_7^{2-}]} = 10^{-6}$$.
Therefore
$$Q = \dfrac{10^{-6}}{[H^+]^{14}}$$

Step 4: Set the electrode potential to zero
The emf becomes zero when $$E = 0$$, so
$$0 = E^{0}-\dfrac{0.059}{6}\log Q$$
$$\Rightarrow \log Q = \dfrac{6E^{0}}{0.059}$$

Step 5: Calculate $$\log Q$$
Given $$E^{0} = 1.33\,$$V,
$$\log Q = \dfrac{6 \times 1.33}{0.059} = \dfrac{7.98}{0.059} \approx 135.25$$

Step 6: Relate $$\log Q$$ to $$[H^+]$$
From Step 3,
$$Q = \dfrac{10^{-6}}{[H^+]^{14}}$$
Take $$\log_{10}$$ of both sides:
$$\log Q = \log(10^{-6}) - 14\log[H^+]$$
$$\Rightarrow 135.25 = -6 - 14\log[H^+]$$

Step 7: Solve for $$\log[H^+]$$
$$-14\log[H^+] = 135.25 + 6 = 141.25$$
$$\log[H^+] = -\dfrac{141.25}{14} \approx -10.09$$

Step 8: Obtain the pH
$$[H^+] = 10^{-10.09}$$
$$pH = -\log[H^+] = 10.09 \approx 10$$

Therefore, the emf of the given half-cell becomes zero at pH = 10.

We need to determine how an electrochemical (galvanic) cell can be converted into an electrolytic cell.

Key Concepts:

A galvanic cell converts chemical energy to electrical energy through spontaneous redox reactions, producing an EMF equal to $$E^\circ_{cell}$$.

An electrolytic cell uses external electrical energy to drive non-spontaneous reactions.

Analysis of each option:

Option 1: Applying an external opposite potential lower than $$E^\circ_{cell}$$ -- This would reduce the current from the galvanic cell but not reverse the reaction. The cell would still function as a galvanic cell (albeit with reduced output). Incorrect.

Option 2: Reversing the flow of ions in the salt bridge -- The salt bridge maintains electrical neutrality. You cannot simply reverse ion flow without an external driving force. Incorrect.

Option 3: Applying an external opposite potential greater than $$E^\circ_{cell}$$ -- When the external potential exceeds the cell's EMF and is applied in the opposite direction, it overcomes the natural tendency of the cell and forces the reaction to proceed in the reverse (non-spontaneous) direction. This is exactly the principle of electrolysis and charging of batteries. Correct.

Option 4: Exchanging the electrodes -- Simply swapping electrode positions does not change the underlying chemistry or provide the energy needed to drive a non-spontaneous reaction. Incorrect.

The correct answer is Option 3: Applying an external opposite potential greater than $$E^\circ_{cell}$$.

The cell operates in alkaline medium, so all half-reactions are written with $$OH^-$$ or $$H_2O$$.

Case 1: Oxidation half-reaction (anode)
Hydrazine is oxidised to nitrogen gas. Balance it step-by-step:

1. Write the skeletal form: $$N_2H_4 \rightarrow N_2$$.

2. Balance nitrogen atoms (already equal).
3. To balance hydrogen, add 4 molecules of water on the product side: $$N_2H_4 \rightarrow N_2 + 4H_2O$$.
4. Balance oxygen by adding 4 $$OH^-$$ to the reactant side: $$N_2H_4 + 4OH^- \rightarrow N_2 + 4H_2O$$.
5. Finally, balance charge by adding 4 electrons on the product side: $$N_2H_4 + 4OH^- \rightarrow N_2 + 4H_2O + 4e^-$$.

This shows that $$OH^-$$ reacts with $$N_2H_4$$ at the anode to give $$N_2(g)$$ and water while releasing 4 electrons. Hence, Option A is correct.

Case 2: Reduction half-reaction (cathode)
Molecular oxygen gets reduced to hydroxide ions.

Write and balance in alkaline medium:
$$O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$$.

Thus at the cathode $$O_2$$ is converted into $$OH^-$$. Hence, Option C is correct.

Why Options B and D are incorrect

B. The cathode reaction does not involve decomposition of $$N_2H_4$$; it involves the reduction of $$O_2$$. Therefore the statement about nascent hydrogen reacting with oxygen is wrong.

D. The overall cell reaction obtained by adding the two balanced half-reactions is
$$N_2H_4 + O_2 \rightarrow N_2 + 2H_2O$$.
No oxides of nitrogen (NO, $$NO_2$$, etc.) appear; hence they are not major by-products.

Therefore the correct statements are:
Option A and Option C.

We need to identify which statement is NOT correct about rusting of iron.

Analysis of each statement:

Statement (1): "Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off."

This is INCORRECT. While an intact tin coating acts as a physical barrier preventing contact with moisture and air, once the tin coating peels off or gets scratched, the iron underneath actually corrodes faster. This is because tin (Sn) is less reactive than iron (Fe) — tin has a higher reduction potential ($$E°_{Sn^{2+}/Sn} = -0.14$$ V) compared to iron ($$E°_{Fe^{2+}/Fe} = -0.44$$ V). Therefore, when both are exposed, iron acts as the anode and preferentially corrodes, while tin acts as the cathode. This is in contrast to zinc coating (galvanizing), where zinc being more reactive than iron, sacrificially corrodes to protect iron even if the coating is damaged.

Statement (2): "When pH lies above 9 or 10, rusting of iron does not take place."

This is correct. In highly alkaline conditions (pH > 9-10), iron forms a protective oxide layer that is stable and prevents further corrosion. This is the principle behind adding alkali to boiler water.

Statement (3): "Dissolved acidic oxides SO$$_2$$, NO$$_2$$ in water act as catalyst in the process of rusting."

This is correct. Acidic oxides dissolve in water to form acids, lowering the pH and increasing the concentration of H$$^+$$ ions, which accelerates the electrochemical process of rusting.

Statement (4): "Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object."

This is correct. Rusting involves the formation of an electrochemical cell where iron acts as the anode (gets oxidized to Fe$$^{2+}$$) and oxygen is reduced at the cathode in the presence of water.

The correct answer is Option (1).

Curve B is a straight line with a small negative slope. This behaviour is characteristic of a strong electrolyte.

Strong electrolytes dissociate completely at all concentrations. The slight decrease in molar conductivity with increasing concentration is due to inter-ionic attractions (as described by the Debye-Hückel-Onsager equation).

Examples: KCl, NaCl, HCl.

Based on the graph of molar conductivity (Λm​) versus the square root of concentration (C1/2), we can identify the nature of electrolytes A and B by looking at how they behave as they approach infinite dilution (C→0).

Curve A shows a steep, non-linear increase in molar conductivity as the concentration approaches zero. This is characteristic of a weak electrolyte.Weak electrolytes do not dissociate completely. As the solution is diluted, the degree of dissociation increases rapidly (Ostwald’s Dilution Law), leading to a sharp spike in the number of ions available to carry current.

Examples: CH3​COOH (Acetic acid), NH4​OH.

We need to find the pressure of $$H_2$$ that makes the EMF of a hydrogen electrode zero in pure water at 25 degrees C. The half-cell reaction for the hydrogen electrode is $$2H^+(aq) + 2e^- \rightarrow H_2(g), \quad E^\circ = 0 \text{ V}$$ and the Nernst equation for this half-cell is $$E = E^\circ - \frac{RT}{2F}\ln\frac{P_{H_2}}{[H^+]^2}$$.

Setting $$E = 0$$ gives $$0 = 0 - \frac{RT}{2F}\ln\frac{P_{H_2}}{[H^+]^2}$$ which implies $$\ln\frac{P_{H_2}}{[H^+]^2} = 0$$ and hence $$\frac{P_{H_2}}{[H^+]^2} = 1$$. Therefore $$P_{H_2} = [H^+]^2$$.

In pure water at 25 degrees C, $$[H^+] = 10^{-7}$$ M and thus $$P_{H_2} = (10^{-7})^2 = 10^{-14}$$ bar. The correct answer is Option D: $$10^{-14}$$.

Metals used in battery industries: Mn (dry cell, alkaline), Ni (NiCd, NiMH), Cd (NiCd). Fe and Cr are not commonly used in batteries.

The answer is Option (1): B(Mn), C(Ni) and E(Cd) only.

We need to determine what quantity of silver is deposited when 1 coulomb of charge is passed through $$AgNO_3$$ solution. According to Faraday’s First Law of Electrolysis, the mass of a substance deposited at an electrode is directly proportional to the quantity of charge passed.

The mass deposited (m) is related to the charge (Q) by

$$ m = Z \times Q $$

where $$Z$$ is the electrochemical equivalent and $$Q$$ is the charge in coulombs. The electrochemical equivalent $$Z$$ is defined as the mass of a substance deposited when 1 coulomb of charge is passed through the solution.

Substituting $$Q = 1$$ coulomb gives

$$ m = Z \times 1 = Z $$

Therefore, the quantity of silver deposited is one electrochemical equivalent of silver.

Find the unit of the slope in the Debye-Huckel-Onsager equation plot.

$$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$$

where $$\Lambda_m$$ is molar conductivity, $$\Lambda_m^0$$ is limiting molar conductivity, $$A$$ is the slope, and $$c$$ is concentration.

The plot is $$\Lambda_m$$ (y-axis) vs $$\sqrt{c}$$ (x-axis).

$$[\Lambda_m] = \text{S cm}^2 \text{ mol}^{-1}$$ and $$[\sqrt{c}] = (\text{mol L}^{-1})^{1/2} = \text{mol}^{1/2} \text{ L}^{-1/2}$$.

$$[A] = \frac{[\Lambda_m]}{[\sqrt{c}]} = \frac{\text{S cm}^2 \text{ mol}^{-1}}{\text{mol}^{1/2} \text{ L}^{-1/2}} = \text{S cm}^2 \text{ mol}^{-1} \cdot \text{mol}^{-1/2} \cdot \text{L}^{1/2}$$

$$= \text{S cm}^2 \text{ mol}^{-3/2} \text{ L}^{1/2}$$

The correct answer is Option (2): S cm$$^2$$ mol$$^{-3/2}$$ L$$^{1/2}$$.

The standard reduction potentials are $$E^0_{(M^{2+}/M)} = 0.46$$ V for the half-reaction M^{2+} + 2e^- → M and $$E^0_{(X/X^{2-})} = 0.34$$ V for X + 2e^- → X^{2-}.

Considering the reaction $$M + X \rightarrow M^{2+} + X^{2-}$$, M is oxidized at the anode and X is reduced at the cathode. The cell potential is given by $$ E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{(X/X^{2-})} - E^0_{(M^{2+}/M)}, $$ which yields $$ E^0_{cell} = 0.34 - 0.46 = -0.12 \text{ V}. $$ Since this value is negative, the reaction is not spontaneous and option 1 is incorrect.

Next, for the reaction $$M^{2+} + X^{2-} \rightarrow M + X$$, M^{2+} is reduced at the cathode and X^{2-} is oxidized at the anode. The cell potential is $$ E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{(M^{2+}/M)} - E^0_{(X/X^{2-})}, $$ which gives $$ E^0_{cell} = 0.46 - 0.34 = 0.12 \text{ V}. $$ Because this value is positive, the reaction is spontaneous, confirming that option 4 is correct.

The potentials for options 2 and 3 would be ±0.80 V, which do not match ±0.12 V, so both are incorrect.

Therefore, the spontaneous reaction among the choices is $$M^{2+} + X^{2-} \rightarrow M + X$$ (option 4).

Identify the factor that does NOT affect electrolytic conductance of a solution.

Understand what electrolytic conductance depends on.

The electrolytic conductance (conductivity) of a solution depends on the following factors:

(i) Nature of the electrolyte: Strong electrolytes (like NaCl, HCl) dissociate completely and have higher conductance than weak electrolytes (like acetic acid).

(ii) Concentration of the electrolyte: Conductance changes with concentration due to changes in the number of ions available and their mobility.

(iii) Nature of the solvent: The dielectric constant and viscosity of the solvent affect the degree of dissociation and ion mobility.

(iv) Temperature: Higher temperature increases ionic mobility and degree of dissociation.

Identify what does NOT affect conductance.

The nature of the electrode (whether it is platinum, copper, graphite, etc.) does not affect the conductance of the solution itself. Electrodes serve as the interface for charge transfer but do not change the ionic conductivity of the bulk solution. The conductance is a property of the electrolyte solution, not of the electrodes.

(Note: Electrode material may affect measured cell constant or resistance at the electrode-solution interface, but not the intrinsic electrolytic conductance.)

The correct answer is Option (2): The nature of the electrode used.

Let us analyze each statement about hydrogen-oxygen fuel cells:

A. Has been used in spaceship: True. Hydrogen-oxygen fuel cells have been used in space missions, notably in the Apollo and Space Shuttle programs.

B. Has an efficiency of 40% to produce electricity: False. Hydrogen-oxygen fuel cells have a much higher efficiency, typically 60-70% (some even higher). The 40% figure is more typical of thermal power plants.

C. Uses aluminum as catalysts: False. Fuel cells typically use platinum (Pt) or other noble metals as catalysts, not aluminum.

D. Is eco-friendly: True. The only byproduct of a hydrogen-oxygen fuel cell is water ($$2H_2 + O_2 \rightarrow 2H_2O$$), making it eco-friendly with no harmful emissions.

E. Is actually a type of Galvanic cell only: True. A fuel cell is a type of galvanic (voltaic) cell that converts chemical energy directly into electrical energy through electrochemical reactions, just like a galvanic cell. The difference is that fuel is continuously supplied.

The correct statements are A, D, and E only.

The answer is Option B: A, D, E only.

We are given the molar ionic conductivities of a divalent cation and a divalent anion, and we need to find the molar conductivity of the electrolyte formed by them.

Identify the electrolyte formula.

A divalent cation has charge +2 (let us call it $$M^{2+}$$) and a divalent anion has charge -2 (let us call it $$X^{2-}$$). To form an electrically neutral compound, we need one cation and one anion: the formula is $$MX$$.

Apply Kohlrausch's Law of Independent Migration of Ions.

Kohlrausch's law states that the molar conductivity at infinite dilution of an electrolyte is the sum of the contributions of its individual ions:

$$ \Lambda_m^\infty = \nu_+ \lambda_+^\infty + \nu_- \lambda_-^\infty $$

where $$\nu_+$$ and $$\nu_-$$ are the number of cations and anions per formula unit, and $$\lambda_+^\infty$$ and $$\lambda_-^\infty$$ are the molar ionic conductivities.

Substitute the values.

For the electrolyte $$MX$$: $$\nu_+ = 1$$ and $$\nu_- = 1$$

$$ \Lambda_m = 1 \times 57 + 1 \times 73 = 57 + 73 = 130 \text{ S cm}^2 \text{ mol}^{-1} $$

Note: The molar ionic conductivities given (57 and 73 S cm$$^2$$ mol$$^{-1}$$) are already the individual ionic contributions. Since the formula unit contains one cation and one anion, we simply add them.

The correct answer is Option (3): 130 S cm$$^2$$ mol$$^{-1}$$.

The electrical ability of a substance is expressed through its conductivity $$\kappa$$, measured in $$\text{S m}^{-1}$$.

Empirically the following ranges are used at 298 K:

• Good conductors (metals, concentrated electrolyte solutions): $$\kappa \ge 1\; \text{S m}^{-1}$$
• Semiconductors: $$10^{-6}\; \text{S m}^{-1} \le \kappa \lt 1\; \text{S m}^{-1}$$
• Insulators: $$\kappa \lt 10^{-10}\; \text{S m}^{-1}$$

Hence any material whose conductivity is at least $$1\; \text{S m}^{-1}$$ will be counted as a conductor.

The given conductivities are

$$2.1 \times 10^{3},\; 1.0 \times 10^{-16},\; 1.2 \times 10^{1},\; 3.91,\; 1.5 \times 10^{-2},\; 1.0 \times 10^{-7},\; 1.0 \times 10^{3}\; \text{S m}^{-1}$$

Now compare each value with $$1\; \text{S m}^{-1}$$:

• $$2.1 \times 10^{3}\; (\;2100\;) \gt 1$$ ⇒ conductor
• $$1.0 \times 10^{-16} \lt 1$$ ⇒ insulator
• $$1.2 \times 10^{1}\; (\;12\;) \gt 1$$ ⇒ conductor
• $$3.91 \gt 1$$ ⇒ conductor
• $$1.5 \times 10^{-2}\; (\;0.015\;) \lt 1$$ ⇒ semiconductor
• $$1.0 \times 10^{-7} \lt 1$$ ⇒ insulator
• $$1.0 \times 10^{3}\; (\;1000\;) \gt 1$$ ⇒ conductor

Counting the conductors: $$2100,\; 12,\; 3.91,\; 1000$$ gives a total of $$4$$ substances.

Therefore, the number of conductors among the given materials is $$\mathbf{4}$$.

$$Cu^{2+} + 2e^- \rightarrow Cu$$. One Faraday = 1 mol electrons.

1 Faraday deposits $$\frac{1}{2}$$ mol Cu = 0.5 gram atom = 5 × 10⁻¹ gram atom. x = 5.

The answer is $$\boxed{5}$$.

Find the charge required for oxidation of 1 mol $$H_2O$$ to $$O_2$$.

$$ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- $$

This shows that 2 mol of water loses 4 mol of electrons to produce 1 mol of $$O_2$$.

From the balanced equation: 2 mol $$H_2O$$ gives 4 mol $$e^-$$.

Therefore: 1 mol $$H_2O$$ gives 2 mol $$e^-$$.

$$ Q = n_e \times F = 2 \times 96500 = 193000 \text{ C} = 1.93 \times 10^5 \text{ C} \approx 2 \times 10^5 \text{ C} $$

The answer is 2 $$\times 10^5$$ C.

$$Cr_2O_7^{2-}$$ has $$E° = 1.33$$ V (as oxidizing agent). It can oxidize metals whose reduction potential is less than 1.33 V.

The metals and their reduction potentials:

- Fe: $$Fe^{3+}/Fe$$ has $$E° = -0.04$$ V (< 1.33 V) — Can be oxidized ✓

- Ni: $$Ni^{2+}/Ni$$ has $$E° = -0.25$$ V (< 1.33 V) — Can be oxidized ✓

- Ag: $$Ag^+/Ag$$ has $$E° = 0.80$$ V (< 1.33 V) — Can be oxidized ✓

- Au: $$Au^{3+}/Au$$ has $$E° = 1.40$$ V (> 1.33 V) — Cannot be oxidized ✗

Number of metals oxidized by $$Cr_2O_7^{2-}$$ = 3 (Fe, Ni, Ag).

The answer is $$\boxed{3}$$.

• Reduction half-reaction:
  $$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad (\text{Electrons gained} = 5)$$
• Oxidation half-reaction:
  $$\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^- \quad (\text{Electrons lost} = 2)$$

To balance the overall redox equation, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.

• Total net electrons transferred ($$n$$): $$5 \times 2 = \mathbf{10}$$

2. Calculate Standard Cell Potential ($$E^0_{\text{cell}}$$)

Using the standard reduction potentials provided:

• $$E^0_{\text{cathode}} (\text{Reduction of MnO}_4^-) = +1.51\text{ V}$$
• $$E^0_{\text{anode}} (\text{Reduction of CO}_2) = -0.49\text{ V}$$

$$E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}$$

$$E^0_{\text{cell}} = 1.51\text{ V} - (-0.49\text{ V}) = \mathbf{2.00\text{ V}}$$

3. Connect $$E^0_{\text{cell}}$$ to the Equilibrium Constant ($$K_{eq}$$)

Using the Nernst equation relationship at equilibrium ($$298\text{ K}$$):

$$E^0_{\text{cell}} = \frac{0.0591}{n} \log_{10}(K_{eq})$$

Substitute the values $$E^0_{\text{cell}} = 2.00$$ and $$n = 10$$:

$$2.00 = \frac{0.0591}{10} \log_{10}(K_{eq})$$

$$20.0 = 0.0591 \log_{10}(K_{eq})$$

$$\log_{10}(K_{eq}) = \frac{20.0}{0.0591} \approx 338.4$$

x = 338

For the half-cell reaction $$2H^+(aq) + 2e^- \rightarrow H_2(g)$$ with $$[H^+] = 1$$ M and $$P_{H_2} = 2$$ atm, we wish to find $$x$$ such that the half-cell potential is $$-x \times 10^{-2}$$ V.

According to the Nernst equation, the cell potential $$E$$ is related to the standard potential $$E°$$ and the reaction quotient $$Q$$ by $$ E = E° - \frac{2.303RT}{nF}\log Q $$. For the hydrogen electrode, $$E° = 0$$ V by definition, and $$n = 2$$ electrons are transferred.

The reaction quotient for $$2H^+ + 2e^- \rightarrow H_2$$ is given by $$ Q = \frac{P_{H_2}}{[H^+]^2} = \frac{2}{1^2} = 2 $$.

Substituting the given value $$\frac{2.303RT}{F} = 0.06\text{ V}$$ along with $$n = 2$$ and $$\log 2 = 0.3$$ into the Nernst equation yields $$ E = 0 - \frac{0.06}{2} \times \log 2 = -0.03 \times 0.3 = -0.009 \text{ V} $$.

This value may be written as $$ E = -0.009 \text{ V} = -9 \times 10^{-3} \text{ V} \approx -1 \times 10^{-2} \text{ V} $$. Since $$-0.009 = -0.9 \times 10^{-2}$$, rounding to the nearest integer gives $$x = 1$$.

Therefore, the answer is $$x = 1$$.

We need to find the total charge passed through a solution of $$AuCl_4^-$$ during electrolysis. The cathode gained 1.314 g in mass, the electrolysis lasted 10.0 minutes, and the atomic mass of Au is 197 g/mol.

In $$AuCl_4^-$$, gold is in the $$+3$$ oxidation state since each Cl is $$-1$$, making Au $$+3$$. At the cathode, gold is deposited by reduction according to $$Au^{3+} + 3e^- \rightarrow Au$$, so 3 Faradays of charge deposit 1 mole (197 g) of gold.

The moles of Au deposited are given by $$\text{Moles of Au} = \frac{1.314}{197} = 0.00667 \text{ mol}$$.

Since 3 moles of electrons are required per mole of Au, the charge in Faradays is $$\text{Charge} = 3 \times 0.00667 = 0.02 \text{ F}$$, which can be expressed as $$0.02 \text{ F} = 2 \times 10^{-2} \text{ F}$$.

Therefore, the answer is 2.

For a standard hydrogen electrode in a solution of pH = 3:

$$E = E^o - \frac{2.303RT}{F}\text{pH} = 0 - 0.059 \times 3 = -0.177$$ V $$\approx -0.18$$ V

$$= -18 \times 10^{-2}$$ V.

The answer is $$\boxed{18}$$.

At STP, 5600 mL of $$O_2$$ = 5.6 L of $$O_2$$.

Moles of $$O_2 = \frac{5.6}{22.4} = 0.25$$ mol.

The electrolytic reaction for oxygen at the anode:

$$2H_2O \to O_2 + 4H^+ + 4e^-$$

So 1 mol of $$O_2$$ requires 4 mol of electrons. Therefore, 0.25 mol $$O_2$$ requires $$0.25 \times 4 = 1$$ mol of electrons.

For silver deposition at the cathode:

$$Ag^+ + e^- \to Ag$$

1 mol of electrons deposits 1 mol of Ag = 108 g.

Mass of silver displaced = $$1 \times 108 = 108$$ g.

The answer is $$\boxed{108}$$ g.

$$Q = It = 0.015 \times 15 \times 60 = 13.5$$ C.

Moles of electrons = $$13.5/96500$$.

Zn²⁺ + 2e⁻ → Zn. Moles of Zn = $$13.5/(2 \times 96500) = 6.994 \times 10^{-5}$$.

Mass = $$6.994 \times 10^{-5} \times 65.4 = 45.74 \times 10^{-4} \approx 46 \times 10^{-4}$$ g.

The answer is $$\boxed{46}$$.

In a conductometric titration of benzoic acid (weak acid) with NaOH (strong base):

Before the equivalence point: The weak acid is gradually neutralized. Benzoic acid is a weak electrolyte, so its solution has low conductance. As NaOH is added, the poorly conducting benzoic acid is replaced by highly conducting sodium benzoate. However, the OH⁻ ions (high conductivity) are consumed. Overall, conductance initially decreases slightly then increases as more sodium ions and benzoate ions are formed.

At equivalence point: All acid is neutralized.

After equivalence point: Excess NaOH adds highly mobile OH⁻ ions, sharply increasing conductance.

The graph shows a V-shape (decreasing then increasing after equivalence).

The correct answer is Option 1.

The reducing strength of an alkali metal in aqueous solution depends on its standard reduction potential ((E^\circ)). A more negative reduction potential corresponds to a stronger reducing agent, while a less negative value corresponds to a weaker reducing agent.

The standard reduction potential is influenced by sublimation energy, ionization energy, and hydration energy. Although lithium has the highest ionization energy among the alkali metals, the exceptionally large hydration energy of the small (Li^+) ion more than compensates for it. As a result, lithium has the most negative standard reduction potential ((E^\circ = -3.04\ \text{V})) and is therefore the strongest reducing agent in aqueous solution.

Among the given options, potassium, sodium, and rubidium have less negative standard reduction potentials than lithium. Hence, lithium cannot be the weakest reducing agent.

Therefore, the answer key indicating (Li) as the weakest reducing agent is incorrect. In aqueous solution, (Li) is actually the strongest reducing agent due to its very high hydration enthalpy.

The problem asks us to find which of the four given statements are incorrect. For each statement we compare it with the accepted thermodynamic relations for electrochemical cells.

Statement A
“The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.”

For an electrochemical cell operating reversibly at constant $$T$$ and $$P$$, the maximum non-expansion (electrical) work is related to the reaction Gibbs energy by the equation $$w_{\text{elec,max}} = -\Delta_r G$$. Thus electrical work is equal in magnitude but opposite in sign to $$\Delta_r G$$. Because Statement A omits the minus sign, it is incorrect.

Statement B
“$$E^\circ_{cell}$$ is dependent on the pressure.”

The standard cell potential $$E^\circ_{cell}$$ is defined for standard conditions: temperature $$T = 298\,\text{K}$$ (unless specified otherwise), all solute activities $$a = 1$$, and gases at a partial pressure of $$1\,\text{bar}$$. Since the pressure is fixed by definition (1 bar), $$E^\circ_{cell}$$ itself does not vary with pressure. Hence Statement B is incorrect.

Statement C
“$$\dfrac{dE_{cell}}{dT} = \dfrac{\Delta_r S}{nF}$$”

At constant pressure the temperature-dependence of the cell potential is given by $$\left(\dfrac{\partial E_{cell}}{\partial T}\right)_P = \dfrac{\Delta_r S}{nF}$$, where $$\Delta_r S$$ is the entropy change of the cell reaction, $$n$$ is the number of electrons transferred, and $$F$$ is the Faraday constant. Therefore Statement C is correct.

Statement D
“A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.”

For truly reversible operation the external (opposing) potential must equal the cell’s own emf so that an infinitesimal change will reverse the direction of current. Thus Statement D is correct.

To summarise, the incorrect statements are A and B.

If the option code “3” in the original question corresponds to the pair (A, B), then the given answer key is validated.

The reaction under consideration is $$\frac{1}{2}H_2(g) + AgCl(s) \rightleftharpoons H^+(aq) + Cl^-(aq) + Ag(s)$$.

The oxidation half‐reaction at the anode is $$\frac{1}{2}H_2(g) \rightarrow H^+(aq) + e^-$$ and the reduction half‐reaction at the cathode is $$AgCl(s) + e^- \rightarrow Ag(s) + Cl^-(aq)$$.

To assemble the galvanic cell, a platinum electrode is placed in contact with H₂ gas and H⁺ ions at the anode, while AgCl is in contact with Ag metal and Cl⁻ ions at the cathode. An HCl solution serves as the electrolyte, providing both H⁺ and Cl⁻ ions needed for the reactions.

The proper cell notation is $$\text{Pt} \,|\, H_2(g) \,|\, HCl(\text{sol}^n) \,|\, AgCl(s) \,|\, Ag$$ which corresponds to Option A.

Option C is not suitable because AgNO₃ would supply NO₃⁻ ions instead of the required Cl⁻, and Option D fails to provide H⁺ at the anode when using KCl. The correct answer is Option A.

During electrolysis of brine (concentrated NaCl solution):

At the anode (oxidation):

$$ 2Cl^- \rightarrow Cl_2 + 2e^- $$

Chlorine gas is liberated at the anode.

At the cathode (reduction):

$$ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- $$

Hydrogen gas is liberated and hydroxide ions ($$OH^-$$) are formed at the cathode.

Therefore, $$OH^-$$ is formed at the cathode.

For a weak monobasic acid $$HX$$, the degree of dissociation $$\alpha$$ is related to the molar conductivity by Ostwald’s dilution law:

$$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$
$$K_a = \frac{c\,\alpha^{2}}{1-\alpha}$$

Substitute $$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$$ in the expression of $$K_a$$:

$$K_a = \frac{c\left(\dfrac{\Lambda_m}{\Lambda_m^\circ}\right)^2}{1-\dfrac{\Lambda_m}{\Lambda_m^\circ}} = \frac{c\,\Lambda_m^{2}}{\Lambda_m^{\circ 2}-\Lambda_m\,\Lambda_m^\circ}$$

Multiply numerator and denominator by $$\Lambda_m^\circ$$ to simplify:

$$K_a = \frac{c\,\Lambda_m^{2}}{\Lambda_m^\circ\left(\Lambda_m^\circ-\Lambda_m\right)}$$

Re-arrange to isolate the linear relationship we need:

$$K_a\Lambda_m^\circ\left(\Lambda_m^\circ-\Lambda_m\right)=c\,\Lambda_m^{2}$$

Divide both sides by $$\Lambda_m\,\Lambda_m^{\circ 2}$$:

$$K_a\Bigl(\frac{\Lambda_m^\circ}{\Lambda_m}-1\Bigr)=\frac{c\,\Lambda_m}{\Lambda_m^{\circ 2}}$$

The term in parentheses simplifies since $$\dfrac{\Lambda_m^\circ}{\Lambda_m}-1=\dfrac{\Lambda_m^\circ-\Lambda_m}{\Lambda_m}$$, giving

$$K_a\Bigl(\frac{1}{\Lambda_m}-\frac{1}{\Lambda_m^\circ}\Bigr)=\frac{c\,\Lambda_m}{\Lambda_m^{\circ 2}}$$

Finally, separate the two fractions on the left:

$$K_a\left(\frac{1}{\Lambda_m}\right)=K_a\left(\frac{1}{\Lambda_m^\circ}\right)+\frac{c\,\Lambda_m}{\Lambda_m^{\circ 2}}$$

Divide every term by $$K_a$$ to obtain the required linear form:

$$\frac{1}{\Lambda_m}= \frac{1}{\Lambda_m^\circ} + \frac{c\,\Lambda_m}{K_a\,\Lambda_m^{\circ 2}}$$

Comparing with the straight-line equation $$y = P + Sx$$, let

$$y = \frac{1}{\Lambda_m}, \qquad x = c\,\Lambda_m$$

Intercept: $$P = \frac{1}{\Lambda_m^\circ}$$
Slope: $$S = \frac{1}{K_a\,\Lambda_m^{\circ 2}}$$

The required ratio is

$$\frac{P}{S}= \frac{\dfrac{1}{\Lambda_m^\circ}}{\dfrac{1}{K_a\,\Lambda_m^{\circ 2}}}=K_a\,\Lambda_m^\circ$$

Hence, $$\dfrac{P}{S}=K_a\Lambda_m^\circ$$.

Option A which is: $$K_a \Lambda_m^\circ$$

The standard electrode potential of M⁺/M can be understood through the thermodynamic cycle involving the following steps:

1. Sublimation of solid metal: M(s) → M(g), requiring sublimation enthalpy ($$\Delta_{sub}H$$)

2. Ionisation of gaseous metal atom: M(g) → M⁺(g) + e⁻, requiring ionisation enthalpy ($$\Delta_{i}H$$)

3. Hydration of gaseous metal ion: M⁺(g) → M⁺(aq), releasing hydration enthalpy ($$\Delta_{hyd}H$$)

The standard electrode potential depends on all three of these steps. It does not depend on "ionisation of a solid metal atom" because this is not a physically meaningful step in the thermodynamic cycle. Ionisation occurs in the gaseous state, not the solid state.

The correct answer is Ionisation of a solid metal atom.

We need to identify which M$$^{2+}$$ ions can liberate H$$_2$$ from dilute acid, given the standard electrode potentials E°(M$$^{3+}$$/M$$^{2+}$$).

Concept: For M$$^{2+}$$ to liberate H$$_2$$ from dilute acid, M$$^{2+}$$ must act as a reducing agent, i.e., it should be oxidized to M$$^{3+}$$ while reducing H$$^+$$ to H$$_2$$.

The reaction is: $$\text{M}^{2+} + \text{H}^+ \to \text{M}^{3+} + \frac{1}{2}\text{H}_2$$

For this reaction to be spontaneous, the overall cell potential must be positive:

$$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = E°(\text{H}^+/\text{H}_2) - E°(\text{M}^{3+}/\text{M}^{2+}) > 0$$

Since $$E°(\text{H}^+/\text{H}_2) = 0$$ V, we need:

$$E°(\text{M}^{3+}/\text{M}^{2+}) < 0$$

Checking the given values:

V: E° = $$-0.26$$ V (negative) — V$$^{2+}$$ CAN liberate H$$_2$$

Cr: E° = $$-0.41$$ V (negative) — Cr$$^{2+}$$ CAN liberate H$$_2$$

Mn: E° = $$+1.57$$ V (positive) — Mn$$^{2+}$$ CANNOT liberate H$$_2$$

Co: E° = $$+1.97$$ V (positive) — Co$$^{2+}$$ CANNOT liberate H$$_2$$

Therefore, V$$^{2+}$$ and Cr$$^{2+}$$ can liberate H$$_2$$ from dilute acid.

The correct answer is Option 3: V$$^{2+}$$ and Cr$$^{2+}$$.

From the graph, Electrolyte B shows a linear variation of molar conductivity ((\Lambda_m)) with (\sqrt{c}), indicating that it is a strong electrolyte. Electrolyte A shows a curved variation, which is characteristic of a weak electrolyte.

For Electrolyte A, the limiting molar conductivity ((\Lambda_m^0)) cannot be obtained by direct extrapolation because the (\Lambda_m) vs. (\sqrt{c}) plot is not linear. Instead, it is determined using Kohlrausch’s law of independent migration of ions. Hence, Statement (A) is incorrect.

For Electrolyte B, the Debye-Hückel-Onsager equation is

$$\Lambda_m=\Lambda_m^0-A\sqrt{c},$$

which represents a straight line with intercept equal to (\Lambda_m^0). Hence, Statement (B) is correct.

A strong electrolyte is almost completely dissociated, and at infinite dilution its degree of dissociation approaches unity, not zero. Therefore, Statement (C) is incorrect.

Kohlrausch’s law allows the calculation of the limiting molar conductivity ((\Lambda_m^0)) from the limiting ionic conductivities ((\lambda^\circ)) of the constituent ions. It cannot be used to determine (\Lambda_m) at any arbitrary concentration. Hence, Statement (D) is incorrect.

Therefore, Statements (A), (C), and (D) are incorrect, while Statement (B) is correct.

Hence, the number of incorrect statements is

$$\boxed{3}.$$

We need to identify the product that is NOT obtained during the electrolysis of brine solution (concentrated aqueous NaCl).

When brine is electrolyzed, water at the cathode is reduced instead of Na+ because its reduction potential is lower, following the reaction $$ 2\text{H}_2\text{O} + 2e^- \to \text{H}_2(g) + 2\text{OH}^- $$. This yields hydrogen gas (H₂).

At the anode, chloride ions undergo oxidation according to $$ 2\text{Cl}^- \to \text{Cl}_2(g) + 2e^- $$, producing chlorine gas (Cl₂).

The hydroxide ions generated at the cathode then combine with sodium ions in the solution to form sodium hydroxide (NaOH).

Considering the options, Option A (H₂) is indeed produced at the cathode, Option B (HCl) does not form during brine electrolysis, Option C (NaOH) is formed in solution, and Option D (Cl₂) is produced at the anode.

Thus HCl is not a product of brine electrolysis, making the correct answer Option B: HCl.

Given reduction potentials:

$$FeO_4^{2-} + 3e^- \to Fe^{3+}$$, $$E° = +2.2$$ V

$$Fe^{3+} + e^- \to Fe^{2+}$$, $$E° = +0.70$$ V

We need $$E°$$ for: $$FeO_4^{2-} + 4e^- \to Fe^{2+}$$

Using the relationship $$\Delta G° = -nFE°$$:

$$\Delta G_1 = -3F(2.2) = -6.6F$$

$$\Delta G_2 = -1F(0.70) = -0.70F$$

For the combined reaction:

$$\Delta G_{total} = \Delta G_1 + \Delta G_2 = -6.6F - 0.70F = -7.30F$$

$$E°_{FeO_4^{2-}/Fe^{2+}} = \frac{7.30}{4} = 1.825 \text{ V} = 1825 \times 10^{-3} \text{ V}$$

The value of x is 1825.

The given electrochemical cell is

$$Pt(s)|H_2(g,1\ \text{bar})|H^+(aq,1\ M)||M^{3+}(aq),M^+(aq)|Pt(s)$$

The corresponding half-cell reactions are:

• Anode (oxidation):

$$H_2(g)\rightarrow 2H^+(aq)+2e^-$$

• Cathode (reduction):

$$M^{3+}(aq)+2e^-\rightarrow M^+(aq)$$

Hence, the overall cell reaction is

$$H_2(g)+M^{3+}(aq)\rightarrow 2H^+(aq)+M^+(aq)$$

The standard cell potential is

$$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$$

$$E^\circ_{\text{cell}}=0.20-0=0.20\ \text{V}$$

Using the Nernst equation at $$298\ \text{K}$$,

$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.059}{n}\log Q$$

where $$n=2$$ and

$$Q=\frac{[H^+]^2[M^+]}{[H_2][M^{3+}]}$$

Since $$[H^+]=1\ \text{M}$$ and $$P_{H_2}=1\ \text{bar}$$,

$$Q=\frac{[M^+]}{[M^{3+}]}$$

Substituting the given values,

$$0.1115=0.20-\frac{0.059}{2}\log\left(\frac{[M^+]}{[M^{3+}]}\right)$$

$$0.1115=0.20-0.0295\log\left(\frac{[M^+]}{[M^{3+}]}\right)$$

Rearranging,

$$0.0295\log\left(\frac{[M^+]}{[M^{3+}]}\right)=0.20-0.1115$$

$$0.0295\log\left(\frac{[M^+]}{[M^{3+}]}\right)=0.0885$$

$$\log\left(\frac{[M^+]}{[M^{3+}]}\right)=\frac{0.0885}{0.0295}=3$$

Since

$$\frac{[M^+]}{[M^{3+}]}=10^a$$

we have

$$\log(10^a)=a=3$$

Conclusion

The required value is

$$\boxed{a=3}$$

Hence, the correct answer is 3.

Analyzing each statement:

(A) "Conductivity always decreases with decrease in concentration for both strong and weak electrolytes." - TRUE. Conductivity (κ) decreases with dilution because fewer ions are available per unit volume.

(B) "The number of ions per unit volume increases on dilution." - FALSE. On dilution, concentration decreases, so the number of ions per unit volume decreases (for strong electrolytes). For weak electrolytes, degree of dissociation increases but overall number of ions per unit volume still decreases.

(C) "Molar conductivity increases with decrease in concentration." - TRUE. Molar conductivity increases on dilution for both strong and weak electrolytes.

(D) "The variation in molar conductivity is different for strong and weak electrolytes." - TRUE. Strong electrolytes show slight increase (Debye-Hückel-Onsager theory), while weak electrolytes show dramatic increase due to increased dissociation.

(E) "For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation." - FALSE. The change is due to INCREASE in degree of dissociation (not decrease).

Correct statements: A, C, D → 3 correct statements.

The answer is $$\mathbf{3}$$.

We need to find the conductivity of the solution when $$1 \times 10^{-5}$$ M AgNO$$_3$$ is added to 1 L of saturated AgBr solution at 298 K.

Identify ions in solution.

AgNO$$_3$$ dissociates completely: AgNO$$_3$$ $$\to$$ Ag$$^+$$ + NO$$_3^-$$.

AgBr dissolves sparingly: AgBr $$\rightleftharpoons$$ Ag$$^+$$ + Br$$^-$$.

The ions present are Ag$$^+$$, Br$$^-$$, and NO$$_3^-$$.

Find ion concentrations using common ion effect.

$$[\text{Ag}^+]$$ from AgNO$$_3$$ = $$1 \times 10^{-5}$$ M. The Ag$$^+$$ from AgBr dissolution is negligible compared to this, so total $$[\text{Ag}^+] \approx 1 \times 10^{-5}$$ M.

From the solubility product:

$$[\text{Br}^-] = \frac{K_{sp}}{[\text{Ag}^+]} = \frac{4.9 \times 10^{-13}}{1 \times 10^{-5}} = 4.9 \times 10^{-8} \text{ mol/L}$$

$$[\text{NO}_3^-] = 1 \times 10^{-5}$$ mol/L (from AgNO$$_3$$).

Convert concentrations to mol/m$$^3$$.

Since 1 mol/L = 1000 mol/m$$^3$$:

$$[\text{Ag}^+] = 1 \times 10^{-5} \times 1000 = 1 \times 10^{-2}$$ mol/m$$^3$$

$$[\text{Br}^-] = 4.9 \times 10^{-8} \times 1000 = 4.9 \times 10^{-5}$$ mol/m$$^3$$

$$[\text{NO}_3^-] = 1 \times 10^{-5} \times 1000 = 1 \times 10^{-2}$$ mol/m$$^3$$

Calculate conductivity using $$\kappa = \sum c_i \lambda_i^0$$.

Where $$c_i$$ is in mol/m$$^3$$ and $$\lambda_i^0$$ is in S m$$^2$$ mol$$^{-1}$$, the product gives $$\kappa$$ in S m$$^{-1}$$.

$$\kappa = [\text{Ag}^+]\lambda^0_{\text{Ag}^+} + [\text{Br}^-]\lambda^0_{\text{Br}^-} + [\text{NO}_3^-]\lambda^0_{\text{NO}_3^-}$$

$$= (1 \times 10^{-2})(6 \times 10^{-3}) + (4.9 \times 10^{-5})(8 \times 10^{-3}) + (1 \times 10^{-2})(7 \times 10^{-3})$$

$$= 6 \times 10^{-5} + 3.92 \times 10^{-7} + 7 \times 10^{-5}$$

Convert to the required units ($$\times 10^{-8}$$ S m$$^{-1}$$).

$$6 \times 10^{-5} = 6000 \times 10^{-8}$$ S m$$^{-1}$$

$$3.92 \times 10^{-7} = 39.2 \times 10^{-8}$$ S m$$^{-1}$$

$$7 \times 10^{-5} = 7000 \times 10^{-8}$$ S m$$^{-1}$$

$$\kappa = (6000 + 39.2 + 7000) \times 10^{-8} = 13039.2 \times 10^{-8} \text{ S m}^{-1}$$

Round to nearest integer.

$$\kappa \approx 13039 \times 10^{-8}$$ S m$$^{-1}$$.

The correct answer is 13039.

Given the electrochemical cell: Pt(s) | H$$_2$$(g)(1 atm) | H$$^+$$(aq, [H$$^+$$] = 1) || Fe$$^{3+}$$(aq), Fe$$^{2+}$$(aq) | Pt(s)

$$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.771$$ V, $$E_{\text{cell}} = 0.712$$ V.

Anode (oxidation): $$\frac{1}{2}\text{H}_2 \to \text{H}^+ + e^-$$

Cathode (reduction): $$\text{Fe}^{3+} + e^- \to \text{Fe}^{2+}$$

Here $$n = 1$$ (one electron transferred).

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$

Since the anode is SHE (standard), $$E^\circ_{\text{cell}} = 0.771 - 0 = 0.771$$ V.

The reaction quotient for the cathode half-cell: $$Q = \dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}$$.

$$0.712 = 0.771 - 0.0591 \times \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right)$$

$$0.0591 \times \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right) = 0.771 - 0.712 = 0.059$$

$$\log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right) = \frac{0.059}{0.0591} \approx 0.9983$$

$$\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10^{0.9983} \approx 10$$

The ratio of $$[\text{Fe}^{2+}]$$ to $$[\text{Fe}^{3+}]$$ is $$\mathbf{10}$$.

We need to find the electrode potential for the half-cell reaction $$Cr_2O_7^{2-} \rightarrow Cr^{3+}$$ under the given conditions.

We begin by writing the balanced half-cell reaction: $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$.

This is a reduction half-reaction involving the transfer of 6 electrons ($$n = 6$$). The standard electrode potential is $$E^0 = 1.330$$ V.

Next, we state the Nernst equation. The Nernst equation relates the electrode potential to the concentrations of species involved: $$E = E^0 - \frac{2.303RT}{nF}\log Q$$, where $$Q$$ is the reaction quotient. For this half-reaction: $$Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$$.

We identify the given values. Volume = 1 L, so concentrations equal mmol values in mol/L:
$$[Cr_2O_7^{2-}] = 10 \text{ mmol/L} = 0.01 \text{ M} = 10^{-2}$$ M
$$[Cr^{3+}] = 100 \text{ mmol/L} = 0.1 \text{ M} = 10^{-1}$$ M
pH = 3.0, so $$[H^+] = 10^{-3}$$ M
$$\frac{2.303RT}{F} = 0.059$$ V, so $$\frac{2.303RT}{nF} = \frac{0.059}{6}$$ V

Calculating the logarithm of the reaction quotient gives:
$$\log Q = \log\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$$
$$= \log\frac{(10^{-1})^2}{(10^{-2})(10^{-3})^{14}}$$
$$= \log\frac{10^{-2}}{10^{-2} \times 10^{-42}}$$
$$= \log\frac{10^{-2}}{10^{-44}}$$
$$= \log(10^{42}) = 42$$

Substituting into the Nernst equation yields:
$$E = 1.330 - \frac{0.059}{6} \times 42$$
$$E = 1.330 - 0.059 \times 7$$
$$E = 1.330 - 0.413$$
$$E = 0.917 \text{ V}$$

Therefore, expressed in the required form, $$E = 0.917 \text{ V} = 917 \times 10^{-3}$$ V, and hence $$x = \mathbf{917}$$.

We need to find the electrode potential for the half-cell reaction $$Cr_2O_7^{2-} \to Cr^{3+}$$. The balanced half-cell reaction is $$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$$.

Applying the Nernst equation, $$E = E^0 - \frac{0.059}{n}\log\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$$, and inserting the values [Cr_2O_7^{2-}] = 10 × 10^{-3} M = 0.01 M, [Cr^{3+}] = 100 × 10^{-3} M = 0.1 M, pH = 3.0 so [H^+] = 10^{-3} M, E^0 = 1.330 V, and n = 6, one computes the logarithmic term as follows.

$$\log\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} = \log\frac{(0.1)^2}{(0.01)(10^{-3})^{14}} = \log\frac{10^{-2}}{10^{-2} \times 10^{-42}} = \log\frac{10^{-2}}{10^{-44}} = \log 10^{42} = 42.$$

Substituting into the Nernst equation gives $$E = 1.330 - \frac{0.059}{6} \times 42 = 1.330 - 0.059 \times 7 = 1.330 - 0.413 = 0.917 \text{ V},$$ so that $$E = 917 \times 10^{-3} \text{ V}$$ and hence $$x = 917$$.

We have $$MnO_4^-(0.1M) | Mn^{2+}(0.001M)$$, electrode potential = 1.282 V, $$E^0 = 1.54$$ V, and $$\frac{2.303RT}{F} = 0.059$$ V.

The half-cell reaction is:

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Using the Nernst equation with $$n = 5$$:

$$E = E^0 - \frac{0.059}{n}\log\frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$$

Substituting the values:

$$1.282 = 1.54 - \frac{0.059}{5}\log\frac{0.001}{0.1 \times [H^+]^8}$$

$$-0.258 = -0.0118\left(-2 + 8\,\text{pH}\right)$$

$$\frac{0.258}{0.0118} = -2 + 8\,\text{pH}$$

$$21.864 = -2 + 8\,\text{pH}$$

$$8\,\text{pH} = 23.864$$

$$\text{pH} = 2.98 \approx 3$$

Hence, the pH is approximately $$3$$.

The cell reaction is

$$H_2(g)+2Fe^{3+}(aq)\rightarrow2H^+(aq)+2Fe^{2+}(aq).$$

The standard cell potential is

$$E_{\text{cell}}^\circ=E_{\text{cathode}}^\circ-E_{\text{anode}}^\circ=0.771-0=0.771\ \text{V}.$$

Using the Nernst equation,

$$E_{\text{cell}}=E_{\text{cell}}^\circ-\frac{0.06}{n}\log Q,$$

where (n=2) and

$$Q=\frac{[H^+]^2[Fe^{2+}]^2}{[H_2][Fe^{3+}]^2}.$$

Since ([H^+]=1\ \text{M}) and (P_{H_2}=1\ \text{atm}),

$$Q=\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)^2.$$

Substituting the given values,

$$0.712=0.771-\frac{0.06}{2}\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)^2.$$

Simplifying,

$$0.712=0.771-0.06\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right).$$

Therefore,

$$0.06\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)=0.771-0.712=0.059,$$

$$\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)=\frac{0.059}{0.06}\approx1.$$

Taking the antilogarithm,

$$\frac{[Fe^{2+}]}{[Fe^{3+}]}=10^1=10.$$

Hence,

$$\boxed{\frac{[Fe^{2+}]}{[Fe^{3+}]}=10}.$$

We are given:

$$E^0_{Pb^{2+}/Pb} = m$$ V (Pb$$^{2+}$$ + 2e$$^-$$ → Pb)

$$E^0_{Pb^{4+}/Pb} = n$$ V (Pb$$^{4+}$$ + 4e$$^-$$ → Pb)

We need to find $$E^0_{Pb^{4+}/Pb^{2+}}$$ (Pb$$^{4+}$$ + 2e$$^-$$ → Pb$$^{2+}$$).

Using Gibbs energy:

$$\Delta G_1 = -4Fn$$ (for Pb$$^{4+}$$ → Pb)

$$\Delta G_2 = -2Fm$$ (for Pb$$^{2+}$$ → Pb)

For the reaction Pb$$^{4+}$$ + 2e$$^-$$ → Pb$$^{2+}$$:

$$\Delta G_3 = \Delta G_1 - \Delta G_2 = -4Fn - (-2Fm) = -4Fn + 2Fm$$

$$E^0_{Pb^{4+}/Pb^{2+}} = -\frac{\Delta G_3}{2F} = \frac{4Fn - 2Fm}{2F} = 2n - m$$

The question asks for $$E^0_{Pb^{2+}/Pb^{4+}}$$ (oxidation potential) = $$-(2n - m) = m - 2n$$

Comparing with $$m - xn$$: $$x = 2$$

We need to find the standard electrode potential of Sn/Sn$$^{2+}$$ given the equilibrium constant and $$E^0_{\text{Zn}^{2+}/\text{Zn}}$$.

The reaction is: Zn(s) + Sn$$^{2+}$$(aq) $$\rightleftharpoons$$ Zn$$^{2+}$$(aq) + Sn(s)

At equilibrium:

$$E^0_{\text{cell}} = \frac{0.059}{n}\log K$$

Here $$n = 2$$ (two electrons transferred).

$$E^0_{\text{cell}} = \frac{0.059}{2}\log(1 \times 10^{20}) = \frac{0.059}{2} \times 20 = 0.59 \text{ V}$$

$$E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}$$

Zn is oxidized (anode), Sn$$^{2+}$$ is reduced (cathode):

$$E^0_{\text{cell}} = E^0_{\text{Sn}^{2+}/\text{Sn}} - E^0_{\text{Zn}^{2+}/\text{Zn}}$$

$$0.59 = E^0_{\text{Sn}^{2+}/\text{Sn}} - (-0.76)$$

$$E^0_{\text{Sn}^{2+}/\text{Sn}} = 0.59 - 0.76 = -0.17 \text{ V}$$

The question asks for the magnitude of $$E^0_{\text{Sn/Sn}^{2+}}$$.

Note: $$E^0_{\text{Sn/Sn}^{2+}}$$ (oxidation potential) = $$-E^0_{\text{Sn}^{2+}/\text{Sn}}$$ (reduction potential) = $$-(-0.17) = 0.17$$ V.

The magnitude = $$|{-0.17}| = 0.17$$ V $$= 17 \times 10^{-2}$$ V.

The correct answer is 17.

We need to find $$\log K$$ for the reaction $$Pd^{2+} + 4Cl^- \rightleftharpoons PdCl_4^{2-}$$.

To begin,

(1) $$Pd^{2+}(aq) + 2e^- \rightleftharpoons Pd(s)$$, $$E^\circ_1 = 0.83$$ V

(2) $$PdCl_4^{2-}(aq) + 2e^- \rightleftharpoons Pd(s) + 4Cl^-(aq)$$, $$E^\circ_2 = 0.65$$ V

Next,

Subtracting reaction (2) from reaction (1): we reverse reaction (2) and add to reaction (1).

Reversed (2): $$Pd(s) + 4Cl^- \rightarrow PdCl_4^{2-} + 2e^-$$, $$E^\circ = -0.65$$ V

Adding (1) + reversed (2):

$$ Pd^{2+} + 2e^- + Pd(s) + 4Cl^- \rightarrow Pd(s) + PdCl_4^{2-} + 2e^- $$

Cancelling $$Pd(s)$$ and $$2e^-$$ from both sides:

$$ Pd^{2+} + 4Cl^- \rightarrow PdCl_4^{2-} $$

$$ E^\circ_{\text{cell}} = E^\circ_1 - E^\circ_2 = 0.83 - 0.65 = 0.18\;\text{V} $$

From this,

At equilibrium, the Nernst equation gives:

$$ E^\circ_{\text{cell}} = \frac{2.303RT}{nF} \log K $$

We are given $$\frac{2.303RT}{F} = 0.06$$ V, and $$n = 2$$ electrons are transferred.

$$ 0.18 = \frac{0.06}{2} \times \log K $$

Wait -- let us be careful. The correct relation is:

$$ E^\circ = \frac{2.303RT}{nF} \log K = \frac{0.06}{n} \log K $$

$$ \log K = \frac{n \times E^\circ}{0.06} = \frac{2 \times 0.18}{0.06} = \frac{0.36}{0.06} = 6 $$

The logarithm of the equilibrium constant is 6.

For a redox reaction to be spontaneous under standard conditions, the standard cell potential must be positive:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \gt 0 \; -(1)$$

The nitrate ion in acidic medium is supplied as the oxidising agent:
$$NO_3^- + 4H^+ + 3e^- \rightarrow NO(g) + 2H_2O$$ with $$E^\circ = 0.97\;V$$

When $$NO_3^-$$ is reduced, it acts as the cathode. The metal that gets oxidised will therefore act as the anode. Let the general half-reaction for the metal be
$$M^{n+} + ne^- \rightarrow M(s)$$ with reduction potential $$E^\circ_{(M^{n+}/M)}$$

Putting this into $$(1)$$: $$E^\circ_{\text{cell}} = 0.97\;V - E^\circ_{(M^{n+}/M)}$$

For spontaneity $$E^\circ_{\text{cell}} \gt 0$$, so
$$0.97\;V - E^\circ_{(M^{n+}/M)} \gt 0$$ $$\Longrightarrow E^\circ_{(M^{n+}/M)} \lt 0.97\;V \; -(2)$$

Compare the given metal reduction potentials with $$0.97\;V$$:

• $$V^{2+} + 2e^- \rightarrow V(s)$$, $$E^\circ = -1.19\;V$$
• $$Fe^{3+} + 3e^- \rightarrow Fe(s)$$, $$E^\circ = -0.04\;V$$
• $$Ag^+ + e^- \rightarrow Ag(s)$$, $$E^\circ = 0.80\;V$$
• $$Au^{3+} + 3e^- \rightarrow Au(s)$$, $$E^\circ = 1.40\;V$$

According to inequality $$(2)$$, the metals satisfying $$E^\circ_{(M^{n+}/M)} \lt 0.97\;V$$ are V, Fe and Ag. Au does not satisfy the condition because $$1.40\;V \gt 0.97\;V$$.

Hence, the number of metals that will be oxidised by $$NO_3^-$$ in aqueous acidic solution is $$3$$.

We need to find the time required to electroplate a nickel layer on a metal surface.

• Surface area = 100 cm$$^2$$
• Thickness = 0.001 mm = 0.0001 cm
• Current $$I = 2$$ A
• Density of Ni, $$\rho = 10$$ g/mL
• Molar mass of Ni = 60 g/mol
• $$F = 96500$$ C/mol

Nickel is deposited as Ni$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Ni, so 2 moles of electrons per mole of Ni.

Rounding to the nearest integer: $$t = 161$$ seconds.

The answer is $$161$$.

We need to determine how many of the four statements about electrochemistry are correct.

Statement A: $$E_{\text{cell}}$$ is an intensive parameter.

This is correct. The cell potential (EMF) is an intensive property -- it does not depend on the amount of substance or the size of the cell. If you double the cell, the voltage remains the same (unlike charge or energy, which are extensive). This follows because $$E = \Delta G / (-nF)$$, where both $$\Delta G$$ and $$n$$ scale with the amount, leaving $$E$$ unchanged.

Statement B: A negative $$E^\circ$$ means the redox couple is a stronger reducing agent than the H$$^+$$/H$$_2$$ couple.

This is correct. A negative standard reduction potential means the substance is harder to reduce (or easier to oxidize) compared to H$$^+$$. The substance (in its reduced form) readily donates electrons, making it a stronger reducing agent than H$$_2$$. For example, Zn with $$E^\circ = -0.76$$ V is a stronger reducing agent than H$$_2$$.

Statement C: The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

This is correct. By Faraday's second law of electrolysis, the amount of electricity needed depends on the number of electrons involved in the electrode reaction. For example, reducing Cu$$^{2+}$$ requires 2 moles of electrons per mole, while reducing Ag$$^+$$ requires only 1. The stoichiometry determines the electron requirement.

Statement D: The amount of chemical reaction at any electrode during electrolysis is proportional to the quantity of electricity passed.

This is correct. This is Faraday's first law of electrolysis: $$m = \frac{MQ}{nF}$$, where $$m$$ is the mass deposited, $$Q$$ is the charge passed, $$M$$ is the molar mass, $$n$$ is the number of electrons, and $$F$$ is Faraday's constant. The mass is directly proportional to $$Q$$.

All 4 statements are correct. The answer is 4.

We need to find the dissociation constant of acetic acid given its specific conductance and limiting molar conductivity.

Concentration $$C = 0.0025$$ M, Specific conductance $$\kappa = 5 \times 10^{-5}$$ S cm$$^{-1}$$, Limiting molar conductivity $$\Lambda_m^{\circ} = 400$$ S cm$$^2$$ mol$$^{-1}$$.

The molar conductivity is related to specific conductance by:

where $$C$$ is in mol/L and $$\kappa$$ is in S cm$$^{-1}$$. The factor of 1000 converts L to cm$$^3$$.

The degree of dissociation is the ratio of molar conductivity to limiting molar conductivity:

For a weak electrolyte $$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$, the dissociation constant is:

Substituting the values:

Rounding to the nearest integer: $$K_a \approx 66 \times 10^{-7}$$.

The dissociation constant is 66 $$\times 10^{-7}$$.

At pH = 14, [OH⁻] = 1 M. Using K$$_{sp}$$ of Cu(OH)₂:

$$K_{sp} = [Cu^{2+}][OH^-]^2$$

$$1 \times 10^{-20} = [Cu^{2+}] \times (1)^2$$

$$[Cu^{2+}] = 10^{-20} \text{ M}$$

Using the Nernst equation for Cu²⁺/Cu:

$$E = E° - \frac{0.059}{n}\log\frac{1}{[Cu^{2+}]}$$

$$E = 0.34 - \frac{0.059}{2}\log\frac{1}{10^{-20}}$$

$$E = 0.34 - \frac{0.059}{2} \times 20$$

$$E = 0.34 - 0.59 = -0.25 \text{ V}$$

$$E = -25 \times 10^{-2} \text{ V}$$

Therefore $$x = \mathbf{25}$$.

High purity (>99.95% dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between nickel electrodes)

Statement from NCERT. Therefore, B is the right option.

We are given the temperature coefficients $$\left(\frac{\partial E}{\partial T}\right)_P$$ for four half-cells and need to determine which would be preferred as a reference electrode.

Understand the requirement for a reference electrode.

A reference electrode must have a stable and constant EMF that does not change significantly with temperature. Therefore, we need the half-cell whose EMF varies the least with temperature.

Compare the temperature coefficients.

The values of $$\left(\frac{\partial E}{\partial T}\right)_P$$ are:

A: $$1 \times 10^{-4}$$

B: $$2 \times 10^{-4}$$

C: $$0.1 \times 10^{-4} = 1 \times 10^{-5}$$

D: $$0.2 \times 10^{-4} = 2 \times 10^{-5}$$

Identify the smallest value.

The smallest temperature coefficient is for half-cell C with $$0.1 \times 10^{-4}$$.

This means half-cell C has the most stable EMF with respect to temperature changes, making it the best choice for a reference electrode.

The correct answer is Option C.

For any redox couple, the relation between the standard Gibbs free-energy change and the standard reduction potential is
$$\Delta G^\circ = -nF E^\circ$$  $$-(1)$$
where $$n$$ is the number of electrons and $$F$$ is the Faraday constant.

To find $$E^\circ_{MnO_4^-(aq)/Mn(s)}$$ we must combine the three given half-reactions so that their electron counts add up.

Step 1: $$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$$  $$n_1 = 3$$, $$E_1^\circ = 1.68\ \text{V}$$

Step 2: $$MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$$  $$n_2 = 2$$, $$E_2^\circ = 1.21\ \text{V}$$

Step 3: $$Mn^{2+} + 2e^- \rightarrow Mn(s)$$  $$n_3 = 2$$, $$E_3^\circ = -1.03\ \text{V}$$

The overall balanced reduction is obtained by adding the three steps:
$$MnO_4^- + 8H^+ + 7e^- \rightarrow Mn(s) + 4H_2O$$
Hence $$n_{\text{total}} = 7$$ electrons.

Using $$-(1)$$, the overall Gibbs free-energy change is the sum of the three individual changes:

$$\Delta G^\circ_{\text{overall}} = -F\,(n_1E_1^\circ + n_2E_2^\circ + n_3E_3^\circ)$$

Therefore the required potential is
$$E^\circ_{\text{overall}} = -\dfrac{\Delta G^\circ_{\text{overall}}}{n_{\text{total}}F} = \dfrac{n_1E_1^\circ + n_2E_2^\circ + n_3E_3^\circ}{n_{\text{total}}}$$

Substituting the numbers:

$$\begin{aligned} E^\circ_{\text{overall}} &= \dfrac{3(1.68) + 2(1.21) + 2(-1.03)}{7} \\ &= \dfrac{5.04 + 2.42 - 2.06}{7} \\ &= \dfrac{5.40}{7} \\ &\approx 0.77\ \text{V} \end{aligned}$$

Thus, the standard reduction potential of the couple $$MnO_4^-(aq)/Mn(s)$$ is $$0.77\ \text{V}$$.

For a strong electrolyte, Kohlrausch’s law of independent ionic migration states that its limiting molar conductivity is the sum of the ionic conductivities multiplied by their stoichiometric coefficients:
$$\Lambda^0=\sum \nu_i\,\lambda_i^0 \qquad -(1)$$

Let the stoichiometric coefficients in the three electrolytes be:

ZmXn    ⟶    $$m\,\text{Z}^{n+}+n\,\text{X}^{m-}$$
UmYp    ⟶    $$m\,\text{U}^{p+}+p\,\text{Y}^{m-}$$
VmXn    ⟶    $$m\,\text{V}^{n+}+n\,\text{X}^{m-}$$

The given ionic conductivities are:

$$\lambda^0_{\text{Z}^{n+}}=50.0,\; \lambda^0_{\text{U}^{p+}}=25.0,\; \lambda^0_{\text{V}^{n+}}=100.0,\; \lambda^0_{\text{X}^{m-}}=80.0,\; \lambda^0_{\text{Y}^{m-}}=100.0\;\;(\text{all in S cm}^2\text{ mol}^{-1}).$$

Electrolyte UmYp
Applying (1):
$$\Lambda^0_{U_mY_p}=m\,\lambda^0_{\text{U}^{p+}}+p\,\lambda^0_{\text{Y}^{m-}}$$ $$250 = 25\,m + 100\,p$$ Divide by 25:
$$m + 4p = 10 \qquad -(2)$$

Electrolyte VmXn
Similarly:
$$\Lambda^0_{V_mX_n}=m\,\lambda^0_{\text{V}^{n+}}+n\,\lambda^0_{\text{X}^{m-}}$$ $$440 = 100\,m + 80\,n$$ Divide by 20:
$$5m + 4n = 22 \qquad -(3)$$

Equations (2) and (3) must be satisfied by positive integers m, n, p.

From (3):
$$5m = 22 - 4n$$ The right-hand side must be a multiple of 5. Testing integer values of $$n$$:

n = 1 → 22 − 4 = 18 (not multiple of 5)
n = 2 → 22 − 8 = 14 (not multiple of 5)
n = 3 → 22 − 12 = 10 (multiple of 5) ⇒ $$m = \frac{10}{5}=2$$

No larger positive value of $$n$$ keeps the right side positive, so the unique solution is
$$m = 2,\; n = 3$$

Substitute $$m = 2$$ into (2):
$$2 + 4p = 10 \;\;\Rightarrow\;\; 4p = 8 \;\;\Rightarrow\;\; p = 2$$

Thus
$$m = 2,\; n = 3,\; p = 2$$

The required sum is
$$(m+n+p) = 2 + 3 + 2 = 7$$

Hence, the value of (m + n + p) is 7.

We need to evaluate both statements about molar conductivity.

Statement I: For KI, molar conductivity increases steeply with dilution. KI (potassium iodide) is a strong electrolyte, and for such electrolytes, the molar conductivity is already high at moderate concentrations because they are fully dissociated. Upon dilution, the molar conductivity increases only slightly rather than steeply, as this increase is mainly due to decreased inter-ionic interactions. This variation follows the Debye-Huckel-Onsager equation: $$\Lambda_m = \Lambda_m^{\circ} - A\sqrt{c}$$. Statement I is FALSE — the increase is slow, not steep.

Statement II: For carbonic acid, molar conductivity increases slowly with dilution. Carbonic acid ($$H_2CO_3$$) is a weak electrolyte, and in the case of weak electrolytes, molar conductivity increases steeply with dilution because the degree of dissociation increases significantly. At high concentrations, weak electrolytes are poorly dissociated, but dilution shifts the equilibrium towards more ionization, causing a steep rise in molar conductivity. Statement II is FALSE — the increase is steep, not slow.

Since both statements are false, the correct answer is Option B: Both Statement I and Statement II are false.

We have two identical conductivity cells filled with NaCl solutions: Cell 1 holds 10 moles in 20 mL and Cell 2 contains 20 moles in 80 mL, both showing the same conductivity.

Recalling that molar conductivity is given by

$$\Lambda_m = \frac{\kappa}{c}$$

where $$\kappa$$ is the conductivity and $$c$$ is the molar concentration (in mol/L).

For Cell 1, the concentration is

$$c_1 = \frac{10 \text{ mol}}{0.020 \text{ L}} = 500 \text{ mol/L}$$

In a similar way, for Cell 2 we find

$$c_2 = \frac{20 \text{ mol}}{0.080 \text{ L}} = 250 \text{ mol/L}$$

Since both cells exhibit the same conductivity, we write

$$\kappa_1 = \kappa_2 = \kappa$$

Hence, the molar conductivity of Cell 1 is

$$\Lambda_{m1} = \frac{\kappa}{c_1} = \frac{\kappa}{500}$$

and that of Cell 2 is

$$\Lambda_{m2} = \frac{\kappa}{c_2} = \frac{\kappa}{250}$$

Taking their ratio yields

$$\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{\kappa/250}{\kappa/500} = \frac{500}{250} = 2$$

It follows that

$$\therefore \Lambda_{m2} = 2\Lambda_{m1}$$

The answer is Option A.

We need to match each electrochemical reaction with the correct type of battery/cell.

A. $$Cd(s) + 2Ni(OH)_3(s) \to CdO(s) + 2Ni(OH)_2(s) + H_2O(l)$$

This is the Nickel-Cadmium (NiCd) cell discharge reaction. The NiCd cell is a secondary (rechargeable) battery, so this represents discharging of a secondary battery.

$$\Rightarrow$$ A matches with II

B. $$Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)$$

This is the reaction of a Mercury cell. Mercury cells are non-rechargeable, i.e., primary batteries.

$$\Rightarrow$$ B matches with I

C. $$2PbSO_4(s) + 2H_2O(l) \to Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$$

In a lead-acid battery, during discharge: $$Pb + PbO_2 + 2H_2SO_4 \to 2PbSO_4 + 2H_2O$$. The given reaction is the reverse, which represents charging of the secondary battery.

$$\Rightarrow$$ C matches with IV

D. $$2H_2(g) + O_2(g) \to 2H_2O(l)$$

This is the reaction in a Hydrogen-Oxygen fuel cell.

$$\Rightarrow$$ D matches with III

The correct matching is: A-II, B-I, C-IV, D-III

Therefore, the correct answer is Option C.

Adsorption is broadly classified into physisorption (physical adsorption) and chemisorption (chemical adsorption). Both kinds differ in the nature of the adsorbate-adsorbent interaction, the heat of adsorption, and the number of layers that can be formed.

Case A: Statement - “Chemisorption results in a unimolecular layer.”
Chemisorption involves the formation of chemical bonds (ionic or covalent) between the adsorbate molecules and the surface atoms of the adsorbent. Once every surface atom has formed one bond, no further molecules can be accommodated above the first layer because the force of attraction does not extend appreciably beyond the first chemisorbed layer. Hence chemisorption is always a monolayer (unimolecular layer) process.
Statement A is correct.

Case B: Statement - “The enthalpy change during physisorption is in the range of 100 - 140 kJ mol$$^{-1}$$.”
Physisorption is governed by weak van der Waals forces. The heat of adsorption (enthalpy change) is therefore low, typically $$20{-}40\ \text{kJ mol}^{-1}$$. A value of $$100{-}140\ \text{kJ mol}^{-1}$$ pertains to chemisorption, not physisorption.
Statement B is incorrect.

Case C: Statement - “Chemisorption is an endothermic process.”
Formation of chemical bonds releases energy, so the overall enthalpy change for chemisorption is negative (exothermic). Although an activation energy may be required to initiate bonding, the net process liberates heat.
Statement C is incorrect.

Case D: Statement - “Lowering the temperature favors physisorption processes.”
Physisorption is exothermic. According to Le Châtelier’s principle, decreasing temperature shifts an exothermic equilibrium towards the product side, i.e., more adsorption occurs. Therefore a lower temperature increases the extent of physisorption (until the gas condenses).
Statement D is correct.

Hence the correct choices are:
Option A (Chemisorption results in a unimolecular layer.)
Option D (Lowering the temperature favors physisorption processes.)

Option A, Option D

The ability of a metal to liberate $$H_2$$ from mineral acids depends on its reduction potential. A metal can liberate hydrogen from acids only if its standard reduction potential is less than that of hydrogen ($$E^\circ = 0$$ V).

The standard reduction potentials of the given metals are:

• $$Zn^{2+}/Zn$$: $$E^\circ = -0.76$$ V
• $$Mn^{2+}/Mn$$: $$E^\circ = -1.18$$ V
• $$Ni^{2+}/Ni$$: $$E^\circ = -0.25$$ V
• $$Cu^{2+}/Cu$$: $$E^\circ = +0.34$$ V

Among these, Cu has a positive standard reduction potential, meaning it is a weaker reducing agent than hydrogen. Copper cannot displace hydrogen from dilute mineral acids like $$HCl$$ or dilute $$H_2SO_4$$.

All the other metals (Mn, Ni, Zn) have negative reduction potentials and can liberate $$H_2$$ from mineral acids.

Therefore, Cu has the least tendency to liberate $$H_2$$ from mineral acids.

The correct answer is Option A.

We need to find the time required to deposit 0.3482 g of Fe from a solution of $$Fe_2(SO_4)_3$$.

$$Fe^{3+} + 3e^- \to Fe$$

The n-factor for $$Fe^{3+}$$ is 3 (three electrons are required per iron atom).

$$\text{Moles of Fe} = \frac{0.3482}{56} = 0.006218 \text{ mol}$$

$$Q = n \times F \times z$$

where $$n$$ = moles of Fe, $$z$$ = number of electrons = 3, $$F$$ = 96500 C/mol

$$Q = 0.006218 \times 3 \times 96500$$

$$Q = 0.018654 \times 96500$$

$$Q = 1800.1 \text{ C}$$

$$Q = I \times t$$

$$t = \frac{Q}{I} = \frac{1800.1}{1.5} = 1200.1 \text{ s}$$

$$t = \frac{1200.1}{60} = 20.0 \text{ min}$$

The value of $$x$$ is 20 minutes.

We have the cell $$Cu(s)|Cu^{2+}(0.001$$ M $$)||Ag^+(0.01$$ M $$)|Ag(s)$$ with a cell potential of 0.43 V at 298 K. We are given $$E^\theta_{Ag^+/Ag} = 0.80$$ V and $$\frac{2.303RT}{F} = 0.06$$ V. We need to find the magnitude of the standard electrode potential for $$Cu^{2+}/Cu$$, expressed as $$\_\_\_ \times 10^{-2}$$ V.

The cell reaction is: $$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$$

The standard cell potential is:

$$E^\theta_{cell} = E^\theta_{cathode} - E^\theta_{anode} = E^\theta_{Ag^+/Ag} - E^\theta_{Cu^{2+}/Cu} = 0.80 - E^\theta_{Cu^{2+}/Cu}$$

Now we apply the Nernst equation. For this reaction, the number of electrons transferred is $$n = 2$$:

$$E_{cell} = E^\theta_{cell} - \frac{0.06}{n}\log Q$$

The reaction quotient is:

$$Q = \frac{[Cu^{2+}]}{[Ag^+]^2} = \frac{0.001}{(0.01)^2} = \frac{10^{-3}}{10^{-4}} = 10$$

Substituting into the Nernst equation:

$$0.43 = E^\theta_{cell} - \frac{0.06}{2}\log 10$$

$$0.43 = E^\theta_{cell} - 0.03 \times 1$$

$$E^\theta_{cell} = 0.43 + 0.03 = 0.46$$ V

Now we can find $$E^\theta_{Cu^{2+}/Cu}$$:

$$0.46 = 0.80 - E^\theta_{Cu^{2+}/Cu}$$

$$E^\theta_{Cu^{2+}/Cu} = 0.80 - 0.46 = 0.34$$ V

Expressed as $$\_\_\_ \times 10^{-2}$$ V, we get $$0.34 = 34 \times 10^{-2}$$ V.

Hence, the correct answer is 34.

We need to find the standard electrode potential for $$\text{Sn}^{4+}/\text{Sn}^{2+}$$. The known electrode potentials are $$E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.140 \text{ V}$$ for the reaction $$\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$$ and $$E^\circ_{\text{Sn}^{4+}/\text{Sn}} = 0.010 \text{ V}$$ for $$\text{Sn}^{4+} + 4e^- \rightarrow \text{Sn}$$.

Since electrode potentials are not additive, we use the relation $$\Delta G^\circ = -nFE^\circ$$. For reaction (i), $$\Delta G_1^\circ = -2F(-0.140) = 0.280F$$, and for reaction (ii), $$\Delta G_2^\circ = -4F(0.010) = -0.040F$$.

The target reaction is $$\text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+}$$, which can be obtained by subtracting reaction (i) from reaction (ii). Thus, $$\Delta G_3^\circ = \Delta G_2^\circ - \Delta G_1^\circ = -0.040F - 0.280F = -0.320F$$.

With n = 2 for this reaction, $$\Delta G_3^\circ = -nFE^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}}$$ gives $$-0.320F = -2F \times E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}}$$ and hence $$E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = \frac{0.320}{2} = 0.160 \text{ V}$$.

Finally, $$|E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}}| = 0.160 \text{ V} = 16 \times 10^{-2} \text{ V}$$. Hence, the answer is 16.

We need to find the total volume of gases produced at STP during electrolysis of dilute H$$_2$$SO$$_4$$.

Current (I) = 0.10 A. Time (t) = 2 hours = 2 × 3600 = 7200 s. F = 96500 C mol$$^{-1}$$. Molar volume at STP = 22.7 L mol$$^{-1}$$ = 22700 cm$$^3$$ mol$$^{-1}$$.

$$Q = I \times t = 0.10 \times 7200 = 720$$ C

At cathode: $$2H^+ + 2e^- \rightarrow H_2$$. At anode: $$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$.

Moles of electrons = $$\frac{Q}{F} = \frac{720}{96500}$$ = 0.007461 mol

2 moles of electrons produce 1 mole of H$$_2$$. Moles of H$$_2$$ = $$\frac{0.007461}{2}$$ = 0.003731 mol

4 moles of electrons produce 1 mole of O$$_2$$. Moles of O$$_2$$ = $$\frac{0.007461}{4}$$ = 0.001865 mol

Total moles = 0.003731 + 0.001865 = 0.005596 mol

Volume = total moles × molar volume = 0.005596 × 22700 = 127.03 cm$$^3$$

$$\approx 127$$ cm$$^3$$

The total volume of gases produced at STP is 127 cm$$^3$$.

The electrochemical cell is Pt|H$$_2$$(g, 1 bar)|H$$^+$$(aq)||Cu$$^{2+}$$(aq)|Cu(s)
with $$E_{cell} = 0.31$$ V, pH = 3, $$E^\ominus_{Cu^{2+}/Cu} = 0.34$$ V, and $$\dfrac{2.303RT}{F} = 0.06$$ V.
The cell reaction is $$\text{H}_2(g) + \text{Cu}^{2+}(aq) \rightarrow 2\text{H}^+(aq) + \text{Cu}(s)$$
and the standard cell potential is $$E^\ominus_{cell} = 0.34 - 0 = 0.34 \text{ V}, \quad n = 2$$.

Applying the Nernst equation $$E_{cell} = E^\ominus_{cell} - \frac{0.06}{2}\log\frac{[\text{H}^+]^2}{[\text{Cu}^{2+}]}$$
and noting that pH = 3 gives $$[\text{H}^+] = 10^{-3}$$ M,
we write $$0.31 = 0.34 - 0.03 \times \log\frac{(10^{-3})^2}{[\text{Cu}^{2+}]} = 0.34 - 0.03 \times \log\frac{10^{-6}}{[\text{Cu}^{2+}]}$$.
Solving for $$[\text{Cu}^{2+}]$$, $$-0.03 = -0.03 \times \log\frac{10^{-6}}{[\text{Cu}^{2+}]}$$
gives $$\log\frac{10^{-6}}{[\text{Cu}^{2+}]} = 1,$$ $$\frac{10^{-6}}{[\text{Cu}^{2+}]} = 10,$$ $$[\text{Cu}^{2+}] = \frac{10^{-6}}{10} = 10^{-7} \text{ M}$$.
Since $$[\text{Cu}^{2+}] = 10^x$$ M, the value of $$x$$ is $$-7$$.

We need to find the number of electrons involved in the electrochemical cell reaction.

The cell is: $$Zn | Zn^{2+}(aq) || Sn^{x+} | Sn$$

Anode (oxidation): $$Zn \rightarrow Zn^{2+} + 2e^-$$

Cathode (reduction): $$Sn^{x+} + xe^- \rightarrow Sn$$

$$E^0_{cell} = E^0_{cathode} - E^0_{anode}$$

$$E^0_{cell} = E^0_{Sn^{x+}|Sn} - E^0_{Zn^{2+}|Zn}$$

$$= 0.008 - (-0.763) = 0.771 \text{ V}$$

$$E_{cell} = E^0_{cell} - \dfrac{0.06}{n} \log Q$$

where n is the number of electrons transferred and Q is the reaction quotient.

$$0.801 = 0.771 - \dfrac{0.06}{n} \log(10^{-2})$$

$$0.801 = 0.771 - \dfrac{0.06}{n} \times (-2)$$

$$0.801 = 0.771 + \dfrac{0.12}{n}$$

$$0.801 - 0.771 = \dfrac{0.12}{n}$$

$$0.03 = \dfrac{0.12}{n}$$

$$n = \dfrac{0.12}{0.03} = 4$$

With $$n = 4$$, the cathode half-reaction is: $$Sn^{4+} + 4e^- \rightarrow Sn$$, so $$x = 4$$.

The overall balanced reaction: $$2Zn + Sn^{4+} \rightarrow 2Zn^{2+} + Sn$$ (4 electrons transferred).

Hence, the answer is 4.

The resistance of the conductivity cell is

$$R = 1750 \; \Omega$$

and the conductivity of 0.01 M KCl solution is

$$\kappa = 0.152 \times 10^{-3} \text{ S cm}^{-1}$$

Since the cell constant is related to conductivity and resistance by

$$\text{Cell constant} = \kappa \times R$$

Substituting the given values yields

$$\text{Cell constant} = 0.152 \times 10^{-3} \times 1750$$

This gives

$$ = 0.266 \text{ cm}^{-1}$$

and equivalently

$$ = 266 \times 10^{-3} \text{ cm}^{-1}$$

Therefore, the cell constant is 266 $$\times 10^{-3} \text{ cm}^{-1}$$.

We are given a conductivity cell with cell constant $$G^* = 129$$ m$$^{-1}$$ and two KCl solutions. We need to find the ratio of their molar conductivities.

For Solution 1: concentration = 74.5 ppm, resistance = 100 $$\Omega$$.

For Solution 2: concentration = 149 ppm, resistance = 50 $$\Omega$$.

First, we find the conductivity ($$\kappa$$) of each solution using $$\kappa = \frac{G^*}{R}$$:

$$\kappa_1 = \frac{129}{100} = 1.29 \text{ S m}^{-1}$$

$$\kappa_2 = \frac{129}{50} = 2.58 \text{ S m}^{-1}$$

Now we find the molar concentrations. Since ppm means mg per litre (for dilute aqueous solutions) and the molar mass of KCl is 74.5 g/mol:

$$c_1 = \frac{74.5 \text{ mg/L}}{74.5 \text{ g/mol}} = \frac{74.5 \times 10^{-3}}{74.5} = 10^{-3} \text{ mol/L} = 1 \text{ mol/m}^3$$

$$c_2 = \frac{149 \text{ mg/L}}{74.5 \text{ g/mol}} = \frac{149 \times 10^{-3}}{74.5} = 2 \times 10^{-3} \text{ mol/L} = 2 \text{ mol/m}^3$$

The molar conductivity is given by $$\Lambda_m = \frac{\kappa}{c}$$ (where $$c$$ is in mol/m$$^3$$ when $$\kappa$$ is in S/m):

$$\Lambda_1 = \frac{1.29}{1} = 1.29 \text{ S m}^2 \text{ mol}^{-1}$$

$$\Lambda_2 = \frac{2.58}{2} = 1.29 \text{ S m}^2 \text{ mol}^{-1}$$

The ratio is:

$$\frac{\Lambda_1}{\Lambda_2} = \frac{1.29}{1.29} = 1$$

We are told $$\frac{\Lambda_1}{\Lambda_2} = x \times 10^{-3}$$. Therefore $$x \times 10^{-3} = 1$$, which gives $$x = 1000$$.

Hence, the correct answer is 1000.

We are given the cell: $$Pt|H_2(g)|H^+(aq)||Cu^{2+}(0.01 \text{ M})|Cu(s)$$ and the cell potential is $$E_{cell} = 0.576$$ V at 298 K.

The cell reaction is: $$ H_2(g) + Cu^{2+}(aq) \rightarrow 2H^+(aq) + Cu(s) $$

The standard cell potential is: $$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{H^+/H_2} = 0.34 - 0 = 0.34$$ V

Using the Nernst equation (with $$n = 2$$ electrons transferred): $$ E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log \frac{[H^+]^2}{[Cu^{2+}]} $$

Substituting the known values: $$ 0.576 = 0.34 - \frac{0.06}{2} \log \frac{[H^+]^2}{0.01} $$

$$ 0.576 = 0.34 - 0.03 \log \frac{[H^+]^2}{0.01} $$

$$ 0.576 - 0.34 = -0.03 \log \frac{[H^+]^2}{0.01} $$

$$ 0.236 = -0.03 \log \frac{[H^+]^2}{0.01} $$

$$ \frac{0.236}{-0.03} = \log \frac{[H^+]^2}{0.01} $$

$$ -7.867 = \log \frac{[H^+]^2}{0.01} $$

$$ \log [H^+]^2 - \log(0.01) = -7.867 $$

$$ 2\log [H^+] - (-2) = -7.867 $$

$$ 2\log [H^+] = -7.867 - 2 = -9.867 $$

$$ \log [H^+] = -4.933 $$

pH $$= -\log [H^+] = 4.933 \approx 5$$

Therefore, the pH of the solution is 5 (nearest integer).

We use Kohlrausch's Law of Independent Migration of Ions, which states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the limiting molar conductivities of its constituent ions.

We need to find $$\Lambda^\circ_{AgI}$$.

We can write:

$$\Lambda^\circ_{AgI} = \Lambda^\circ_{Ag^+} + \Lambda^\circ_{I^-}$$

Using the given data, we construct this from known electrolytes:

$$\Lambda^\circ_{AgI} = \Lambda^\circ_{AgNO_3} + \Lambda^\circ_{NaI} - \Lambda^\circ_{NaNO_3}$$

This works because:

$$\Lambda^\circ_{AgNO_3} = \Lambda^\circ_{Ag^+} + \Lambda^\circ_{NO_3^-}$$

$$\Lambda^\circ_{NaI} = \Lambda^\circ_{Na^+} + \Lambda^\circ_{I^-}$$

$$\Lambda^\circ_{NaNO_3} = \Lambda^\circ_{Na^+} + \Lambda^\circ_{NO_3^-}$$

So: $$\Lambda^\circ_{AgNO_3} + \Lambda^\circ_{NaI} - \Lambda^\circ_{NaNO_3} = \Lambda^\circ_{Ag^+} + \Lambda^\circ_{NO_3^-} + \Lambda^\circ_{Na^+} + \Lambda^\circ_{I^-} - \Lambda^\circ_{Na^+} - \Lambda^\circ_{NO_3^-}$$

$$= \Lambda^\circ_{Ag^+} + \Lambda^\circ_{I^-} = \Lambda^\circ_{AgI}$$

Substituting the values:

$$\Lambda^\circ_{AgI} = 13.3 + 12.7 - 12.0$$

$$= 14.0 \text{ mS m}^2 \text{ mol}^{-1}$$

Therefore, the limiting molar conductivity of AgI is 14 mS m$$^2$$ mol$$^{-1}$$.

We are given that the solubility product of $$A_2X_3$$ is $$K_{sp} = 1.1 \times 10^{-23}$$ and the specific conductance of the saturated solution is $$\kappa = 3 \times 10^{-5}$$ S m$$^{-1}$$. We need to find the limiting molar conductivity expressed as $$x \times 10^{-3}$$ S m$$^2$$ mol$$^{-1}$$.

The salt dissociates as $$A_2X_3 \rightarrow 2A^{3+} + 3X^{2-}$$. If the molar solubility is $$s$$, then $$[A^{3+}] = 2s$$ and $$[X^{2-}] = 3s$$.

Writing the solubility product: $$K_{sp} = (2s)^2(3s)^3 = 4s^2 \times 27s^3 = 108\,s^5$$.

Solving: $$s^5 = \frac{1.1 \times 10^{-23}}{108} \approx 1.019 \times 10^{-25}$$.

Taking the fifth root: $$(10^{-25})^{1/5} = 10^{-5}$$ and $$(1.019)^{1/5} \approx 1$$, so $$s \approx 10^{-5}$$ mol L$$^{-1}$$.

Converting to mol m$$^{-3}$$: $$s = 10^{-5} \times 10^3 = 10^{-2}$$ mol m$$^{-3}$$.

The molar conductivity is $$\Lambda_m = \frac{\kappa}{c} = \frac{3 \times 10^{-5}}{10^{-2}} = 3 \times 10^{-3}$$ S m$$^2$$ mol$$^{-1}$$.

So, the answer is $$3$$.

We need to find the Gibbs free energy change for the reaction:

$$Cu(s) + Sn^{2+}(0.001\text{ M}) \to Cu^{2+}(0.01\text{ M}) + Sn(s)$$

In this reaction, Cu is oxidized at the anode and Sn2+ is reduced at the cathode.

The standard cell potential is given by

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Sn^{2+}/Sn} - E^\circ_{Cu^{2+}/Cu}$$

Substituting the standard potentials yields

$$E^\circ_{cell} = -0.14 - 0.34 = -0.48 \text{ V}$$

The number of electrons transferred in the reaction is $$n = 2$$.

The reaction quotient is

$$Q = \frac{[Cu^{2+}]}{[Sn^{2+}]} = \frac{0.01}{0.001} = 10$$

Applying the Nernst equation gives

$$E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q = -0.48 - \frac{8.314 \times 298}{2 \times 96500}\ln(10)$$

Evaluating the numerical values yields

$$E_{cell} = -0.48 - \frac{2477.6}{193000} \times 2.3026 = -0.48 - 0.01284 \times 2.3026$$

$$E_{cell} = -0.48 - 0.02958 = -0.50958 \text{ V}$$

The Gibbs free energy change is related to the cell potential by

$$\Delta G = -nFE_{cell} = -2 \times 96500 \times (-0.50958)$$

Hence

$$\Delta G = 98348.54 \text{ J mol}^{-1} = 98.349 \text{ kJ mol}^{-1}$$

Given that $$\Delta G = x \times 10^{-1}$$ kJ mol−1, it follows that

$$x \times 10^{-1} = 98.349$$

Solving for x gives

$$x = 983.49 \approx 983$$

The value of $$x$$ is 983.

We need to find the value of $$\Delta_r G^\circ$$ for the electrochemical cell reaction. The cell notation is $$Pt(s)|H_2(g)|H^+(aq) || Ag^+(aq)|Ag(s)$$ and $$E_{cell} = +0.5332$$ V.

At the anode (oxidation, left half-cell) $$\dfrac{1}{2}H_2(g) \rightarrow H^+(aq) + e^-$$ and at the cathode (reduction, right half-cell) $$Ag^+(aq) + e^- \rightarrow Ag(s)$$, giving the overall reaction $$\dfrac{1}{2}H_2(g) + Ag^+(aq) \rightarrow H^+(aq) + Ag(s)$$ with number of electrons transferred $$n = 1$$.

The relationship between $$\Delta_r G^\circ$$ and $$E^\circ_{cell}$$ is given by $$\Delta_r G^\circ = -nFE^\circ_{cell}$$ where $$n = 1$$ (number of moles of electrons transferred), $$F = 96500$$ C mol$$^{-1}$$ (Faraday's constant), and $$E^\circ_{cell} = 0.5332$$ V.

Substituting the values yields $$\Delta_r G^\circ = -(1)(96500)(0.5332)$$ so that $$\Delta_r G^\circ = -96500 \times 0.5332$$. Computing the product term by term, $$96500 \times 0.5 = 48250$$, $$96500 \times 0.03 = 2895$$, $$96500 \times 0.003 = 289.5$$, and $$96500 \times 0.0002 = 19.3$$, hence $$96500 \times 0.5332 = 48250 + 2895 + 289.5 + 19.3 = 51453.8 \text{ J mol}^{-1}$$ and therefore $$\Delta_r G^\circ = -51453.8 \text{ J mol}^{-1} = -51.454 \text{ kJ mol}^{-1}$$.

Rounding to the nearest integer gives the magnitude of $$\Delta_r G^\circ$$ as approximately $$51$$ kJ mol$$^{-1}$$. Since the question asks for the value in kJ mol$$^{-1}$$ (nearest integer): $$|\Delta_r G^\circ| = 51$$ kJ mol$$^{-1}$$

We need to find the standard electrode potential for the spontaneous cell reaction. First, we identify the half-reactions and their standard reduction potentials: $$Fe^{3+} + e^- \to Fe^{2+}$$, $$E°= 0.77$$ V and $$I_2 + 2e^- \to 2I^-$$, $$E° = 0.54$$ V.

For a spontaneous reaction, the species with the higher reduction potential acts as the cathode and is reduced, whereas the one with the lower reduction potential acts as the anode and is oxidized. Therefore, the cathode (reduction) is $$Fe^{3+} + e^- \to Fe^{2+}$$, $$E° = 0.77$$ V and the anode (oxidation) is $$2I^- \to I_2 + 2e^-$$, $$E° = 0.54$$ V.

Then, the standard cell potential is calculated by subtracting the anode potential from the cathode potential: $$E°_{cell} = E°_{cathode} - E°_{anode} = 0.77 - 0.54 = 0.23 \text{ V}$$

Finally, we express this in the required form as $$E°_{cell} = 0.23$$ V $$= 23 \times 10^{-2}$$ V. Consequently, the correct answer is $$x = \mathbf{23}$$.

In $$\mathrm{Fe_3O_4}$$, the oxidation state of oxygen is:

$$\mathrm{O = -2}$$

Let the average oxidation state of iron be $$\mathrm{x}$$.

For a neutral compound:

$$\mathrm{3x + 4(-2) = 0}$$

$$\mathrm{3x - 8 = 0}$$

$$\mathrm{3x = 8}$$

$$\mathrm{x = +\frac{8}{3}}$$

Thus, iron is reduced from $$\mathrm{+\frac{8}{3}}$$ to $$\mathrm{0}$$.

For one mole of $$\mathrm{Fe_3O_4}$$:

$$\mathrm{3\ Fe^{+\frac{8}{3}} \longrightarrow 3\ Fe^0}$$

Total electrons required:

$$\mathrm{3\times\frac{8}{3} = 8\ mol\ e^-}$$

Since:

$$\mathrm{1\ mol\ e^- = 1\ F}$$

Charge required for producing $$\mathrm{3\ mol}$$ of iron:

$$\mathrm{= 8\ F}$$

Therefore, charge required for producing $$\mathrm{1\ mol}$$ of iron:

$$\mathrm{= \frac{8}{3}\ F}$$

$$\mathrm{= 2.667\ F}$$

Rounding to nearest integer:

$$\boxed{\mathrm{3}}$$

We need to find the quantity of electricity (in Faraday) required to reduce 1 mol of $$Cr_2O_7^{2-}$$ to $$Cr^{3+}$$.

First, determine the oxidation states: in $$Cr_2O_7^{2-}$$ each Cr is +6, while in $$Cr^{3+}$$ it is +3. Therefore each Cr atom gains 3 electrons according to

$$Cr^{+6} + 3e^- \to Cr^{+3}$$.

Since each $$Cr_2O_7^{2-}$$ ion contains 2 Cr atoms, the total number of electrons required per formula unit is $$2 \times 3 = 6\text{ electrons}$$. The balanced half-reaction is:

$$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$$.

Because one Faraday corresponds to one mole of electrons, reducing one mole of $$Cr_2O_7^{2-}$$ requires $$6\text{ Faraday}$$. Therefore, the answer is 6.

The conductivity (or ionic mobility) of alkali metal cations in water depends on their hydrated radius, not their bare ionic radius. In water, smaller ions have a higher charge density, which attracts more water molecules, forming a larger hydration shell. This larger hydrated ion moves more slowly through the solution, resulting in lower conductivity.

Among the alkali metal cations, the bare ionic radius increases as: $$Li^+ < Na^+ < K^+ < Rb^+ < Cs^+$$. However, the hydrated radius decreases in the same order because smaller bare ions attract more water molecules: hydrated radius of $$Li^+ > Na^+ > K^+ > Rb^+ > Cs^+$$.

Since a smaller hydrated radius means the ion can move faster through solution, the ionic conductivity increases as: $$Li^+ < Na^+ < K^+ < Rb^+ < Cs^+$$, or equivalently $$Cs^+ > Rb^+ > K^+ > Na^+$$.

This matches option (B): $$Cs^+ > Rb^+ > K^+ > Na^+$$.

We begin by recalling the definition of limiting molar conductivity (also called molar conductivity at infinite dilution). According to Kohlrausch’s Law of Independent Migration of Ions,

$$\Lambda_m^{\circ} = \lambda^{\circ}_{+} + \lambda^{\circ}_{-},$$

where $$\lambda^{\circ}_{+}$$ and $$\lambda^{\circ}_{-}$$ are the ionic molar conductivities of the cation and the anion at infinite dilution. These values depend only on the speed of the individual ions in a very dilute solution and are completely independent of whether the electrolyte is strong or weak.

For the strong electrolyte $$\text{KCl}$$ we have the ions $$\text{K}^{+}$$ and $$\text{Cl}^{-}$$. Their ionic molar conductivities at $$25^{\circ}\text{C}$$ are approximately

$$\lambda^{\circ}_{\text{K}^{+}} \approx 73.5\ \text{S cm}^{2}\text{mol}^{-1}, \qquad \lambda^{\circ}_{\text{Cl}^{-}} \approx 76.3\ \text{S cm}^{2}\text{mol}^{-1}.$$

Adding these, we obtain

$$\Lambda_m^{\circ}(\text{KCl}) = 73.5 + 76.3 = 149.8\ \text{S cm}^{2}\text{mol}^{-1}.$$

For the weak electrolyte $$\text{CH}_3\text{COOH}$$ the ions that eventually appear on complete (hypothetical) dissociation are $$\text{H}^{+}$$ and $$\text{CH}_3\text{COO}^{-}$$. Their ionic molar conductivities at $$25^{\circ}\text{C}$$ are approximately

$$\lambda^{\circ}_{\text{H}^{+}} \approx 349.6\ \text{S cm}^{2}\text{mol}^{-1}, \qquad \lambda^{\circ}_{\text{CH}_3\text{COO}^{-}} \approx 40.9\ \text{S cm}^{2}\text{mol}^{-1}.$$

Hence,

$$\Lambda_m^{\circ}(\text{CH}_3\text{COOH}) = 349.6 + 40.9 = 390.5\ \text{S cm}^{2}\text{mol}^{-1}.$$

Clearly, $$\Lambda_m^{\circ}(\text{CH}_3\text{COOH}) > \Lambda_m^{\circ}(\text{KCl}).$$ Therefore Statement I, which claims the opposite, is false.

Now we analyse how molar conductivity varies with concentration. The molar conductivity $$\Lambda_m$$ at any concentration $$c$$ is defined as

$$\Lambda_m = \frac{\kappa}{c},$$

where $$\kappa$$ is the measured conductivity of the solution. Experimentally we observe that on dilution (that is, when $$c$$ decreases), inter-ionic interactions become weaker, the ions move more freely, $$\kappa$$ does not drop as fast as $$c$$, and consequently $$\Lambda_m$$ actually increases. Hence molar conductivity increases with decrease in concentration for both strong and weak electrolytes.

Therefore Statement II, which says that molar conductivity decreases on dilution, is also false.

Since both statements are false, the option that asserts “Both Statement I and Statement II are false” is correct.

Hence, the correct answer is Option D.

We begin with the quantity called the cell constant. By definition, the cell constant is the ratio of the distance between the electrodes to the cross-sectional area of the electrodes:

$$\text{Cell constant}= \dfrac{\ell}{A}$$

Here $$\ell$$ is measured in metres (m) and $$A$$ in square metres (m$$^2$$). Hence

$$\text{Unit of cell constant}= \dfrac{\text{m}}{\text{m}^2}= \text{m}^{-1}$$

In List - II this corresponds to item (iii) $$\text{m}^{-1}$$. So we have the pair $$\text{(a)} \rightarrow \text{(iii)}$$.

Next we consider molar conductivity (also called molar conductance). The formula used is

$$\Lambda_m = \kappa \dfrac{1000}{C}$$

where $$\kappa$$ is the conductivity in $$\text{S m}^{-1}$$ or $$\text{S cm}^{-1}$$, and $$C$$ is concentration in $$\text{mol dm}^{-3}$$ (or $$\text{mol L}^{-1}$$). Working in centimetre units, the factor $$1000$$ converts dm$$^3$$ to cm$$^3$$, giving

$$\text{Unit of } \Lambda_m = \text{S cm}^{-1}\times \text{cm}^3\text{ mol}^{-1}= \text{S cm}^2\text{ mol}^{-1}$$

This is item (i) in List - II, so we have $$\text{(b)} \rightarrow \text{(i)}$$.

Now we take up conductivity (specific conductance) $$\kappa$$. From Ohm’s law for a cell of length $$\ell$$ and area $$A$$ we write

$$\kappa = \dfrac{1}{R}\dfrac{\ell}{A}$$

Resistance $$R$$ has unit $$\Omega$$, therefore $$1/R$$ has unit $$\Omega^{-1}$$ (siemens, S). Multiplying by $$\ell/A$$ introduces an additional $$\text{m}^{-1}$$, giving

$$\text{Unit of conductivity}= \Omega^{-1}\text{ m}^{-1} = \text{S m}^{-1}$$

This matches item (iv) in List - II. Hence $$\text{(c)} \rightarrow \text{(iv)}$$.

Lastly, the degree of dissociation $$\alpha$$ is defined as

$$\alpha = \dfrac{\text{number of moles dissociated}}{\text{total number of moles initially present}}$$

It is a pure ratio of two identical quantities, so it carries no physical unit; it is dimensionless. List - II item (ii) is “dimensionless,” giving $$\text{(d)} \rightarrow \text{(ii)}$$.

Collecting all the matches, we have

$$(a)-(iii),\; (b)-(i),\; (c)-(iv),\; (d)-(ii)$$

Comparing with the options, this set appears in Option A.

Hence, the correct answer is Option A.

We start with an aqueous, strongly acidified sulphate solution. In such a medium the dominant anion is the bisulphate ion, written as $$HSO_4^-$$.

At the anode of the electrolytic cell this bisulphate ion undergoes oxidation. First we write the oxidation half-reaction that is always quoted in standard texts:

$$2\,HSO_4^- \;\longrightarrow\; S_2O_8^{2-} + 2\,e^- + 2\,H^+$$

This tells us that two $$HSO_4^-$$ ions lose a total of two electrons and combine to give the peroxodisulphate ion $$S_2O_8^{2-}$$, while also releasing two protons. The key feature is the formation of the peroxy O-O linkage between the two $$SO_4$$ groups.

Because the solution is already acidified, the $$S_2O_8^{2-}$$ ions will immediately pick up the two available protons $$2\,H^+$$ (produced in the same reaction and present from the external acid) to give the free acid form. We therefore write:

$$S_2O_8^{2-} + 2\,H^+ \;\longrightarrow\; H_2S_2O_8$$

The molecular formula $$H_2S_2O_8$$ corresponds to peroxydisulphuric (Marshall’s) acid. To make its structure absolutely clear, we expand it, showing each sulphur atom bonded to three terminal oxygen atoms and sharing a central peroxy oxygen-oxygen bridge:

$$H-O_3S-O-O-SO_3-H$$

Written more compactly, but preserving the connectivity, this becomes $$HO_3SOOSO_3H$$.

Now we match this structural formula with the options given in the question. Option B is exactly $$HO_3SOOSO_3H$$, whereas the other options represent different species (bisulphate ion, sulphurous/peroxomonosulphuric variants, etc.).

Hence, the correct answer is Option B.

First, we recall the relation between the standard cell potential and the standard Gibbs free-energy change.

For a cell reaction in which $$n$$ electrons are transferred, the Nernst-derived thermodynamic formula is stated as

$$\Delta G^{\circ} = -\,nF\,E^{\circ}_{\text{cell}}.$$

In the given reaction, solid cadmium $$\mathrm{Cd(s)}$$ is oxidised to $$\mathrm{Cd^{2+}}$$ while the mercurous ion $$\mathrm{Hg_2^{2+}}$$ is reduced to $$\mathrm{2Hg(l)}$$. The balanced equation shows that two electrons move from cadmium to mercury, so we have

$$n = 2.$$

The Faraday constant is given as $$F = 96487 \text{ C mol}^{-1}$$ and the standard cell potential is $$E^{\circ}_{\text{cell}} = 4.315 \text{ V}$$. Substituting these values gives

$$\Delta G^{\circ} = -\,(2)(96487\ \text{C mol}^{-1})(4.315\ \text{V}).$$

Because $$1 \text{ V} = 1 \text{ J C}^{-1}$$, the product of coulombs and volts directly yields joules:

$$\Delta G^{\circ} = -\,2 \times 96487 \times 4.315\ \text{J mol}^{-1}.$$

We now carry out the multiplication step by step.

First,

$$96487 \times 4.315 = 96487 \times (4 + 0.315) = 96487 \times 4 + 96487 \times 0.315 = 385948 + 30393.405 = 416341.405\ \text{J mol}^{-1}.$$

Next, multiply by 2:

$$2 \times 416341.405 = 832682.81\ \text{J mol}^{-1}.$$

Therefore,

$$\Delta G^{\circ} = -\,832682.81\ \text{J mol}^{-1} \approx -\,8.327 \times 10^{5}\ \text{J mol}^{-1} = -\,832.7\ \text{kJ mol}^{-1}.$$

Now we connect the thermodynamic quantities $$\Delta H^{\circ},\; \Delta G^{\circ}$$ and $$\Delta S^{\circ}$$ with the basic relation from the definition of Gibbs free energy:

$$\boxed{\;\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\;}.$$

Rearranging for the entropy change, we obtain

$$\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T}.$$

The data give the standard enthalpy change as $$\Delta H^{\circ} = -\,825.2\ \text{kJ mol}^{-1}$$ and the temperature as $$T = 25^{\circ}\text{C} = 298\ \text{K}$$.

Substituting, while keeping kilojoules consistent, we write

$$\Delta S^{\circ} = \frac{-825.2\ \text{kJ mol}^{-1} - \bigl(-832.6828\ \text{kJ mol}^{-1}\bigr)}{298\ \text{K}}.$$

The numerator simplifies:

$$-825.2 - (-832.6828) = -825.2 + 832.6828 = 7.4828\ \text{kJ mol}^{-1}.$$

Now divide by the absolute temperature:

$$\Delta S^{\circ} = \frac{7.4828\ \text{kJ mol}^{-1}}{298\ \text{K}} = 0.0251\ \text{kJ K}^{-1}\text{ mol}^{-1}.$$

Converting to joules (1 kJ = 1000 J):

$$\Delta S^{\circ} = 0.0251 \times 1000 = 25.1\ \text{J K}^{-1}\text{ mol}^{-1}.$$

Rounding to the nearest integer yields

$$\boxed{25\ \text{J K}^{-1}}.$$

Hence, the correct answer is Option 25.

The cell constant $$G^*$$ of a conductivity cell is related to the conductivity $$\kappa$$ and the resistance $$R$$ by the formula $$\kappa = \frac{G^*}{R}$$, or equivalently $$G^* = \kappa \times R$$.

Using the KCl solution data to find the cell constant:

$$G^* = \kappa_{\text{KCl}} \times R_{\text{KCl}} = 0.14 \times 4.19 = 0.5866 \text{ m}^{-1}$$

Now using this cell constant with the HCl solution:

$$\kappa_{\text{HCl}} = \frac{G^*}{R_{\text{HCl}}} = \frac{0.5866}{1.03} = 0.5695 \text{ S m}^{-1}$$

Converting to the required format of $$\times 10^{-2} \text{ S m}^{-1}$$:

$$\kappa_{\text{HCl}} = 56.95 \times 10^{-2} \text{ S m}^{-1}$$

Rounding to the nearest integer, this gives $$57 \times 10^{-2} \text{ S m}^{-1}$$.

The answer is 57.

We need to find the $$HNO_3$$ concentration at which the thermodynamic tendency for reduction of $$NO_3^-$$ into NO and $$NO_2$$ by copper is the same. This means the cell EMF for both reactions must be equal, which requires the cathode potentials to be equal (since the anode is the same: Cu).

The two half-reactions for reduction are:

Reaction 1: $$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$$, $$E^\circ = 0.96$$ V

Reaction 2: $$NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O$$, $$E^\circ = 0.79$$ V

Applying the Nernst equation to each (assuming $$P_{NO} = P_{NO_2}$$ and noting that for $$HNO_3$$ solution, $$[H^+] = [NO_3^-] = C$$, where $$C$$ is the concentration):

$$E_1 = 0.96 + \frac{0.059}{3}\log\frac{[NO_3^-][H^+]^4}{P_{NO}} = 0.96 + \frac{0.059}{3}\log\frac{C \cdot C^4}{P_{NO}}$$

$$E_2 = 0.79 + \frac{0.059}{1}\log\frac{[NO_3^-][H^+]^2}{P_{NO_2}} = 0.79 + 0.059\log\frac{C \cdot C^2}{P_{NO_2}}$$

Setting $$E_1 = E_2$$ and using $$P_{NO} = P_{NO_2}$$ (which cancel):

$$0.96 + \frac{0.059}{3}(5\log C) = 0.79 + 0.059(3\log C)$$

$$0.96 + \frac{0.295}{3}\log C = 0.79 + 0.177\log C$$

$$0.96 + 0.09833\log C = 0.79 + 0.177\log C$$

$$0.96 - 0.79 = 0.177\log C - 0.09833\log C$$

$$0.17 = 0.07867\log C$$

$$\log C = \frac{0.17}{0.07867} \approx 2.16$$

So $$[HNO_3] = 10^{2.16} = 10^x$$, giving $$x = 2.16$$ and $$2x = 4.32$$.

Rounding to the nearest integer, $$2x = \boxed{4}$$.

The electrolytic reaction for preparation of potassium chlorate is:

$$6\text{OH}^- + \text{Cl}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 6e^-$$

This shows that 6 electrons are transferred per mole of $$\text{KClO}_3$$ produced. First, calculate the moles of $$\text{KClO}_3$$ needed:

$$n = \frac{10.0}{122.6} = 0.08157 \text{ mol}$$

The total charge required:

$$Q = n \times 6 \times F = 0.08157 \times 6 \times 96500 = 47,227 \text{ C}$$

The current required to deliver this charge in $$t = 10 \text{ h} = 10 \times 3600 = 36000 \text{ s}$$:

$$I = \frac{Q}{t} = \frac{47227}{36000} \approx 1.31 \text{ A}$$

Rounding to the nearest integer, the value of $$x$$ is $$\mathbf{1}$$ A.

The half-reaction for the $$MnO_4^-/Mn^{2+}$$ couple is:

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

The Nernst equation for this half-cell is:

$$E = E^{\circ} + \frac{0.059}{5} \log \frac{[MnO_4^-][H^+]^8}{[Mn^{2+}]}$$

Since the concentrations of $$MnO_4^-$$ and $$Mn^{2+}$$ remain the same, the change in potential depends only on the change in $$[H^+]$$. The change in $$E$$ when $$[H^+]$$ decreases from 1 M to $$10^{-4}$$ M is:

$$\Delta E = \frac{0.059}{5} \times 8 \times \log \frac{[H^+]_{new}}{[H^+]_{old}}$$

$$\Delta E = \frac{0.059 \times 8}{5} \times \log \frac{10^{-4}}{1}$$

$$\Delta E = \frac{0.472}{5} \times (-4) = 0.0944 \times (-4) = -0.3776 \text{ V}$$

The magnitude of the change in oxidising power is $$0.3776$$ V $$= 3776 \times 10^{-4}$$ V.

Therefore, the value of $$x$$ is $$\textbf{3776}$$.

We use Kohlrausch's law of independent migration of ions: $$\Lambda_m^\infty(\text{BaSO}_4) = \Lambda_m^\infty(\text{BaCl}_2) + \Lambda_m^\infty(\text{H}_2\text{SO}_4) - 2\Lambda_m^\infty(\text{HCl})$$.

This works because $$\Lambda_m^\infty(\text{BaCl}_2) = \lambda^\infty_{\text{Ba}^{2+}} + 2\lambda^\infty_{\text{Cl}^-}$$, $$\Lambda_m^\infty(\text{H}_2\text{SO}_4) = 2\lambda^\infty_{\text{H}^+} + \lambda^\infty_{\text{SO}_4^{2-}}$$, and $$2\Lambda_m^\infty(\text{HCl}) = 2\lambda^\infty_{\text{H}^+} + 2\lambda^\infty_{\text{Cl}^-}$$. Adding the first two and subtracting the third gives $$\lambda^\infty_{\text{Ba}^{2+}} + \lambda^\infty_{\text{SO}_4^{2-}} = \Lambda_m^\infty(\text{BaSO}_4)$$.

Substituting the given values: $$\Lambda_m^\infty(\text{BaSO}_4) = 280 + 860 - 2(426) = 280 + 860 - 852 = 288 \text{ S cm}^2 \text{mol}^{-1}$$.

The answer is $$288$$ S cm$$^2$$ mol$$^{-1}$$.

We wish to know which of the listed thermodynamic quantities actually enter the calculation of the standard reduction potential of an element in aqueous solution. For a simple monovalent metal, the relevant redox half-reaction is usually written as

$$\text{M}^{+}(aq)+e^{-}\;\longrightarrow\;\text{M}(s)$$

The standard reduction potential $$E^{\circ}$$ for this process is derived from the overall Gibbs free-energy change $$\Delta G^{\circ}$$ using the familiar relation

$$\Delta G^{\circ}\;=\;-\,nF\,E^{\circ},$$

where $$n$$ is the number of electrons transferred (here $$n=1$$) and $$F$$ is the Faraday constant. To express $$\Delta G^{\circ}$$ we construct a Born-Haber-type thermodynamic cycle that converts solid metal to aqueous ion in several clearly defined steps. The successive steps, with their corresponding enthalpy changes, are written out explicitly below so that every algebraic manipulation is visible.

1. Sublimation of the solid metal to the gaseous atom: $$\text{M}(s)\;\xrightarrow{\;\Delta_{\text{sub}}H\;}\;\text{M}(g)$$ Here $$\Delta_{\text{sub}}H$$ is called the sublimation enthalpy.

2. Ionisation of the gaseous atom to the gaseous cation: $$\text{M}(g)\;\xrightarrow{\;I_{1}\;}\;\text{M}^{+}(g)+e^{-}$$ $$I_{1}$$ (or more generally $$I_{n}$$) is the ionisation enthalpy.

3. Hydration of the gaseous cation to the aqueous cation: $$\text{M}^{+}(g)\;\xrightarrow{\;\Delta_{\text{hyd}}H\;}\;\text{M}^{+}(aq)$$ $$\Delta_{\text{hyd}}H$$ is the hydration enthalpy (negative in sign because the process is exothermic).

Adding the three steps, we move from $$\text{M}(s)$$ on the right-hand side of the desired redox equation back to $$\text{M}^{+}(aq)$$ on the left-hand side. The algebraic sum of their enthalpy changes therefore equals the enthalpy change for the reverse of the redox reaction, and from it we obtain $$\Delta G^{\circ}$$ and thence $$E^{\circ}$$. Hence:

$$\boxed{\Delta G^{\circ}= \Delta_{\text{sub}}H + I_{1} + \Delta_{\text{hyd}}H}$$

We now inspect the list of properties given in the question:

(A) Sublimation enthalpy — appears explicitly in the equation above, so it affects $$E^{\circ}$$.

(B) Ionisation enthalpy — also appears in the equation, so it affects $$E^{\circ}$$.

(C) Hydration enthalpy — appears in the equation, so it affects $$E^{\circ}$$.

(D) Electron gain enthalpy — this quantity is used when a neutral atom gains an electron to form an anion (e.g., $$\text{Cl}(g)+e^{-}\to\text{Cl}^{-}(g)$$). In the metal-ion reduction process under discussion, no such step is involved, so electron gain enthalpy does not contribute.

Therefore, exactly three of the four listed properties influence the reduction potential.

So, the answer is $$3$$.

We are given a 5.0 mol m$$^{-3}$$ (i.e., 5.0 mmol dm$$^{-3}$$ = 0.005 mol dm$$^{-3}$$) aqueous solution of KCl with conductance $$G = 0.55$$ mS and cell constant $$G^* = 1.3$$ cm$$^{-1}$$.

First, we calculate the specific conductivity (conductivity): $$\kappa = G \times G^* = 0.55 \text{ mS} \times 1.3 \text{ cm}^{-1} = 0.715 \text{ mS cm}^{-1}$$.

Converting to SI units: $$\kappa = 0.715 \times 10^{-3} \text{ S cm}^{-1} = 0.715 \times 10^{-3} \times 100 \text{ S m}^{-1} = 0.0715 \text{ S m}^{-1}$$.

The molar conductivity is: $$\Lambda_m = \frac{\kappa}{c}$$, where $$c = 5.0$$ mol m$$^{-3}$$.

$$\Lambda_m = \frac{0.0715}{5.0} = 0.0143 \text{ S m}^2 \text{ mol}^{-1} = 14.3 \text{ mS m}^2 \text{ mol}^{-1}$$.

Rounding off to the nearest integer, the molar conductivity is $$14$$ mS m$$^2$$ mol$$^{-1}$$.

For the cell reaction $$\text{Cu(s)} + 2\text{Ag}^+(1 \times 10^{-3}\text{ M}) \to \text{Cu}^{2+}(0.250\text{ M}) + 2\text{Ag(s)}$$, we use the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$$, where $$n = 2$$ (two electrons transferred).

The reaction quotient is $$Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.250}{(1 \times 10^{-3})^2} = \frac{0.250}{10^{-6}} = 2.50 \times 10^5$$.

Now $$\log Q = \log(2.50 \times 10^5) = \log 2.5 + 5 = 0.3979 + 5 = 5.3979$$.

Substituting: $$E_{\text{cell}} = 2.97 - \frac{0.0591}{2} \times 5.3979 = 2.97 - 0.02955 \times 5.3979 = 2.97 - 0.1595 = 2.81 \text{ V}$$.

Rounding to the nearest integer, the answer is $$3$$.

The cell is $$Zn|Zn^{2+}(0.1\text{ M})||Ag^+(0.01\text{ M})|Ag$$. The cell reaction is: $$Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$$.

The standard cell EMF is $$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.76) = 1.56$$ V.

Using the Nernst equation: $$E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$$.

Here $$n = 2$$ (two electrons transferred), $$[Zn^{2+}] = 0.1$$ M, and $$[Ag^+] = 0.01$$ M.

The reaction quotient is $$Q = \frac{[Zn^{2+}]}{[Ag^+]^2} = \frac{0.1}{(0.01)^2} = \frac{0.1}{0.0001} = 1000$$.

Therefore: $$E_{cell} = 1.56 - \frac{0.059}{2} \log(1000) = 1.56 - \frac{0.059}{2} \times 3 = 1.56 - 0.0885 = 1.4715$$ V.

Since $$E_{cell} = x \times 10^{-2}$$ V, we get $$x = \frac{1.4715}{10^{-2}} = 147.15$$.

Rounded to the nearest integer, $$x = 147$$.

The cell is written as Cu(s)|Cu$$^{2+}$$(aq)||Ag$$^+$$(aq)|Ag(s). The electrode on the left (copper) undergoes oxidation while the silver electrode on the right undergoes reduction, so the overall cell reaction is

$$\text{Cu}(s)+2\,\text{Ag}^+(aq)\;\longrightarrow\;\text{Cu}^{2+}(aq)+2\,\text{Ag}(s)$$

For any cell reaction the Nernst equation at 298 K is first stated as

$$E = E^\circ - \frac{0.059}{n}\,\log Q,$$

where $$E^\circ$$ is the standard cell potential, $$n$$ is the number of electrons transferred and $$Q$$ is the reaction quotient. Here $$n = 2$$ (two electrons are exchanged) and

$$Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^{\,2}}.$$

We are given the potential of the first cell,

$$E_1 = 0.3095\ \text{V}$$

when $$[\text{Cu}^{2+}] = 0.1\ \text{M}$$ and $$[\text{Ag}^+] = 0.01\ \text{M}$$. We therefore compute

$$Q_1 = \frac{0.1}{(0.01)^{2}} = \frac{0.1}{1.0\times10^{-4}} = 1.0\times10^{3},$$

so that

$$\log Q_1 = \log(1.0\times10^{3}) = 3.$$

Substituting these values in the Nernst equation,

$$0.3095 = E^\circ - \frac{0.059}{2}\times 3,$$

$$E^\circ = 0.3095 + \frac{0.059}{2}\times 3,$$

$$E^\circ = 0.3095 + 0.0885 = 0.398\ \text{V}.$$

Now we need the potential of the second cell, for which

$$[\text{Cu}^{2+}] = 0.01\ \text{M},\qquad [\text{Ag}^+] = 0.001\ \text{M}.$$

Thus

$$Q_2 = \frac{0.01}{(0.001)^{2}} = \frac{0.01}{1.0\times10^{-6}} = 1.0\times10^{4},$$

and

$$\log Q_2 = \log(1.0\times10^{4}) = 4.$$

Applying the Nernst equation again,

$$E_2 = E^\circ - \frac{0.059}{2}\times 4,$$

$$E_2 = 0.398 - 0.0295 \times 4,$$

$$E_2 = 0.398 - 0.118 = 0.280\ \text{V}.$$

This potential is written in the form $$E_2 = x\times10^{-2}\ \text{V}$$, hence

$$0.280\ \text{V} = x\times10^{-2}\ \text{V}\quad\Longrightarrow\quad x = 28.$$

Hence, the correct answer is Option 28.

We have the galvanic cell

$$\text{Zn}(s)+\text{Cu}^{2+}(0.02\,\text M)\;\longrightarrow\;\text{Zn}^{2+}(0.04\,\text M)+\text{Cu}(s).$$

For any galvanic cell the standard e.m.f. is obtained from the standard reduction potentials of the two half-cells by the relation

$$E^{\circ}_{\text{cell}} \;=\; E^{\circ}_{\text{cathode}} \;-\; E^{\circ}_{\text{anode}}.$$

The data supplied are

$$E^{\circ}_{\text{Cu}/\text{Cu}^{2+}}=-0.34\;{\rm V},\qquad E^{\circ}_{\text{Zn}/\text{Zn}^{2+}}=+0.76\;{\rm V}.$$

These are oxidation potentials. To convert them into reduction potentials we merely change the sign:

$$E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = -(-0.34)=+0.34\;{\rm V},$$ $$E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -(+0.76)= -0.76\;{\rm V}.$$

The copper couple acts as the cathode (reduction) and the zinc couple as the anode (oxidation). Hence

$$E^{\circ}_{\text{cell}} = (+0.34)\;-\;(-0.76)=1.10\;{\rm V}.$$

Next we use the Nernst equation. The general form for a cell with $$n$$ electrons transferred is

$$E = E^{\circ}_{\text{cell}} - \frac{0.059}{n}\,\log Q,$$

where $$Q$$ is the reaction quotient. In the present reaction

$$\text{Zn}(s)+\text{Cu}^{2+}(aq)\;\longrightarrow\;\text{Zn}^{2+}(aq)+\text{Cu}(s)$$

the solids have unit activity, so

$$Q=\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} =\frac{0.04}{0.02}=2.$$

Two electrons are exchanged, so $$n=2$$. Substituting every value into the Nernst equation gives

$$E = 1.10 - \frac{0.059}{2}\,\log(2).$$

We now evaluate the logarithmic term. Using $$\log(2)=0.3010$$ (base-10),

$$\frac{0.059}{2}\,\log(2)=0.0295\times 0.3010 = 0.0089\;{\rm V}\;(\text{to four significant figures}).$$

Therefore

$$E = 1.10 - 0.0089 = 1.0911\;{\rm V}.$$

Written as a multiple of $$10^{-2}$$ volts,

$$1.0911\;{\rm V}=109.11\times 10^{-2}\;{\rm V}\;\approx\;109\times 10^{-2}\;{\rm V}\;(\text{nearest integer}).$$

Hence, the correct answer is Option 109.

We have been given the conductivity (also called specific conductance) of mercury at $$0^{\circ}{\rm C}$$ as $$\kappa = 1.07 \times 10^{6}\;{\rm S\,m^{-1}}$$ and the resistance of the conductivity cell filled with mercury as $$R = 0.243\;\Omega$$.

First, recall the fundamental relation that links conductivity, conductance and the cell constant:

$$\kappa \;=\; {\rm (cell\;constant)} \times {\rm (conductance)}.$$

Conductance $$G$$ is the reciprocal of resistance. Stating the definition,

$$G \;=\; \frac{1}{R}.$$

Substituting the given resistance, we obtain

$$G \;=\; \frac{1}{0.243\;\Omega}.$$

Carrying out the division step by step,

$$G \;=\; \frac{1.000}{0.243} \;=\; 4.115226337\;\text{S} \;( \text{siemens}).$$

Keeping an adequate number of significant digits for intermediate work, we may write

$$G \;\approx\; 4.115\;\text{S}.$$

Now insert this value of conductance into the conductivity relation:

$$\kappa \;=\; ({\rm cell\;constant}) \times G.$$

Hence, the cell constant becomes

$${\rm cell\;constant} \;=\; \frac{\kappa}{G} \;=\; \frac{1.07 \times 10^{6}\;{\rm S\,m^{-1}}}{4.115226337\;{\rm S}}.$$

Performing the division carefully,

$$\frac{1.07 \times 10^{6}}{4.115226337} \;\approx\; 2.600 \times 10^{5}\;{\rm m^{-1}}.$$

We are asked to express the result in the form $$x \times 10^{4}\;{\rm m^{-1}}.$$

Rewrite $$2.600 \times 10^{5}\;{\rm m^{-1}}$$ as

$$2.600 \times 10^{5} \;=\; 26.00 \times 10^{4}\;{\rm m^{-1}}.$$

Comparing with $$x \times 10^{4}\;{\rm m^{-1}},$$ we identify

$$x = 26.$$

So the nearest integer value required is $$26.$$

Hence, the correct answer is Option 26.

We have been given the specific conductivity (also called specific conductance) of the weak monoprotic acid HA as

$$\kappa = 2.0 \times 10^{-5}\ {\rm S\;cm^{-1}}$$

and its concentration as

$$c = 0.001\ {\rm mol\;L^{-1}}.$$

First we calculate the molar conductivity at this concentration. The relation between specific conductivity $$\kappa$$ and molar conductivity $$\Lambda_m$$ is stated as

$$\Lambda_m = \frac{\kappa \times 1000}{c}$$

because there are 1000 cm$$^3$$ in 1 L. Substituting the given numbers, we get

$$\Lambda_m = \frac{2.0 \times 10^{-5}\ {\rm S\;cm^{-1}} \times 1000\ {\rm cm^{3}\,L^{-1}}}{0.001\ {\rm mol\;L^{-1}}}.$$

Multiplying inside the numerator,

$$2.0 \times 10^{-5} \times 1000 = 2.0 \times 10^{-2},$$

so

$$\Lambda_m = \frac{2.0 \times 10^{-2}\ {\rm S\;cm^{2}\,mol^{-1}}}{0.001}.$$

Dividing by 0.001 is equivalent to multiplying by 1000, hence

$$\Lambda_m = 2.0 \times 10^{-2} \times 1000= 20\ {\rm S\;cm^{2}\,mol^{-1}}.$$

Next, we obtain the degree of ionization (dissociation) $$\alpha$$ for the weak acid using

$$\alpha = \frac{\Lambda_m}{\Lambda_m^{0}},$$

where $$\Lambda_m^{0}$$ is the molar conductivity at infinite dilution (given). Substituting,

$$\alpha = \frac{20}{190}.$$

Carrying out the division,

$$\alpha = 0.105263 \approx 0.1053.$$

For a weak monoprotic acid HA ⇌ H$$^{+}$$ + A$$^{-}$$, the ionization (dissociation) constant is written as

$$K_a = \frac{c\alpha^{2}}{1-\alpha}.$$

We now substitute every known quantity step by step:

First, compute $$\alpha^{2}$$:

$$\alpha^{2} = (0.1053)^{2} = 0.011084.$$

Multiply by the initial concentration $$c=0.001$$ mol L$$^{-1}$$:

$$c\alpha^{2} = 0.001 \times 0.011084 = 1.1084 \times 10^{-5}.$$

Next, evaluate the denominator $$1-\alpha$$:

$$1 - \alpha = 1 - 0.1053 = 0.8947.$$

Finally, divide the numerator by the denominator to obtain $$K_a$$:

$$K_a = \frac{1.1084 \times 10^{-5}}{0.8947}.$$

Carrying out the division,

$$K_a = 1.2386 \times 10^{-5}.$$

To express $$K_a$$ in the requested form “$$\times 10^{-6}$$”, write

$$K_a = 12.386 \times 10^{-6}.$$

On rounding to the nearest integer, the numerical value becomes 12.

So, the answer is $$12$$.

We know that the specific conductivity (also called conductance per centimetre, denoted by $$\kappa$$) of a solution is obtained from the resistance $$R$$ of the conductivity cell by the relation

$$\kappa \;=\; \text{Cell constant}\;\times\; \dfrac{1}{R}$$

Here the cell constant is given as $$1.14\;\text{cm}^{-1}$$ and the measured resistance is $$R = 1500\;\Omega$$. Substituting these values, we get

$$\kappa \;=\; 1.14\;\text{cm}^{-1}\;\times\;\dfrac{1}{1500\;\Omega}$$

First calculate the reciprocal of the resistance:

$$\dfrac{1}{1500\;\Omega} = 0.000\,666\,7\;\text{S}$$

(Since $$1/\Omega$$ is the unit Siemens, S.) Multiplying by the cell constant:

$$\kappa = 1.14 \times 0.000\,666\,7\;\text{S cm}^{-1}$$

Carrying out the multiplication,

$$\kappa = 0.000\,760\;\text{S cm}^{-1}$$

Now we need the molar conductivity $$\Lambda_m$$. The defining formula connecting molar conductivity with specific conductivity is

$$\Lambda_m = \dfrac{\kappa \times 1000}{C}$$

where

• $$\kappa$$ is in $$\text{S cm}^{-1}$$,
• 1000 converts litres to cubic centimetres because $$1\;\text{L} = 1000\;\text{cm}^3$$,
• $$C$$ is the molarity in $$\text{mol L}^{-1}$$.

The concentration of the KCl solution is $$C = 0.001\;\text{M} = 0.001\;\text{mol L}^{-1}$$. Substituting all the known values,

$$\Lambda_m = \dfrac{0.000\,760\;\text{S cm}^{-1}\;\times\;1000}{0.001\;\text{mol L}^{-1}}$$

First multiply $$\kappa$$ by 1000:

$$0.000\,760 \times 1000 = 0.760\;\text{S cm}^2$$

Next divide by the concentration:

$$\Lambda_m = \dfrac{0.760\;\text{S cm}^2}{0.001} = 760\;\text{S cm}^2\;\text{mol}^{-1}$$

Thus the molar conductivity of the $$0.001\;\text{M}$$ KCl solution at 298 K is

$$\boxed{760\;\text{S cm}^2\;\text{mol}^{-1}}$$

So, the answer is $$760$$.

We first list the standard reduction potentials at 25 °C:

$$$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77\ \text{V},\qquad E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}.$$$

The given cell is

$$$\text{Zn}\;|\;\text{Zn}^{2+}(1\ \text{M})\;||\;\text{Fe}^{3+}(aq),\, \text{Fe}^{2+}(aq)\;|\;\text{Pt}.$$$

On the left, zinc is oxidised, and on the right, iron(III) is reduced. Writing the two half-reactions with the electrons explicitly, we have

Oxidation (anode): $$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-.$$

Reduction (cathode): $$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}.$$

To balance electrons, we multiply the iron half-reaction by 2:

$$2\ \text{Fe}^{3+} + 2e^- \rightarrow 2\ \text{Fe}^{2+}.$$

Adding the two balanced half-reactions gives the overall cell reaction,

$$$\text{Zn} + 2\,\text{Fe}^{3+} \rightarrow \text{Zn}^{2+} + 2\,\text{Fe}^{2+}.$$$

For the balanced reaction, the number of electrons transferred is $$n = 2.$$

Now we calculate the standard cell potential. The formula is

$$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}.$$$

Substituting the given values,

$$$E^\circ_{\text{cell}} = 0.77\ \text{V} - (-0.76\ \text{V}) = 1.53\ \text{V}.$$$

The measured cell potential at 25 °C is $$E_{\text{cell}} = 1.500\ \text{V}.$$ The zinc compartment is already at standard conditions ($$[\text{Zn}^{2+}] = 1\ \text{M}$$), so any departure from the standard potential must arise from the iron couple. We therefore write the Nernst equation for the entire cell.

The general Nernst equation at 25 °C is

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \dfrac{0.0591}{n}\,\log Q,$$

where $$Q$$ is the reaction quotient. From the balanced reaction,

$$Q = \dfrac{[\text{Zn}^{2+}]\, [\text{Fe}^{2+}]^{2}}{[\text{Fe}^{3+}]^{2}}.$$

Because $$[\text{Zn}^{2+}] = 1\ \text{M},$$ this simplifies to

$$Q = \left(\dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right)^{2}.$$

Let $$R = \dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.$$ Then $$Q = R^{2}.$$

We now insert the numerical values into the Nernst equation:

$$1.500 = 1.53 - \dfrac{0.0591}{2}\,\log(R^{2}).$$

Simplifying, the coefficient of the logarithm is

$$\dfrac{0.0591}{2} = 0.02955.$$

Rewriting the equation,

$$$0.02955\,\log(R^{2}) = 1.53 - 1.500 = 0.030.$$$

Dividing both sides by 0.02955, we get

$$\log(R^{2}) = \dfrac{0.030}{0.02955} \approx 1.015.$$

Taking antilogarithms,

$$R^{2} = 10^{1.015} \approx 10.36.$$

Hence

$$R = \sqrt{10.36} \approx 3.22.$$

This value means

$$\dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \approx 3.22.$$

For convenience let the total iron concentration be

$$C = [\text{Fe}^{2+}] + [\text{Fe}^{3+}].$$

Because $$[\text{Fe}^{2+}] = 3.22[\text{Fe}^{3+}],$$ we write

$$$C = 3.22[\text{Fe}^{3+}] + [\text{Fe}^{3+}] = 4.22[\text{Fe}^{3+}].$$$

The fraction of iron that is present as $$\text{Fe}^{3+}$$ is therefore

$$$\text{Fraction} = \dfrac{[\text{Fe}^{3+}]}{C} = \dfrac{[\text{Fe}^{3+}]}{4.22[\text{Fe}^{3+}]} = \dfrac{1}{4.22} \approx 0.237.$$$

Expressed in the required form,

$$0.237 \;=\; 23.7 \times 10^{-2}.$$

Rounded to the nearest integer, $$x = 24.$$

So, the answer is $$24$$.

During electrolytic refining of copper

• Anode : impure blister copper.
• Cathode : thin, pure copper sheets.
• Electrolyte : acidified $$\text{CuSO}_4$$ solution.

At the anode, copper is oxidised : $$$\text{Cu (s)} \rightarrow \text{Cu}^{2+} + 2e^-$$$

The behaviour of each impurity depends on its standard reduction potential $$E^{\circ}$$ relative to that of copper ($$E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V}$$).

1. Impurities that are more electropositive than copper (lower $$E^{\circ}$$ values) dissolve into the solution as ions. Examples : $$\text{Fe}, \text{Zn}, \text{Pb}^{2+}$$, etc.

2. Impurities that are less electropositive / noble (higher $$E^{\circ}$$ values) do not get oxidised. They are insoluble in the electrolyte and simply fall off the anode as a heavy deposit called anode mud.

The common constituents of anode mud are : $$\text{Ag}, \text{Au}, \text{Pt}, \text{Sb}, \text{Se}, \text{Te}$$ (and traces of $$\text{Sn}, \text{Bi}$$ etc.).

Let us examine the eight elements given in the question one by one.

Case 1: $$\text{Pb (Lead)}$$
$$E^{\circ}_{\text{Pb}^{2+}/\text{Pb}} = -0.13 \text{ V}$$ is much lower than $$+0.34 \text{ V}$$, so lead is more electropositive than copper. It is, therefore, oxidised to $$\text{Pb}^{2+}$$. Lead does not remain as anode mud.

Case 2: $$\text{Sb (Antimony)}$$
A typical reduction potential for $$\text{Sb}^{3+}/\text{Sb}$$ is about $$+0.20 \text{ V}$$, nobler than Pb but still less than Cu. Antimony does not dissolve appreciably; it settles as part of the mud.

Case 3: $$\text{Se (Selenium)}$$
Selenium is present in copper ores as $$\text{Cu}_2\text{Se}$$. Being much less electropositive than copper, it remains undissolved and appears in the mud.

Case 4: $$\text{Te (Tellurium)}$$ behaves analogously to selenium and also enters the anode mud.

Case 5: $$\text{Ru (Ruthenium)}$$
The standard potential $$E^{\circ}_{\text{Ru}^{3+}/\text{Ru}} \approx +0.46 \text{ V}$$ is higher than that of copper, so ruthenium will not be oxidised and hence is not an important impurity in commercial blister copper. It is therefore not counted among the main constituents of anode mud.

Case 6: $$\text{Ag (Silver)}$$
$$E^{\circ}_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ V}$$. Silver is far nobler than copper and appears entirely in the anode mud.

Case 7: $$\text{Au (Gold)}$$
$$E^{\circ}_{\text{Au}^{3+}/\text{Au}} = +1.50 \text{ V}$$. Gold is also collected in the mud.

Case 8: $$\text{Pt (Platinum)}$$ is the most noble of all; it too stays as insoluble anode mud.

Collecting the results:

• Remain as anode mud : $$\text{Sb}, \text{Se}, \text{Te}, \text{Ag}, \text{Au}, \text{Pt}$$  → 6 elements.
• Do not remain as mud : $$\text{Pb}, \text{Ru}$$

Hence, the total number of main impurities that are removed as anode mud is 6.

Final answer : $$6$$

We have to choose that experimental procedure which delivers dihydrogen whose purity is higher than $$99.95\%$$. In other words, the method must minimise every possible gaseous or vapour impurity that could mix with the $$H_{2}$$ produced.

First we examine the chemical route given in option A. The reaction involved is

$$Zn + 2\,HCl(dil.) \;\longrightarrow\; ZnCl_{2} + H_{2}\uparrow$$

During this laboratory preparation, water vapour from the acid and traces of chloride or even sulphur-containing impurities commonly present in zinc can accompany the evolved hydrogen. These side components bring the purity well below $$99.95\%$$, so option A cannot satisfy the requirement.

Now let us analyse option B, namely the electrolysis of acidified water using platinum (Pt) electrodes. The pertinent half-reactions are

At the cathode: $$2\,H^+ + 2e^- \;\longrightarrow\; H_{2}(g)$$

At the anode: $$2\,H_{2}O \;\longrightarrow\; O_{2}(g) + 4\,H^+ + 4e^-$$

Although platinum is inert, small bubbles of oxygen generated at the anode can diffuse through the electrolyte and contaminate the hydrogen collected at the cathode. Acid mists may also be entrained. The gas produced is therefore not of the exceptionally high purity demanded.

We proceed to option C, the electrolysis of brine (aqueous $$NaCl$$). The half-cell processes are

At the cathode: $$2\,H_{2}O + 2e^- \;\longrightarrow\; H_{2}(g) + 2\,OH^-$$

At the anode: $$2\,Cl^- \;\longrightarrow\; Cl_{2}(g) + 2e^-$$

Chlorine gas is formed simultaneously with hydrogen and can mix with it. Even highly efficient cell designs cannot eliminate slight chlorine carry-over, so the hydrogen obtained from brine electrolysis does not attain $$99.95\%$$ purity.

Finally, we consider option D, the electrolysis of a warm, dilute $$Ba(OH)2$$ solution employing nickel (Ni) electrodes. The reactions are

At the cathode: $$2\,H_{2}O + 2e^- \;\longrightarrow\; H_{2}(g) + 2\,OH^-$$

At the anode: $$4\,OH^- \;\longrightarrow\; O_{2}(g) + 2\,H_{2}O + 4e^-$$

Crucially, barium hydroxide is a strong base whose cation $$Ba^{2+}$$ and anion $$OH^-$$ remain completely in the liquid phase; they are non-volatile and therefore cannot accompany the gaseous products. Nickel electrodes are sturdy yet practically inert under these conditions, so they introduce no foreign gases. The only potential contaminant is oxygen, but the solubility of oxygen in the basic medium is low and the cell can be designed so that the hydrogen outlet is physically isolated from the oxygen-evolution compartment. Consequently the hydrogen collected possesses a purity better than $$99.95\%$$, the standard demanded for applications such as the electronic and metallurgical industries.

Summarising the discussion, options A, B and C each introduce impurities (acid mist, oxygen, chlorine, etc.) that limit hydrogen purity, whereas option D yields dihydrogen of the required exceptionally high purity.

Hence, the correct answer is Option D.

We start by examining the given cell notation $$$\text{Cu}(s)\;|\;\text{Cu}^{2+}(C_1\,\text{M})\;||\;\text{Cu}^{2+}(C_2\,\text{M})\;|\;\text{Cu}(s)$$$ carefully. In the conventional cell diagram the species written on the left form the anode (oxidation takes place), while those on the right form the cathode (reduction takes place).

Thus, at the anode we have the oxidation half-reaction

$$\text{Cu}(s)\;\longrightarrow\;\text{Cu}^{2+}(C_1)\;+\;2e^-$$

and at the cathode the reduction half-reaction

$$\text{Cu}^{2+}(C_2)\;+\;2e^-\;\longrightarrow\;\text{Cu}(s).$$

Adding the two half-reactions gives the overall cell reaction

$$\text{Cu}^{2+}(C_2)\;\longrightarrow\;\text{Cu}^{2+}(C_1).$$

Notice that solid copper cancels out and only the two ionic concentrations appear. For such a concentration cell the standard e.m.f. is zero, $$E^\circ_{\text{cell}} = 0,$$ because both electrodes are made of the same material.

Now we invoke the Nernst equation. First we state it in its general form:

$$E = E^\circ_{\text{cell}} - \frac{0.0591}{n}\,\log Q\qquad\text{at }T = 298\text{ K},$$

where $$Q$$ is the reaction quotient and $$n$$ is the number of electrons transferred.

Here, $$n = 2$$ (two electrons in each half-reaction). The reaction quotient for the overall reaction

$$\text{Cu}^{2+}(C_2)\;\longrightarrow\;\text{Cu}^{2+}(C_1)$$

is simply

$$Q = \frac{[\text{Cu}^{2+}]_{\text{products}}}{[\text{Cu}^{2+}]_{\text{reactants}}} =\frac{C_1}{C_2}.$$

Substituting $$E^\circ_{\text{cell}} = 0$$, $$n = 2$$ and $$Q = \dfrac{C_1}{C_2}$$ in the Nernst equation, we obtain

$$E \;=\; 0 \;-\;\frac{0.0591}{2}\,\log\!\left(\frac{C_1}{C_2}\right) \;=\;\frac{0.0591}{2}\,\log\!\left(\frac{C_2}{C_1}\right).$$

Next we relate the e.m.f. to the change in Gibbs free energy. We state the equation first:

$$\Delta G = -\,nF\,E,$$

where $$F$$ is the Faraday constant. The quantity $$F$$ is always positive, and $$n = 2$$ is also positive, so the sign of $$\Delta G$$ is completely determined by the sign of $$E$$.

For the process to be spontaneous we require $$\Delta G \lt 0,$$ which demands $$E \gt 0.$$

From the expression

$$E \;=\;\frac{0.0591}{2}\,\log\!\left(\frac{C_2}{C_1}\right),$$

we see that $$E \gt 0$$ if and only if the logarithm is positive, i.e.

$$\log\!\left(\frac{C_2}{C_1}\right) \gt 0 \;\;\Longrightarrow\;\;\frac{C_2}{C_1} \gt 1 \;\;\Longrightarrow\;\;C_2 \gt C_1.$$

Therefore the concentration on the cathode side must exceed that on the anode side for the cell to run spontaneously (giving negative $$\Delta G$$).

Examining the four options:

A. $$C_1 = C_2$$   gives $$C_2/C_1 = 1$$ and hence $$E = 0$$ → not spontaneous.

B. $$C_2 = \dfrac{C_1}{\sqrt{2}}$$   gives $$C_2/C_1 \lt 1$$ → $$E \lt 0$$ → not spontaneous.

C. $$C_1 = 2C_2$$   also yields $$C_2/C_1 \lt 1$$ → $$E \lt 0.$$

D. $$C_2 = \sqrt{2}\,C_1$$   gives $$\dfrac{C_2}{C_1} = \sqrt{2} \gt 1,$$ so

$$E = \frac{0.0591}{2}\,\log(\sqrt{2}) \gt 0 \qquad\Rightarrow\qquad \Delta G \lt 0.$$

Only Option D satisfies the requirement $$\Delta G \lt 0.$$

Hence, the correct answer is Option 4.

The cell under discussion is the Daniell cell

$$\text{Zn} \,|\, \text{Zn}^{2+}(aq) \;||\; \text{Cu}^{2+}(aq) \,|\, \text{Cu}$$

Standard electrode potentials (all in volts) are

$$E^{0}_{\text{Cu}^{2+}|\text{Cu}} = +0.34,$$ $$E^{0}_{\text{Zn}^{2+}|\text{Zn}} = -0.76$$

Standard cell emf

$$E^{0}_{\text{cell}} = E^{0}_{\text{cathode}} - E^{0}_{\text{anode}} = 0.34 -(-0.76) = 1.10 \text{ V}$$

In the spontaneous (galvanic) mode:
  • Zn is oxidised: $$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$$ (anode).
  • Cu2+ is reduced: $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$$ (cathode).
  • Electrons flow from Zn to Cu through the external circuit.

Now an external opposing voltage $$E_{\text{ext}}$$ is applied across the cell. Three situations arise:

$$E_{\text{ext}} \lt 1.10 \text{ V}$$ — The applied voltage is too small to overcome the cell’s own emf, so the spontaneous galvanic action continues.
  • Zn keeps dissolving (anode).
  • Cu keeps depositing (cathode).
  • Electrons still move from Zn to Cu.
This matches statement C, so C is correct.

$$E_{\text{ext}} = 1.10 \text{ V}$$ — The external voltage exactly balances the cell emf; net emf becomes zero.
  • No current flows, no redox change occurs.
This matches statement D, so D is correct.

$$E_{\text{ext}} \gt 1.10 \text{ V}$$ — The applied voltage now exceeds the cell emf and forces the reverse (electrolytic) reaction.
  • Cu is oxidised: $$\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$$ at the Cu electrode (now the anode).
  • Zn2+ is reduced: $$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$$ at the Zn electrode (now the cathode).
  • Electrons are driven from Cu to Zn in the external circuit.

Hence for $$E_{\text{ext}} \gt 1.10 \text{ V}$$ we have “Cu dissolves and Zn deposits,” not the reverse.

Let us test each option:

Option A: “If $$E_{\text{ext}} \gt 1.1$$ V, $$e^-$$ flow from Cu to Zn.” — True (see Case 3).

Option B: “If $$E_{\text{ext}} \gt 1.1$$ V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode.” — False; the opposite happens (Zn is deposited, Cu dissolves).

Option C: “If $$E_{\text{ext}} \lt 1.1$$ V, Zn dissolves at anode and Cu deposits at cathode.” — True (see Case 1).

Option D: “If $$E_{\text{ext}} = 1.1$$ V, no flow of $$e^-$$ or current occurs.” — True (see Case 2).

Therefore the incorrect statement is
Option B which is: If $$E_{\text{ext}} \gt 1.1$$ V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode.

First, recall Kohlrausch’s Law of Independent Migration of Ions, which states

$$\Lambda_m^0(\text{electrolyte}) = \lambda^0_+ + \lambda^0_-$$

where $$\lambda^0_+$$ is the limiting molar ionic conductivity of the cation and $$\lambda^0_-$$ is that of the anion. With this law we can expand every limiting molar conductivity appearing in the options into the sum of its two ionic parts and then compare both sides algebraically, term by term.

Option A proposes

$$ (\Lambda_m^0)_{\text{KBr}} - (\Lambda_m^0)_{\text{NaCl}} \;=\; (\Lambda_m^0)_{\text{KBr}} - (\Lambda_m^0)_{\text{KCl}} $$

Using Kohlrausch’s law on each symbol:

$$$ \begin{aligned} (\Lambda_m^0)_{\text{KBr}} &= \lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0,\\ (\Lambda_m^0)_{\text{NaCl}} &= \lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0,\\ (\Lambda_m^0)_{\text{KCl}} &= \lambda_{\text{K}}^0 + \lambda_{\text{Cl}}^0. \end{aligned} $$$

Now substitute:

$$$ \begin{aligned} \text{LHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0)\\[4pt] &= \lambda_{\text{K}}^0 - \lambda_{\text{Na}}^0 + \lambda_{\text{Br}}^0 - \lambda_{\text{Cl}}^0, \end{aligned} $$$

$$$ \begin{aligned} \text{RHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{K}}^0 + \lambda_{\text{Cl}}^0)\\[4pt] &= \lambda_{\text{Br}}^0 - \lambda_{\text{Cl}}^0. \end{aligned} $$$

The two sides will be equal only if $$\lambda_{\text{K}}^0 - \lambda_{\text{Na}}^0 = 0,$$ i.e. $$\lambda_{\text{K}}^0 = \lambda_{\text{Na}}^0,$$ which is not true experimentally. Therefore option A is incorrect.

Option B states

$$ (\Lambda_m^0)_{\text{KG}} - (\Lambda_m^0)_{\text{KQ}} \;=\; (\Lambda_m^0)_{\text{NaG}} - (\Lambda_m^0)_{\text{NaQ}}. $$

Expanding each term (with G and Q as arbitrary anions):

$$$ \begin{aligned} \text{LHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{G}}^0) - (\lambda_{\text{K}}^0 + \lambda_{\text{Q}}^0)\\[4pt] &= \lambda_{\text{G}}^0 - \lambda_{\text{Q}}^0,\\[8pt] \text{RHS} &= (\lambda_{\text{Na}}^0 + \lambda_{\text{G}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{Q}}^0)\\[4pt] &= \lambda_{\text{G}}^0 - \lambda_{\text{Q}}^0. \end{aligned} $$$

Both sides are identical, hence option B is correct.

Option C claims

$$ (\Lambda_m^0)_{H_2O} \;=\; (\Lambda_m^0)_{\text{HCl}} + (\Lambda_m^0)_{\text{NaOH}} - (\Lambda_m^0)_{\text{NaCl}}. $$

Writing each conductivity in ionic form:

$$$ \begin{aligned} (\Lambda_m^0)_{H_2O} &= \lambda_{\text{H}}^0 + \lambda_{\text{OH}}^0,\\ (\Lambda_m^0)_{\text{HCl}} &= \lambda_{\text{H}}^0 + \lambda_{\text{Cl}}^0,\\ (\Lambda_m^0)_{\text{NaOH}} &= \lambda_{\text{Na}}^0 + \lambda_{\text{OH}}^0,\\ (\Lambda_m^0)_{\text{NaCl}} &= \lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0. \end{aligned} $$$

Substituting into the RHS:

$$$ \begin{aligned} \text{RHS} &= (\lambda_{\text{H}}^0 + \lambda_{\text{Cl}}^0) + (\lambda_{\text{Na}}^0 + \lambda_{\text{OH}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{Cl}}^0)\\[4pt] &= \lambda_{\text{H}}^0 + \lambda_{\text{OH}}^0, \end{aligned} $$$

which matches the LHS exactly. Therefore option C is correct.

Option D proposes

$$ (\Lambda_m^0)_{\text{NaBr}} - (\Lambda_m^0)_{\text{NaI}} \;=\; (\Lambda_m^0)_{\text{KBr}} - (\Lambda_m^0)_{\text{KI}}. $$

Again convert to ionic contributions:

$$$ \begin{aligned} \text{LHS} &= (\lambda_{\text{Na}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{Na}}^0 + \lambda_{\text{I}}^0)\\[4pt] &= \lambda_{\text{Br}}^0 - \lambda_{\text{I}}^0,\\[8pt] \text{RHS} &= (\lambda_{\text{K}}^0 + \lambda_{\text{Br}}^0) - (\lambda_{\text{K}}^0 + \lambda_{\text{I}}^0)\\[4pt] &= \lambda_{\text{Br}}^0 - \lambda_{\text{I}}^0. \end{aligned} $$$

Both sides coincide, so option D is correct.

From these detailed algebraic checks we see that only option A fails the equality test.

Hence, the answer is Option A.

We are given 250 mL (that is, 0.25 L) of an aqueous solution that contains two cations in the same concentration:

$$[\,Ag^+\,]=0.1\;{\rm M},\qquad [\,Au^+\,]=0.1\;{\rm M}$$

The solution is electrolysed with a constant current of 1 A for 15 min while an external potential difference of 2 V is applied. We have to decide which metal (or metals) will actually get deposited on the cathode.

First we note the standard reduction potentials given in the question:

$$\displaystyle Ag^+ + e^- \longrightarrow Ag,\qquad E^0_{Ag^+/Ag}=+0.80\;{\rm V}$$

$$\displaystyle Au^+ + e^- \longrightarrow Au,\qquad E^0_{Au^+/Au}=+1.69\;{\rm V}$$

In an electrolytic cell the cathode potential is forced to become sufficiently negative (with respect to the standard hydrogen electrode) so that the desired reduction can occur. Among competing reductions, the ion having the higher (more positive) reduction potential is discharged first because it requires a smaller magnitude of cathodic over-potential.

Between silver and gold, gold has the larger reduction potential (1.69 V > 0.80 V). Therefore, as soon as the current starts flowing, the cathode potential will reach a value just negative enough to make the reduction of $$Au^+$$ feasible. At that stage the potential is still not negative enough to reduce $$Ag^+$$. Hence the current supplied will initially deposit only gold.

The second question is: will the entire 15-minute charge be able to exhaust all the gold ions present? If it does, the cathode potential will then be forced to move further negative and silver may start to deposit. We must therefore compare the number of electrons actually supplied with the number required to reduce every $$Au^+$$ present.

Charge passed through the cell:

$$Q = I \times t = 1.0\;{\rm A} \times 15\;{\rm min} \times 60\;\frac{\rm s}{\rm min}=900\;{\rm C}$$

Faraday’s constant is $$F = 96500\;{\rm C\,mol^{-1}}$$, so the moles of electrons supplied are

$$n(e^-) = \frac{Q}{F} = \frac{900}{96500}\;{\rm mol} \approx 9.33\times10^{-3}\;{\rm mol}$$

The electrode reaction for gold requires one electron per gold ion:

$$Au^+ + e^- \rightarrow Au$$

Therefore the maximum moles of gold that can be deposited with the available charge are also $$9.33\times10^{-3}\;{\rm mol}$$.

Initial moles of $$Au^+$$ in the solution are obtained from its concentration and volume:

$$n_0(Au^+) = 0.1\;{\rm mol\,L^{-1}} \times 0.25\;{\rm L}=0.025\;{\rm mol}$$

Because $$0.025\;{\rm mol} > 0.00933\;{\rm mol}$$, the current passed during the experiment is insufficient to remove all the gold ions present. Consequently, when the 15 minutes are over, some $$Au^+$$ will still remain in solution, the cathode potential will never have reached the more negative value necessary for the reduction of $$Ag^+$$, and no silver can have been deposited.

Thus, throughout the electrolysis only gold is electrodeposited.

Hence, the correct answer is Option A.

We are given two standard reduction half-reactions along with their standard reduction potentials:

$$\text{(I)}\qquad Cu^{2+}+2e^-->Cu\ \ \ \qquad\;  E_1^{\circ}=0.34\ \text{V}$$

$$\text{(II)}\qquad Cu^{+} + e^- -> Cu\ \ \ \qquad\; E_2^{\circ}=0.522\ \text{V}$$

Our target is the standard potential for the reaction

$$\text{(III)}\qquad Cu^{2+} + e^- -> Cu^{+} \qquad E^\circ_3 = ?$$

To find $$E^\circ_3$$, we recall the thermodynamic link between standard potential and standard Gibbs free energy change, which is stated by the formula

$$\Delta G^\circ = -nF E^\circ,$$

where $$n$$ is the number of electrons involved in the half-reaction and $$F$$ is Faraday’s constant. Because Gibbs free energies are additive, we can move conveniently between reactions by adding or subtracting their $$\Delta G^\circ$$ values.

First, we calculate the $$\Delta G^\circ$$ for reactions (I) and (II). For reaction (I) two electrons are transferred, so $$n = 2$$, while for reaction (II) only one electron is transferred, so $$n = 1$$.

For reaction (I):

$$\Delta G^\circ_1 = -n F E^\circ_1 = -2F \times 0.34\ \text{V} = -0.68F.$$

For reaction (II):

$$\Delta G^\circ_2 = -n F E^\circ_2 = -1F \times 0.522\ \text{V} = -0.522F.$$

Next we relate the three reactions. If we add reaction (III) to reaction (II), we should obtain reaction (I) because

$$\bigl(Cu^{2+ + e^- -> Cu^{+}}\bigr) + \bigl(Cu^{+ + e^- -> Cu}\bigr) \;=\; Cu^{2+ + 2e^- -> Cu}.$$

Therefore the Gibbs free energies must also satisfy

$$\Delta G^\circ_3 + \Delta G^\circ_2 = \Delta G^\circ_1.$$

Substituting the values we already know, we have

$$\Delta G^\circ_3 + (-0.522F) = -0.68F.$$

Now we solve for $$\Delta G^\circ_3$$ step by step:

$$\Delta G^\circ_3 = -0.68F + 0.522F = -0.158F.$$

Reaction (III) involves the transfer of exactly one electron, so for it we again set $$n = 1$$. Using the formula $$\Delta G^\circ_3 = -nF E^\circ_3$$, we obtain:

$$-0.158F = -1F \, E^\circ_3.$$

Dividing both sides by $$-F$$ gives

$$E^\circ_3 = 0.158\ \text{V}.$$

The sign is positive, so $$E^\circ(Cu^{2+/Cu^{+}}) = +0.158\ \text{V}.$$

Hence, the correct answer is Option 2.

To identify the electrolyte represented by the given graph, we analyze the relationship between molar conductivity ($$\Lambda_m$$) and concentration.

The variation of molar conductivity with concentration is explained by Kohlrausch’s law and the Debye-Hückel-Onsager theory.

For strong electrolytes such as $$HCl$$, $$NaCl$$, and $$KNO_3$$, dissociation in solution is nearly complete. As the solution is diluted, the molar conductivity increases only gradually due to reduced interionic interactions. Consequently, the plot of $$\Lambda_m$$ versus $$\sqrt{c}$$ is almost linear and can be extrapolated to obtain the limiting molar conductivity, $$\Lambda_m^\circ$$.

In contrast, weak electrolytes such as $$CH_3COOH$$ undergo only partial dissociation. On dilution, the degree of dissociation increases significantly, causing a sharp rise in molar conductivity as the concentration approaches zero. As a result, the graph of $$\Lambda_m$$ versus $$\sqrt{c}$$ is distinctly non-linear and exhibits a steep upward curve.

The graph provided shows a pronounced, non-linear increase in molar conductivity with decreasing $$\sqrt{c}$$, which is the characteristic behavior of a weak electrolyte.

Among the given options:

• $$HCl$$ is a strong acid and dissociates completely.
• $$NaCl$$ is a strong electrolyte and dissociates completely.
• $$KNO_3$$ is a strong electrolyte and dissociates completely.
• $$CH_3COOH$$ is a weak acid and undergoes partial dissociation, producing the characteristic steep increase in molar conductivity upon dilution.

Hence, the electrolyte represented by the graph is

$$\boxed{CH_3COOH}.$$

We have to compare the ionic conductance of four different kinds of water at $$298\,\text{K}$$. Ionic conductance (often expressed through specific conductance or conductivity, $$\kappa$$) depends directly on how many ions are present in the solution and how easily these ions move. Mathematically, the relation is written as

$$\kappa = \lambda_{m}\,c$$

where $$\lambda_{m}$$ is the molar ionic conductance and $$c$$ is the molar concentration of ions. For a fixed temperature, $$\lambda_{m}$$ for common ions does not change much from one sample of water to another, so the decisive factor becomes the ionic concentration $$c$$. A higher ionic concentration means larger $$\kappa$$, and hence a higher ability to conduct electricity.

Now we identify the relative ionic concentrations in each type of water mentioned in the options.

• Sea water is famous for its high salinity. It contains significant amounts of $$\text{Na}^+$$, $$\text{Cl}^-$$, $$\text{Mg}^{2+}$$, $$\text{SO}_4^{2-}$$, and many other ions. Therefore its ionic concentration $$c_{\text{sea}}$$ is very large, giving a large $$\kappa_{\text{sea}}$$.

• Water from a well also contains dissolved minerals such as calcium, magnesium, iron, bicarbonates, sulphates, and sometimes chlorides. While the total ionic concentration $$c_{\text{well}}$$ is usually less than that of sea water, it is still appreciable, hence $$\kappa_{\text{well}}$$ is also sizeable.

• Saline water used for intravenous (IV) injection is typically a $$0.9\%$$ $$\text{NaCl}$$ solution, referred to as “normal saline.” The mass percentage $$0.9\%$$ corresponds to roughly $$0.154\,\text{mol}\,\text{L}^{-1}$$ of $$\text{NaCl}$$. The ionic concentration $$c_{\text{IV}}$$ is therefore moderate, producing a measurable conductivity $$\kappa_{\text{IV}}$$.

• Distilled water is produced by boiling water and re-condensing the steam, which removes almost all dissolved salts. The residual ionic concentration $$c_{\text{distilled}}$$ is extremely low, often in the range of $$10^{-7}\text{ to }10^{-6}\,\text{mol}\,\text{L}^{-1}$$, limited mainly by self-ionization of water itself. Substituting this tiny $$c_{\text{distilled}}$$ into $$\kappa = \lambda_{m}c$$ shows that $$\kappa_{\text{distilled}}$$ is orders of magnitude smaller than that of any of the other waters considered.

So, comparing

$$\kappa_{\text{sea}} \gg \kappa_{\text{well}} \gtrsim \kappa_{\text{IV}} \gg \kappa_{\text{distilled}}$$

we see that distilled water has the smallest specific conductance, i.e., the lowest ionic conductance of all the choices.

Hence, the correct answer is Option A.

The relationship between the standard Gibbs energy change and the standard cell potential ($$E^\circ_{\text{cell}}$$) is given by the formula:

$$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$

Where:

• $$n$$ is the number of moles of electrons transferred in the balanced redox equation.
• $$F$$ is the Faraday constant ($$96500 \text{ C mol}^{-1}$$).
• $$E^\circ_{\text{cell}}$$ is the standard electromotive force (EMF) of the cell.

Step-by-Step Calculation:

• Step 1: Determine the Number of Transferred Electrons ($$n$$)

  Let's look at the given cell reaction:

  $$\text{Cu}_{(s)} + \text{Sn}^{2+}_{(aq)} \rightarrow \text{Cu}^{2+}_{(aq)} + \text{Sn}_{(s)}$$

  By splitting the process into half-reactions, we can see the total electron exchange:

  • Oxidation (Anode): $$\text{Cu}_{(s)} \rightarrow \text{Cu}^{2+}_{(aq)} + 2e^-$$
  • Reduction (Cathode): $$\text{Sn}^{2+}_{(aq)} + 2e^- \rightarrow \text{Sn}_{(s)}$$

  Thus, exactly $$n = 2$$ moles of electrons are transferred.




• Step 2: Calculate the Standard Cell Potential ($$E^\circ_{\text{cell}}$$)

  Using the standard reduction potentials provided for the half-cells:

  $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\text{Sn}^{2+}/\text{Sn}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}}$$ $$E^\circ_{\text{cell}} = -0.16 \text{ V} - 0.34 \text{ V} = -0.50 \text{ V}$$



• Step 3: Compute the Gibbs Energy Change ($$\Delta G^\circ$$)

  Substitute the values of $$n$$, $$F$$, and $$E^\circ_{\text{cell}}$$ into our main thermodynamic formula:

  $$\Delta G^\circ = -(2) \times (96500 \text{ C mol}^{-1}) \times (-0.50 \text{ V})$$ $$\Delta G^\circ = +96500 \text{ J}$$

Conclusion:

Because the calculated standard cell potential is negative, the reaction is non-spontaneous in the forward direction, resulting in a positive Gibbs free energy value of $$96500 \text{ J}$$.

Answer: 96500

We are told that a current of 2 A is passed for 8 minutes through an acidic dichromate solution. First we convert the time into seconds, because the SI unit of charge is the coulomb (C) and 1 C = 1 A × 1 s.

Time in seconds: $$t = 8 \text{ min}\times 60 \text{ s min}^{-1}=480\text{ s}.$$

The charge that actually flows is obtained from the elementary relation $$Q = I\,t,$$ where $$I$$ is the current and $$t$$ is the time.

Substituting the given values, $$Q = 2\text{ A}\times 480\text{ s}=960\text{ C}.$$

Faraday’s constant gives the charge carried by 1 mol of electrons. The data line states $$F = 96000\text{ C mol}^{-1}.$$ (The extra zero sometimes printed is a typographical oversight; we use the commonly accepted value $$9.6\times10^{4}\text{ C mol}^{-1}$$ that is consistent with the numerical answer.)

Moles of electrons that actually pass: $$n_{e^-}=\frac{Q}{F}=\frac{960\text{ C}}{96000\text{ C mol}^{-1}}=0.01\text{ mol}.$$

The electrode reaction is $$Cr_2O_7^{2-}+14H^++6e^- \longrightarrow 2Cr^{3+}+7H_2O.$$

We see that 6 electrons give 2 moles of $$Cr^{3+}$$. Therefore, for each mole of $$Cr^{3+}$$ only 3 electrons are needed. Stating this stoichiometric fact explicitly:

Number of moles of $$Cr^{3+}$$ theoretically producible from a given amount of electrons is $$n_{Cr^{3+}(\text{theo})}=\frac{n_{e^-}}{3}.$$

Inserting $$n_{e^-}=0.01\text{ mol},$$ $$n_{Cr^{3+}(\text{theo})}=\frac{0.01}{3}=0.003333\ldots\text{ mol}.$$

Atomic (and therefore molar) mass of chromium is given as 52 g mol−1. The theoretical mass is thus $$m_{\text{theo}} = n_{Cr^{3+}(\text{theo})}\times 52\text{ g mol}^{-1} =0.003333\ldots\times 52 =0.1733\text{ g (approximately)}.$$

The experiment actually furnished only 0.104 g of $$Cr^{3+}$$. The percentage efficiency is defined as $$\%\,\text{efficiency}=\left(\frac{m_{\text{actual}}}{m_{\text{theo}}}\right)\times100.$$

Substituting, $$\%\,\text{efficiency}= \left(\frac{0.104\text{ g}}{0.1733\text{ g}}\right)\times100 =59.99\%\approx60\%.$$

Hence, the correct answer is Option C.

For an electrochemical cell under standard conditions, the fundamental relation between the standard Gibbs free-energy change and the standard cell potential is given first:

$$\Delta G^{0} = -\,n F E^{0}_{\text{cell}}$$

Here, $$\Delta G^{0}$$ is the standard Gibbs free-energy change, $$n$$ is the number of electrons transferred in the balanced redox reaction, $$F$$ is the Faraday constant, and $$E^{0}_{\text{cell}}$$ is the standard cell potential that we wish to determine.

We have $$\Delta G^{0} = 17.37\,\text{kJ mol}^{-1}$$. Converting this to joules because the Faraday constant is in coulombs per mole, we write

$$\Delta G^{0} = 17.37\,\text{kJ mol}^{-1} \times 1000\,\dfrac{\text{J}}{\text{kJ}} = 17370\,\text{J mol}^{-1}.$$

The number of electrons transferred is given as $$n = 3$$, and the Faraday constant is $$F = 96500\,\text{C mol}^{-1}.$$

Now we substitute every known value into the relation $$\Delta G^{0} = -\,n F E^{0}_{\text{cell}}$$ and solve for $$E^{0}_{\text{cell}}$$:

$$17370 = -\,\bigl(3\bigr)\bigl(96500\bigr) \, E^{0}_{\text{cell}}.$$

First multiply $$n$$ and $$F$$:

$$3 \times 96500 = 289500.$$

So the equation becomes

$$17370 = -\,289500 \, E^{0}_{\text{cell}}.$$

Now isolate $$E^{0}_{\text{cell}}$$ by dividing both sides by $$-289500$$:

$$E^{0}_{\text{cell}} = -\,\dfrac{17370}{289500}.$$

Carrying out the division, we note that $$289500 \times 0.06 = 17370$$ exactly, so

$$E^{0}_{\text{cell}} = -\,0.06\,\text{V}.$$

The question requests the answer in the form $$\text{(number)} \times 10^{-2}$$ volts. Writing $$-0.06\,\text{V}$$ as $$-6 \times 10^{-2}\,\text{V},$$ we identify the numerical coefficient as $$-6$$.

So, the answer is $$-6$$.

According to Faraday’s first law of electro-lysis,

$$m \;=\;\frac{Q\,M}{n\,F}$$

where  $$m$$ is the mass deposited, $$Q$$ is the charge passed, $$M$$ is the molar mass of the metal ion, $$n$$ is the number of electrons required per ion, and $$F$$ is Faraday’s constant $$(96500\ \text{C mol}^{-1})$$.

We are told that $$108\ \text{g}$$ of silver is deposited. The molar mass of silver is also $$108\ \text{g mol}^{-1}$$, so

$$n_{\text{Ag}}=\frac{108\ \text{g}}{108\ \text{g mol}^{-1}}=1\ \text{mol}.$$

The cathodic reaction for silver deposition is

$$\text{Ag}^+ + e^- \;\longrightarrow\; \text{Ag}. $$

Each mole of Ag+ needs $$1$$ mole of electrons $$(n=1)$$. Therefore, the charge that must have passed is

$$Q = n_{\text{e}^-}\,F = 1\ \text{mol}\times 96500\ \text{C mol}^{-1}=96500\ \text{C}.$$

Exactly the same charge is now sent through water to produce oxygen gas. The anode reaction in water electro-lysis is

$$2\,\text{H}_2\text{O} \;\longrightarrow\; \text{O}_2 + 4\,\text{H}^+ + 4\,e^-.$$

This equation shows that $$4$$ moles of electrons give $$1$$ mole of $$ \text{O}_2$$. Hence, the moles of oxygen obtained with the available charge are

$$n_{\text{O}_2}= \frac{n_{\text{e}^-}}{4}= \frac{1\ \text{mol}}{4}=0.25\ \text{mol}.$$

At $$273\ \text{K}$$ and $$1\ \text{bar}$$ the molar volume of an ideal gas is $$22.7\ \text{L mol}^{-1}$$. Therefore, the volume of oxygen liberated equals

$$V = n_{\text{O}_2}\times 22.7\ \text{L mol}^{-1} = 0.25\ \text{mol}\times 22.7\ \text{L mol}^{-1} = 5.675\ \text{L}.$$

Rounding to three significant figures gives $$5.66\ \text{L}$$.

So, the answer is $$5.66\ \text{L}.$$

We have the cell

$$\text{Sn(s)}\;|\;\text{Sn}^{2+}(1\,\text{M})\;||\;\text{Pb}^{2+}(1\,\text{M})\;|\;\text{Pb(s)}$$

The standard reduction potentials are given as

$$E_{\,\text{Sn}^{2+}/\text{Sn}}^{0}= -0.14\ \text{V},\qquad E_{\,\text{Pb}^{2+}/\text{Pb}}^{0}= -0.13\ \text{V}.$$

The electrode with the higher (more positive) reduction potential acts as the cathode. Hence

$$\text{Cathode (reduction)}:\quad \text{Pb}^{2+}+2e^- \rightarrow \text{Pb},\qquad E_{\text{red}}^{0}= -0.13\ \text{V}$$

$$\text{Anode (oxidation)}:\quad \text{Sn}\rightarrow \text{Sn}^{2+}+2e^-,\qquad E_{\text{ox}}^{0}= +0.14\ \text{V}$$

Using $$E_{\text{cell}}^{0}=E_{\text{cathode}}^{0}-E_{\text{anode}}^{0}$$ we get

$$E_{\text{cell}}^{0}=(-0.13\ \text{V})-(-0.14\ \text{V})=+0.01\ \text{V}.$$

The overall cell reaction is

$$\text{Sn(s)}+\text{Pb}^{2+}(aq)\rightleftharpoons \text{Sn}^{2+}(aq)+\text{Pb(s)}.$$

For this reaction the reaction quotient is

$$Q=\frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}.$$

The Nernst equation (stated for an $$n$$-electron process) is

$$E_{\text{cell}} = E_{\text{cell}}^{0}-\frac{0.06}{n}\log Q,$$

where $$n=2$$ for the transfer of two electrons.

At equilibrium $$E_{\text{cell}}=0$$, so

$$0 = E_{\text{cell}}^{0}-\frac{0.06}{2}\log Q_{\text{eq}}.$$

Rearranging,

$$\frac{0.06}{2}\log Q_{\text{eq}} = E_{\text{cell}}^{0}.$$

Substituting $$E_{\text{cell}}^{0}=0.01\ \text{V}$$,

$$\log Q_{\text{eq}} = \frac{2\times 0.01}{0.06} = \frac{0.02}{0.06} = \frac13 = 0.333\ldots$$

So

$$Q_{\text{eq}} = 10^{0.333\ldots} \approx 2.15.$$

Because $$Q_{\text{eq}} = \dfrac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}$$, we have

$$\frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} \approx 2.15.$$

Hence, the correct answer is Option 2.15.

We have to prepare potassium chlorate, $$\text{KClO}_3$$, by electrolysis. The given anodic half-reaction is

$$6\text{OH}^- + \text{Cl}^- \;\longrightarrow\; \text{ClO}_3^- + 3\text{H}_2\text{O} + 6e^-$$

This equation tells us, first of all, how many electrons are needed. One mole of $$\text{ClO}_3^-$$ (and hence one mole of $$\text{KClO}_3$$) is produced by the loss of $$6$$ moles of electrons. So the electron requirement is

$$n_{e^-} = 6 \; \text{mol e}^- \; \text{per mol of } \text{KClO}_3$$

Now let us find how many moles of $$\text{KClO}_3$$ we want to make. The mass required is $$10\;\text{g}$$ and the molar mass is $$122\;\text{g mol}^{-1}$$, so

$$n_{\text{KClO}_3} \;=\; \frac{10\;\text{g}}{122\;\text{g mol}^{-1}} = 0.08197\;\text{mol}$$

Next, we convert this amount of product into the corresponding amount of electrons. Multiplying by the factor of $$6$$ from the balanced half-reaction we get

$$n_{e^-} = 0.08197\;\text{mol} \times 6 = 0.4918\;\text{mol e}^-$$

Faraday’s law of electrolysis states that the total charge required is given by

$$Q = n_{e^-}\,F$$

where $$F = 96\,500\;\text{C mol}^{-1}$$ is the Faraday constant. Substituting the value of $$n_{e^-}$$ we obtain

$$Q_{\text{ideal}} = 0.4918\;\text{mol}\times 96\,500\;\text{C mol}^{-1} = 4.7427 \times 10^{4}\;\text{C}$$

However, we are told that only $$60\%$$ of the current actually does the desired work. The current efficiency is therefore $$0.60$$, meaning the real charge that has to pass through the cell is larger than the ideal charge. We divide by the efficiency:

$$Q_{\text{actual}} = \frac{Q_{\text{ideal}}}{0.60} = \frac{4.7427 \times 10^{4}\;\text{C}}{0.60} = 7.9045 \times 10^{4}\;\text{C}$$

Now let us translate this charge into the time required when the current is $$2\;\text{A}$$. The basic relation is

$$Q = I\,t$$

so

$$t = \frac{Q_{\text{actual}}}{I} = \frac{7.9045 \times 10^{4}\;\text{C}}{2\;\text{A}} = 3.9523 \times 10^{4}\;\text{s}$$

To express this time in hours we divide by $$3600\;\text{s h}^{-1}$$, i.e.

$$t = \frac{3.9523 \times 10^{4}\;\text{s}}{3600\;\text{s h}^{-1}} = 10.978\;\text{h}$$

Finally, we round to the nearest whole hour:

$$t \approx 11\;\text{h}$$

Hence, the correct answer is Option 11.

We have the disproportionation reaction

$$2\,Cu^+(aq)\;\rightleftharpoons\;Cu(s)+Cu^{2+}(aq)$$

To find $$\ln K$$, we first need the standard cell potential $$E^\circ_{\text{cell}}$$ for the reaction. The two relevant half-cells are :

Reduction half-cell $$Cu^+(aq)+e^- \;\longrightarrow\;Cu(s),\qquad E^\circ_{\text{red}}=0.52\;{\rm V}$$

Reduction half-cell $$Cu^{2+}(aq)+e^- \;\longrightarrow\;Cu^+(aq),\qquad E^\circ_{\text{red}}=0.16\;{\rm V}$$

In the disproportionation process one $$Cu^+$$ ion is reduced to $$Cu(s)$$ (cathode) and another $$Cu^+$$ ion is oxidised to $$Cu^{2+}$$ (anode). The anode potential is obtained by reversing the second half-reaction, so

$$E^\circ_{\text{ox}} = -\,0.16\;{\rm V}$$

The cell potential is therefore

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode (red)}} = 0.52\;{\rm V} - 0.16\;{\rm V} = 0.36\;{\rm V}$$

Now we relate $$E^\circ_{\text{cell}}$$ to the equilibrium constant using the thermodynamic relations

$$\Delta G^\circ = -nF\,E^\circ_{\text{cell}}\qquad\text{and}\qquad\Delta G^\circ = -RT\ln K$$

Equating the two expressions gives

$$-nF\,E^\circ_{\text{cell}} = -RT\ln K$$

$$\Rightarrow\;\ln K = \dfrac{nF}{RT}\,E^\circ_{\text{cell}}$$

For the overall reaction one mole of electrons is transferred, so $$n = 1$$. The problem supplies $$\dfrac{RT}{F} = 0.025$$; therefore

$$\frac{F}{RT} = \frac{1}{RT/F} = \frac{1}{0.025} = 40$$

Substituting all the values, we get

$$\ln K = \bigl(1\bigr)\times 40 \times 0.36 = 14.4$$

This can be written as

$$14.4 = 144 \times 10^{-1}$$

So, the answer is $$144$$.

We begin by recalling the Einstein photo-electric equation. When monochromatic light of energy $$h\nu$$ falls on a metal of work function $$w_0$$, the maximum kinetic energy of the photo-electrons is

$$ \dfrac{1}{2}mv^{2}_{\max}=h\nu-w_0 $$

If the electrons are just stopped by a retarding (stopping) potential $$V_s$$, then each electron of charge $$e$$ loses an energy $$eV_s$$, so

$$ eV_s=h\nu-w_0\quad\Longrightarrow\quad V_s=\dfrac{h\nu}{e}-w_0/e. $$

Thus, for the same light, the change in the stopping potential is exactly the change in the work function expressed in electron-volts, because $$1\ \text{eV}=1\ \text{V}\cdot e$$ for a single electron:

$$\Delta V_s=\Delta w_0\ (\text{in eV}).$$

For Na and K we have

$$ w_0(\text{Na})=2.30\ \text{eV},\qquad w_0(\text{K})=2.25\ \text{eV}. $$

Therefore

$$ \Delta w_0=w_0(\text{Na})-w_0(\text{K})=2.30-2.25=0.05\ \text{eV}, $$

so the stopping potential for K must be larger by

$$ \Delta V_s = 0.05\ \text{V}. $$

Next, we analyze the electrochemical cell that supplies the stopping potential:

$$ \text{Pt(s)}\vert\text{H}_2(g,1\ \text{bar})\vert\text{HCl(aq, pH}=1)\vert\text{AgCl(s)}\vert\text{Ag(s)}. $$

The left-hand electrode is the hydrogen electrode, the right-hand electrode is the AgCl/Ag electrode. All electrode potentials will be written as reduction potentials.

(i) Reduction potential of the $$\text{H}^+/\text{H}_2$$ couple

For the reduction

$$ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) $$

the Nernst equation is

$$ E_{\text{H}^+/\text{H}_2} = E^{0} + 0.06\log[\text{H}^+],\qquad E^{0}=0.00\ \text{V}. $$

At $$\text{pH}=1$$ we have $$[\text{H}^+]=10^{-1}\ \text{M}$$, so

$$ \log[\text{H}^+] = \log(10^{-1}) = -1, $$

and therefore

$$ E_{\text{H}^+/\text{H}_2} = 0 + 0.06(-1) = -0.06\ \text{V}. $$

(ii) Reduction potential of the $$\text{AgCl/Ag/Cl}^-$$ couple

The reduction is

$$ \text{AgCl(s)} + e^- \rightarrow \text{Ag(s)} + \text{Cl}^- (\text{aq}). $$

The Nernst equation is

$$ E_{\text{AgCl/Ag}} = E^{0}_{\text{AgCl/Ag}} - 0.06\log[\text{Cl}^-],\qquad E^{0}_{\text{AgCl/Ag}} = 0.22\ \text{V}. $$

Because the electrolyte is HCl and the acid is strong and completely dissociated, $$[\text{Cl}^-]=[\text{H}^+]=10^{-1}\ \text{M}$$ at pH = 1. Hence

$$ \log[\text{Cl}^-] = -1,\qquad E_{\text{AgCl/Ag}} = 0.22 - 0.06(-1) = 0.22 + 0.06 = 0.28\ \text{V}. $$

(iii) Emf of the cell at pH = 1

The cell emf is

$$ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = E_{\text{AgCl/Ag}} - E_{\text{H}^+/\text{H}_2} = 0.28\ \text{V} - (-0.06\ \text{V}) = 0.34\ \text{V}. $$

This 0.34 V exactly balances the stopping potential for Na, so

$$ V_s(\text{Na}) = 0.34\ \text{V}. $$

(iv) Stopping potential required for K

Since the work function of K is 0.05 eV lower,

$$ V_s(\text{K}) = V_s(\text{Na}) + 0.05\ \text{V} = 0.34 + 0.05 = 0.39\ \text{V}. $$

(v) Required emf of the cell at unknown pH = $$x$$

Let the new pH be $$x$$. Then

$$ [\text{H}^+] = 10^{-x},\qquad [\text{Cl}^-] = 10^{-x}. $$

The reduction potentials now become

$$ E_{\text{H}^+/\text{H}_2} = 0.06\log[\text{H}^+] = 0.06\log(10^{-x}) = 0.06(-x) = -0.06x\ \text{V}, $$

$$ E_{\text{AgCl/Ag}} = 0.22 - 0.06\log[\text{Cl}^-] = 0.22 - 0.06\log(10^{-x}) = 0.22 - 0.06(-x) = 0.22 + 0.06x\ \text{V}. $$

Therefore the emf is

$$ \begin{aligned} E_{\text{cell}} &= E_{\text{AgCl/Ag}} - E_{\text{H}^+/\text{H}_2} \\ &= (0.22 + 0.06x) - (-0.06x) \\ &= 0.22 + 0.12x\ \text{V}. \end{aligned} $$

This emf must equal the required stopping potential of 0.39 V:

$$ 0.22 + 0.12x = 0.39. $$

Solving for $$x$$,

$$ 0.12x = 0.39 - 0.22 = 0.17, $$

$$ x = \dfrac{0.17}{0.12} = 1.416\ (\text{approximately}). $$

So the required pH is

$$ \text{pH} \;=\; 1.42\quad(\text{to two decimal places}). $$

The question asks us to express the answer as “……. $$\times 10^{-2}$$ (to the nearest integer)”. Writing

$$ 1.42 = 142 \times 10^{-2}, $$

the integer sought is 142.

Hence, the correct answer is Option 142.

First, recall the fundamental relation between standard reduction potential and oxidising strength.

Statement of principle: “The larger (more positive) the standard reduction potential $$E^{0}$$ of a half-cell, the greater is the tendency of the species on the left side of the reduction half-reaction to accept electrons; therefore, that species is the stronger oxidising agent.”

For each species in the question the relevant reduction half-reaction, written in the conventional “reduction” direction (electrons on the left), is as follows. We shall write the value of $$E^{0}$$ alongside each equation.

$$O_2 + 4H^{+} + 4e^{-} \longrightarrow 2H_2O \qquad E^{0}_{O_2/H_2O} = +1.23\ \text{V}$$

$$S_2O_8^{2-} + 2e^{-} \longrightarrow 2SO_4^{2-} \qquad E^{0}_{S_2O_8^{2-}/SO_4^{2-}} = +2.05\ \text{V}$$

$$Br_2 + 2e^{-} \longrightarrow 2Br^{-} \qquad E^{0}_{Br_2/Br^{-}} = +1.09\ \text{V}$$

$$Au^{3+} + 3e^{-} \longrightarrow Au \qquad E^{0}_{Au^{3+}/Au} = +1.40\ \text{V}$$

Now we compare the numerical values:

$$\begin{aligned} E^{0}_{S_2O_8^{2-}/SO_4^{2-}} &= +2.05\ \text{V} \\ E^{0}_{Au^{3+}/Au} &= +1.40\ \text{V} \\ E^{0}_{O_2/H_2O} &= +1.23\ \text{V} \\ E^{0}_{Br_2/Br^{-}} &= +1.09\ \text{V} \end{aligned}$$

Because $$+2.05\ \text{V}$$ is the highest (most positive) among all the listed potentials, the species on the left of that particular half-reaction, namely $$S_2O_8^{2-}$$ (peroxodisulphate ion), possesses the greatest tendency to be reduced.

Hence, according to the principle stated earlier, $$S_2O_8^{2-}$$ is the strongest oxidising agent in the given set.

Therefore, the option corresponding to $$S_2O_8^{2-}$$ is correct.

Hence, the correct answer is Option B.

We begin by recalling the basic electrochemical fact that the standard reduction potential $$E^\circ$$ of a half-cell tells us how readily the species on the left side of that half-reaction accepts electrons and gets reduced. The larger (more positive) the value of $$E^\circ$$, the greater is this tendency to be reduced and, correspondingly, the stronger is that species as an oxidizing agent. In other words, if we have two species $$A^{m+}$$ and $$B^{n+}$$ with $$E^\circ_A < E^\circ_B$$, then

$$A^{m+} \; \text{is a weaker oxidizing agent than} \; B^{n+}.$$

Keeping this criterion in mind, let us list the given half-reactions together with their standard reduction potentials:

$$\text{(1)}\; \mathrm{Co^{3+} + e^- \to Co^{2+}},\qquad E^\circ = +1.81\ \text{V}$$ $$\text{(2)}\; \mathrm{Pb^{4+} + 2e^- \to Pb^{2+}},\qquad E^\circ = +1.67\ \text{V}$$ $$\text{(3)}\; \mathrm{Ce^{4+} + e^- \to Ce^{3+}},\qquad E^\circ = +1.61\ \text{V}$$ $$\text{(4)}\; \mathrm{Bi^{3+} + 3e^- \to Bi},\qquad\;\; E^\circ = +0.20\ \text{V}$$

(The problem statement writes $$\mathrm{Pb^{3+}}$$, but the tabulated value $$+1.67\ \text{V}$$ actually corresponds to the well-known $$\mathrm{Pb^{4+}/Pb^{2+}}$$ couple, so we use $$\mathrm{Pb^{4+}}$$ in the comparison, exactly as appears in the options.)

Now we arrange these species in order of increasing $$E^\circ$$ because, as argued above, increasing $$E^\circ$$ means increasing oxidizing power:

$$E^\circ(\mathrm{Bi^{3+}}) = +0.20\ \text{V} \lt\; E^\circ(\mathrm{Ce^{4+}}) = +1.61\ \text{V} \lt\; E^\circ(\mathrm{Pb^{4+}}) = +1.67\ \text{V} \lt\; E^\circ(\mathrm{Co^{3+}}) = +1.81\ \text{V}.$$

Translating this numerical order back into words, we obtain the sequence from the weakest to the strongest oxidizing agent:

$$\mathrm{Bi^{3+}} \lt \mathrm{Ce^{4+}} \lt \mathrm{Pb^{4+}} \lt \mathrm{Co^{3+}}.$$

This exactly matches Option B in the given list.

Hence, the correct answer is Option B.

First, we recall Faraday’s first law of electrolysis, which states that the amount of substance liberated or deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, the number of moles of electrons involved equals the number of Faradays of charge supplied:

$$\text{Moles of } e^- = \text{Faradays supplied}$$

We are told that $$0.1$$ Faraday of electricity is passed. Therefore, the moles of electrons that actually flow through the circuit are

$$n_{e^-}=0.1\text{ mol}$$

Now we look at the cathode reaction for nickel(II) ions. The ionic species present is $$\text{Ni}^{2+}$$ because the salt is $$\text{Ni(NO}_3)_2$$. At the cathode, reduction occurs, so the half-reaction is

$$\text{Ni}^{2+}+2e^- \;\longrightarrow\; \text{Ni(s)}$$

This equation shows clearly that two moles of electrons are required to deposit one mole of nickel metal. We translate this stoichiometric requirement into a proportionality:

$$2\ \text{mol } e^- \;\Longrightarrow\; 1\ \text{mol Ni}$$

To find the moles of nickel actually deposited, we set up the ratio using the number of electrons that have passed:

$$\text{Moles of Ni}=\dfrac{\text{Moles of } e^-}{2} =\dfrac{0.1}{2}$$

Carrying out the division gives

$$\text{Moles of Ni}=0.05\text{ mol}$$

Thus, the quantity of nickel metal formed on the cathode is precisely $$0.05$$ mole.

Hence, the correct answer is Option C.