JEE Electrochemistry Questions

The rule for the complete (exhaustive) hydrolysis of an inter-halogen $$XY_n$$ is:
  • Each $$F$$ (or the more electronegative halogen) is replaced by $$OH$$, giving $$HF$$ in the medium.
  • The central halogen $$X$$ keeps the same oxidation number it had in $$XY_n$$ and appears as an oxy-anion $$XO_m^{\,(n-2m+1)-}$$ after loss of the acidic hydrogens.

Let us apply this idea one compound at a time.

$$\textbf{ICl}$$
Oxidation number of I: $$+1$$ (because $$\text{Cl}$$ is more electronegative).
Step-1 (replacement by $$OH$$): $$\mathrm{ICl + H_2O \;\longrightarrow\; HOI + HCl}$$
Step-2 (deprotonation in water): $$\mathrm{HOI \;\longrightarrow\; IO^- + H^+}$$
Thus the stable oxy-anion is $$\mathbf{IO^-}$$ (iodide in the +1 state), together with $$\mathrm{Cl^-}$$ from $$\mathrm{HCl}$$.

$$\textbf{ClF$$_3$$}$$
Oxidation number of Cl: $$+3$$.
Step-1: $$\mathrm{ClF_3 + 3\,H_2O \;\longrightarrow\; Cl(OH)_3 + 3\,HF}$$
Step-2: $$\mathrm{Cl(OH)_3 \;\longrightarrow\; HClO_2 + H_2O}$$ (re-arrangement to the oxy-acid having Cl in +3).
Step-3 (deprotonation): $$\mathrm{HClO_2 \;\longrightarrow\; ClO_2^- + H^+}$$
Hence the oxy-anion present after complete hydrolysis is $$\mathbf{ClO_2^-}$$.

$$\textbf{BrF$$_5$$}$$
Oxidation number of Br: $$+5$$.
Step-1: $$\mathrm{BrF_5 + 5\,H_2O \;\longrightarrow\; Br(OH)_5 + 5\,HF}$$
Step-2: $$\mathrm{Br(OH)_5 \;\longrightarrow\; HBrO_3 + 2\,H_2O}$$ (conversion to the oxy-acid with Br in +5).
Step-3 (deprotonation): $$\mathrm{HBrO_3 \;\longrightarrow\; BrO_3^- + H^+}$$
Therefore the final oxy-anion is $$\mathbf{BrO_3^-}$$.

Collecting the three results:
ICl  ➔ $$IO^-$$
ClF$$_3$$ ➔ $$ClO_2^-$$
BrF$$_5$$ ➔ $$BrO_3^-$$

The sequence matches Option A.

Answer: Option A which is: IO$$^{-}$$, ClO$$_2^{-}$$ and BrO$$_3^{-}$$

Permanganate ion $$\left(MnO_4^- \right)$$ is a strong oxidising agent. The species to which it gets reduced and the product obtained from the reducing agent (here $$I^-$$) depend on the pH of the solution.

Useful facts to remember:
  • In acidic medium: $$MnO_4^- \rightarrow Mn^{2+}$$ and $$I^- \rightarrow I_2$$.
  • In neutral (or weakly basic) medium: $$MnO_4^- \rightarrow MnO_2$$ and $$I^- \rightarrow IO_3^-$$.
  • In strongly alkaline medium: $$MnO_4^- \rightarrow MnO_4^{2-}$$ and $$I^- \rightarrow IO_4^-$$.

Because the question specifically mentions a neutral aqueous medium, we focus on the second line above.

Half-reactions in neutral solution:
Reduction half-reaction for permanganate:
$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$ $$-(1)$$
Oxidation half-reaction for iodide:
$$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$$ $$-(2)$$

To combine $$-(1)$$ and $$-(2)$$, multiply $$-(1)$$ by 2 so that the electrons cancel (total 6 e$$^-$$):
$$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$ $$-(3)$$

Add $$-(2)$$ and $$-(3)$$:

$$2MnO_4^- + I^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$
$$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$$

Electrons cancel. Combine, then simplify water and $$H^+/OH^-$$ to obtain the overall balanced equation in neutral medium:

$$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$$

The iodine species formed is $$IO_3^-$$ (iodate ion).

Hence, in a neutral aqueous medium, the reaction of permanganate ion with iodide ion produces iodate ion.

Option B which is: IO$$_3^-$$

Intermolecular interaction energies are obtained by applying Coulomb’s law to the various charge distributions that can exist on (or be induced in) molecules - permanent charges, permanent dipoles, induced dipoles, etc. The distance dependence and, in some cases, the temperature dependence of the average energy decide whether a given statement is true or false.

Option A
• Potential energy between two point charges (say $$q_1$$ and $$q_2$$) is $$U_{qq}= \dfrac{q_1q_2}{4\pi\varepsilon_0\,r} \propto \dfrac{1}{r}$$.
• Potential energy between a point charge $$q$$ and a point dipole of moment $$\mu$$ (the dipole oriented so that the interaction is maximum) is $$U_{q\mu}= -\dfrac{q\mu}{4\pi\varepsilon_0\,r^{2}} \propto \dfrac{1}{r^{2}}$$.
Because $$1/r^{2}$$ tends to zero faster than $$1/r$$ when $$r\rightarrow\infty$$, the charge-dipole energy approaches zero more rapidly than the charge-charge energy. Option A claims the opposite, hence it is false.

Option B
For two freely rotating permanent dipoles, the orientation-averaged (thermal average) interaction energy has Keesom form
$$\langle U_{\mu\mu}\rangle = -\dfrac{2\mu_1^{2}\mu_2^{2}}{3(4\pi\varepsilon_0)^{2}k_{\mathrm B}T\,r^{6}} \propto -\dfrac{1}{T\,r^{6}}.$$
Thus the energy falls off as $$1/r^{6}$$, not $$1/r^{3}$$. Therefore Option B is false.

Option C
Dipole-induced dipole interaction (Debye interaction): a permanent dipole $$\mu$$ induces a dipole $$\alpha \,E$$ in a neighbouring polarisable molecule (where $$\alpha$$ is its polarisability and $$E$$ the electric field of the permanent dipole). The orientation-averaged energy is
$$\langle U_{\mu\text{-ind}}\rangle = -\dfrac{\mu^{2}\alpha}{(4\pi\varepsilon_0)^{2}\,2\,r^{6}},$$
which contains no temperature term. Hence it is independent of temperature. Option C is true.

Option D
Even molecules with no permanent dipole can attract each other through London dispersion forces. A momentary (instantaneous) dipole in one molecule induces a dipole in the other, giving an average interaction energy
$$\langle U_{\text{disp}}\rangle = -\dfrac{3h\nu\,\alpha_1\alpha_2}{4(4\pi\varepsilon_0)^{2}\,r^{6}},$$
which is always attractive. Hence non-polar molecules do attract one another despite lacking permanent dipoles. Option D is true.

Therefore the correct statements are:
Option C (dipole-induced dipole energy is temperature-independent) and
Option D (non-polar molecules attract each other via dispersion forces).

Final answer: Option C and Option D.

The oxo-acids of phosphorus can be recognised from the general formula $$H_nP_mO_x$$ and by remembering that:

• Every hydrogen attached to oxygen appears as an $$-OH$$ group and is ionisable (acidic).
• Any hydrogen that is written directly next to phosphorus in the condensed formula is attached to phosphorus itself; that bond is a $$P-H$$ bond. Such hydrogens are non-ionisable and take part in reduction reactions.

Let us inspect each option.

Option A : $$H_3PO_4$$ (Orthophosphoric acid)
Expanded structure: $$\;HO-P(=O)(OH)_2$$.
All three hydrogens are bonded to oxygen as $$-OH$$ groups, so there is no $$P-H$$ bond in this molecule.

Option B : $$H_3PO_3$$ (Phosphorous acid)
The correct expanded formula is $$\;P(OH)_3$$.
Here two hydrogens are bonded to oxygen ($$-OH$$ groups) and one hydrogen is directly attached to phosphorus. Therefore the molecule contains one $$P-H$$ bond.

Option C : $$H_4P_2O_7$$ (Pyrophosphoric acid)
This acid is obtained by condensation of two $$H_3PO_4$$ units: $$HO-P(=O)(OH)-O-P(=O)(OH)_2$$.
All four hydrogens appear in $$-OH$$ groups; again, no $$P-H$$ bond is present.

Option D : $$H_3PO_2$$ (Hypophosphorous acid)
Expanded structure: $$\;H_2P(=O)(OH)$$.
One hydrogen forms an $$-OH$$ group, while two hydrogens are directly attached to phosphorus. Thus the molecule contains two $$P-H$$ bonds.

Hence, the acids that contain at least one $$P-H$$ bond are:
Option D which is: $$H_3PO_2$$.

All other options lack a $$P-H$$ bond.

Final answer: Option D (H$$_3$$PO$$_2$$)

For any ion, magnetic behaviour depends on the presence or absence of unpaired electrons.
No unpaired electrons ⇒ diamagnetic, at least one unpaired electron ⇒ paramagnetic.

Lanthanide ground-state configurations are based on the xenon core $$[Xe]$$ followed by $$4f$$, $$5d$$ and $$6s$$ subshells. While writing ionic configurations, electrons are removed first from $$6s$$, then $$5d$$, then $$4f$$.

Case A (La$$^{3+}$$, Ce$$^{4+}$$)

Lanthanum (Z = 57): neutral atom $$[Xe]\,5d^{1}6s^{2}$$.
La$$^{3+}$$ : remove three electrons (two from $$6s$$, one from $$5d$$) ⇒ $$[Xe]\,4f^{0}$$.
All subshells are completely filled; no unpaired electrons ⇒ diamagnetic.

Cerium (Z = 58): neutral atom $$[Xe]\,4f^{1}5d^{1}6s^{2}$$ (the commonly quoted ground state).
Ce$$^{4+}$$ : remove four electrons (two $$6s$$, one $$5d$$, one $$4f$$) ⇒ $$[Xe]\,4f^{0}$$.
Again, no unpaired electrons ⇒ diamagnetic.

Case B (Yb$$^{2+}$$, Lu$$^{3+}$$)

Ytterbium (Z = 70): neutral atom $$[Xe]\,4f^{14}6s^{2}$$ (filled $$4f$$ shell).
Yb$$^{2+}$$ : remove the two $$6s$$ electrons ⇒ $$[Xe]\,4f^{14}$$.
The $$4f$$ subshell is completely filled; no unpaired electrons ⇒ diamagnetic.

Lutetium (Z = 71): neutral atom $$[Xe]\,4f^{14}5d^{1}6s^{2}$$.
Lu$$^{3+}$$ : remove two $$6s$$ and one $$5d$$ electron ⇒ $$[Xe]\,4f^{14}$$.
Filled $$4f^{14}$$ subshell; no unpaired electrons ⇒ diamagnetic.

Case C (La$$^{2+}$$, Ce$$^{3+}$$)

La$$^{2+}$$ : $$[Xe]\,5d^{1}$$ - one electron in $$5d$$ ⇒ unpaired ⇒ paramagnetic.
Ce$$^{3+}$$ : $$[Xe]\,4f^{1}$$ - one electron in $$4f$$ ⇒ unpaired ⇒ paramagnetic.
Hence the pair is not diamagnetic.

Case D (Yb$$^{3+}$$, Lu$$^{2+}$$)

Yb$$^{3+}$$ : $$[Xe]\,4f^{13}$$ - one vacancy (equivalently one unpaired electron) in $$4f$$ ⇒ paramagnetic.
Lu$$^{2+}$$ : $$[Xe]\,4f^{14}5d^{1}$$ - one electron in $$5d$$ ⇒ unpaired ⇒ paramagnetic.
Hence this pair is not diamagnetic.

Therefore, the diamagnetic pairs are:
Option A (La$$^{3+}$$, Ce$$^{4+}$$) and Option B (Yb$$^{2+}$$, Lu$$^{3+}$$).

Option A ( La$$^{3+}$$, Ce$$^{4+}$$ ), Option B ( Yb$$^{2+}$$, Lu$$^{3+}$$ )

The half-reaction for the reduction of dichromate ions in acidic medium is: $$Cr_2O_7^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_2O$$.

From the stoichiometry, $$6$$ electrons (or $$6$$ Faradays) produce $$2$$ moles of $$Cr^{3+}$$.
Therefore, electrons required per mole of $$Cr^{3+}$$:

$$\frac{6\text{ mol e}^{-}}{2\text{ mol }Cr^{3+}}=3\text{ mol e}^{-}\text{ per mol }Cr^{3+}$$.

To obtain $$1$$ mole of $$Cr^{3+}$$, charge needed is:
$$Q = 3F = 3\times 96500\ \text{C mol}^{-1}=289500\ \text{C}$$.

The current flows for $$48.25\ \text{min}$$:
$$t = 48.25\times 60 = 2895\ \text{s}$$.

Current is $$I = \dfrac{Q}{t} = \dfrac{289500\ \text{C}}{2895\ \text{s}} = 100\ \text{A}$$.

Hence, the required current is 100 A.

For any crystalline solid, density is obtained from a single unit cell using the relation

$$\rho = \frac{Z \, M}{N_A \, a^{3}}$$
where
  $$Z$$ = number of atoms per unit cell,
  $$M$$ = molar mass (g mol$$^{-1}$$),
  $$N_A$$ = Avogadro’s constant (mol$$^{-1}$$),
  $$a$$ = edge length of the cubic unit cell (cm).

Step 1: Identify the parameters for the given lattice.
A cubic close-packed (ccp) lattice is face-centred cubic (fcc), so $$Z = 4$$.

Step 2: Convert the edge length to centimetres.
Given $$a = 400 \text{ pm}$$.
$$1 \text{ pm} = 10^{-12} \text{ m} = 10^{-10} \text{ cm}$$.
Hence $$a = 400 \times 10^{-10} \text{ cm} = 4.0 \times 10^{-8} \text{ cm}$$.

Step 3: Cube the edge length.
$$a^{3} = (4.0 \times 10^{-8}\,\text{cm})^{3} = 4^{3} \times 10^{-24}\,\text{cm}^{3} = 64 \times 10^{-24}\,\text{cm}^{3} = 6.4 \times 10^{-23}\,\text{cm}^{3}$$.

Step 4: Substitute all values into the density formula.
Numerator: $$Z \, M = 4 \times 105.6 = 422.4 \text{ g mol}^{-1}$$.
Denominator: $$N_A \, a^{3} = (6 \times 10^{23}) \times (6.4 \times 10^{-23}) = 38.4$$.

Therefore,
$$\rho = \frac{422.4}{38.4} = 11.0 \text{ g cm}^{-3}$$.

The calculated density (to three significant figures) is $$11.0 \text{ g cm}^{-3}$$, which lies in the required range 10.85-11.1.

The van der Waals equation for one mole is$$(P+\frac{a}{V_m^{\,2}})(V_m-b)=RT\;.$$

Expand and rearrange to obtain a cubic polynomial in $$V_m$$:

$$P V_m - P b + \frac{a}{V_m}-\frac{a b}{V_m^{\,2}}=RT$$

Multiply every term by $$V_m^{\,2}$$ to clear the denominators:

$$P V_m^{\,3}-P b V_m^{\,2}+a V_m-a b = R T V_m^{\,2}$$

Collect like terms on one side:

$$P V_m^{\,3}-\left(P b + R T\right) V_m^{\,2}+a V_m-a b = 0\;.$$

Thus, in the cubic $$A V_m^{\,3}+B V_m^{\,2}+C V_m + D=0$$ we have
  • $$A = P$$
  • $$B = -\left(P b + R T\right)$$
  • $$C = a$$
  • $$D = -a b$$

The required ratio is the coefficient of $$V_m^{\,2}$$ divided by the coefficient of $$V_m$$:

$$\frac{B}{C}= \frac{-\left(P b + R T\right)}{a}$$

Insert the given data (units already consistent in dm and atm):
  • $$P = 300\ \text{atm}$$
  • $$b = 0.060\ \text{dm}^3\ \text{mol}^{-1}$$ ⇒ $$P b = 300 \times 0.060 = 18.0$$
  • $$R = 0.082\ \text{dm}^3\ \text{atm}\ \text{mol}^{-1}\ \text{K}^{-1}$$, $$T = 300\ \text{K}$$ ⇒ $$R T = 0.082 \times 300 = 24.6$$
  • $$a = 6.0\ \text{dm}^6\ \text{atm}\ \text{mol}^{-2}$$

Calculate the numerator:
$$P b + R T = 18.0 + 24.6 = 42.6$$

Finally, the ratio is
$$\frac{B}{C}= -\frac{42.6}{6.0}= -7.1\ \text{mol dm}^{-3}$$

Therefore, the required ratio lies in the range −7.2 to −7.

The balanced combustion of butane under standard conditions is
$$C_4H_{10}+ \frac{13}{2}\,O_2 \rightarrow 4\,CO_2 + 5\,H_2O$$

1. Standard Gibbs energy change, $$\Delta_rG^\circ$$
Using $$\Delta_rG^\circ = \sum \Delta_fG^\circ(\text{products})-\sum \Delta_fG^\circ(\text{reactants})$$,

$$\Delta_rG^\circ = \bigl[4(-394) + 5(-237)\bigr] - \bigl[(-18) + 0\bigr] \text{ kJ}$$
$$= (-1576 -1185) + 18 \text{ kJ}$$
$$= -2743 \text{ kJ mol}^{-1}$$

2. Electrons transferred, $$n$$
Each $$O_2$$ molecule is reduced from oxidation state 0 to -2, accepting $$4$$ electrons. Number of $$O_2$$ molecules $$= \dfrac{13}{2}=6.5$$, hence

$$n = 6.5 \times 4 = 26 \text{ mol e}^-$$

3. Cell potential and definition of $$X$$
Maximum electrical work $$= -\Delta_rG^\circ = nF E^\circ$$, so

$$E^\circ = \frac{-\Delta_rG^\circ}{nF}$$

The problem states $$E^\circ = \dfrac{X}{F}\times 10^{3}\ \text{V}$$, that is
$$E^\circ = \frac{1000\,X}{F}$$

Equating the two expressions for $$E^\circ$$ gives

$$\frac{-\Delta_rG^\circ}{nF} = \frac{1000\,X}{F} \Longrightarrow X = \frac{-\Delta_rG^\circ}{1000\,n}$$

Substituting $$\Delta_rG^\circ = -2743 \text{ kJ mol}^{-1}$$ and $$n = 26$$,

$$X = \frac{2743}{26} \text{ kJ} = 105.5 \text{ kJ}$$

The required value therefore lies in the range $$105.4 - 105.6$$.

In qualitative inorganic analysis, cations are separated into groups by adding specific group reagents. Each reagent precipitates only those ions whose analytical product (sulfide, hydroxide, carbonate, etc.) has a very low solubility under the given conditions. We first recall the standard group scheme:

• Group II reagent  : pass $$H_2S$$ in dilute $$HCl$$ (acidic medium) ⇒ precipitates insoluble sulfides such as $$CuS, CdS, PbS$$ …
• Group III reagent : add $$NH_4OH$$ in the presence of excess $$NH_4Cl$$ (buffered weakly basic) ⇒ precipitates gelatinous hydroxides $$Fe(OH)_3, Al(OH)_3, Cr(OH)_3$$ …
• Group IV reagent : pass $$H_2S$$ in basic medium (presence of $$NH_4OH$$) ⇒ precipitates the less‐insoluble sulfides $$ZnS, NiS, CoS, MnS$$ …
• Group V reagent  : add $$(NH_4)_2CO_3$$ in presence of $$NH_4OH$$ ⇒ precipitates carbonates of alkaline-earth metals $$CaCO_3, SrCO_3, BaCO_3$$ …

Now match each List-I reagent with the metal ion of List-II whose salt is precipitated by that reagent.

Reagent : pass $$H_2S$$ in presence of $$NH_4OH$$ (alkaline) ⇒ Group IV.
Among the given ions, $$Mn^{2+}$$ forms the insoluble sulfide $$MnS$$ under these conditions.
Hence P → 3 ($$Mn^{2+}$$).

Reagent : $$(NH_4)_2CO_3$$ in presence of $$NH_4OH$$ ⇒ Group V.
Of the listed ions, $$Ba^{2+}$$ is an alkaline-earth metal and is precipitated as $$BaCO_3$$.
Hence Q → 4 ($$Ba^{2+}$$).

Reagent : $$NH_4OH$$ with $$NH_4Cl$$ buffer ⇒ Group III.
This reagent precipitates trivalent hydroxides; $$Al^{3+}$$ gives $$Al(OH)_3$$.
Hence R → 2 ($$Al^{3+}$$).

Reagent : pass $$H_2S$$ in dilute $$HCl$$ (acidic) ⇒ Group II.
$$Cu^{2+}$$ is a typical group-II cation, precipitated as black $$CuS$$.
Hence S → 1 ($$Cu^{2+}$$).

Collecting the matches: P → 3, Q → 4, R → 2, S → 1.

Therefore the correct option is:
Option A which is: P → 3; Q → 4; R → 2; S → 1.

Manganese has atomic number 25 with ground-state configuration $$[Ar]\,3d^{5}\,4s^{2}$$.

In both complexes the overall charge is $$-3$$ and each ligand carries $$-1$$ charge, so for six ligands:$$x+(-6)=-3\;\Longrightarrow\;x=+3$$Hence Mn is in the $$+3$$ oxidation state, that is $$Mn^{3+}$$ whose electronic configuration is $$3d^{4}$$ (because the two $$4s$$ electrons and one $$3d$$ electron are lost).

For an octahedral $$d^{4}$$ ion two spin arrangements are possible:
• high-spin (weak-field ligand)
• low-spin (strong-field ligand)

Case 1: $$[Mn(Br)_6]^{3-}$$
$$Br^-$$ is a weak-field ligand. Thus the complex is high spin:$$t_{2g}^{3}e_{g}^{1}$$The four electrons occupy different orbitals according to Hund’s rule, giving $$n=4$$ unpaired electrons.
Spin-only magnetic moment:
$$\mu_1=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}\;\text{B.M.}\approx4.90\;\text{B.M.}$$

Case 2:$$[Mn(CN)_6]^{3-}$$
$$CN^-$$ is a strong-field ligand. The complex is low spin:$$t_{2g}^{4}e_{g}^{0}$$The occupancy inside $$t_{2g}$$ is $$\uparrow\downarrow\,\uparrow\,\uparrow$$, leaving $$n=2$$ unpaired electrons.
Spin-only magnetic moment:
$$\mu_2=\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}\;\text{B.M.}\approx2.83\;\text{B.M.}$$

Sum of the magnetic moments
$$\mu_{\text{total}}=\mu_1+\mu_2\approx4.90+2.83=7.73\;\text{B.M.}$$

The required sum lies in the range $$7.5\text{-}7.8\;\text{B.M.}$$ (as specified).

We are given the following standard electrode potentials:

$$FeO_4^{2-} \xrightarrow{+2.0\text{ V}} Fe^{3+} \xrightarrow{+0.8\text{ V}} Fe^{2+} \xrightarrow{-0.5\text{ V}} Fe^0$$

We need to find $$E^0_{FeO_4^{2-}/Fe^{2+}}$$.

We use the relationship between standard Gibbs free energy and electrode potential:

$$ \Delta G^0 = -nFE^0 $$

For $$FeO_4^{2-} \to Fe^{3+}$$: Fe goes from +6 to +3, so $$n_1 = 3$$ and $$ \Delta G_1^0 = -3F(2.0) = -6.0F $$.

For $$Fe^{3+} \to Fe^{2+}$$: Fe goes from +3 to +2, so $$n_2 = 1$$ and $$ \Delta G_2^0 = -1F(0.8) = -0.8F $$.

Adding these gives for $$FeO_4^{2-} \to Fe^{2+}$$ (where Fe changes from +6 to +2, $$n = 4$$): $$ \Delta G^0 = \Delta G_1^0 + \Delta G_2^0 = -6.0F - 0.8F = -6.8F $$.

Thus,

$$ E^0_{FeO_4^{2-}/Fe^{2+}} = \frac{-\Delta G^0}{nF} = \frac{6.8F}{4F} = 1.7 \text{ V} $$. The correct answer is Option 2: 1.7 V.

Let us analyze each case for $$O_2$$ evolution at the anode during electrolysis:

(A) Aqueous $$AgNO_3$$ with silver electrodes: At the anode, the silver electrode dissolves ($$Ag \to Ag^+ + e^-$$) since silver is oxidized preferentially. No $$O_2$$ evolved.

(B) Aqueous $$AgNO_3$$ with platinum electrodes: At the inert Pt anode, water is oxidized: $$2H_2O \to O_2 + 4H^+ + 4e^-$$. $$O_2$$ is evolved. ✓

(C) Dilute $$H_2SO_4$$ with platinum electrodes: At the Pt anode, water is oxidized to give $$O_2$$. ✓

(D) Concentrated $$H_2SO_4$$ with platinum electrodes: At high concentrations, $$SO_4^{2-}$$ or $$HSO_4^-$$ may get oxidized to form peroxodisulphate ($$S_2O_8^{2-}$$) at the anode instead of $$O_2$$. So $$O_2$$ is not the primary product.

The correct answer is Option 2: (B) and (C) only.

We are given two statements about corrosion and need to determine which are correct.

Analysis of Statement I:

Corrosion is an electrochemical process where a metal deteriorates due to reactions with its environment. According to the electrochemical theory of corrosion, a pure metal surface acts as the anode (where oxidation occurs), and impurities or other conducting surfaces act as the cathode (where reduction occurs). For example, in the rusting of iron, pure iron oxidizes at the anode: $$Fe -> Fe^{2+ + 2e^-}$$, while reduction occurs at the cathode on impurities. Thus, Statement I is true.

Analysis of Statement II:

The statement claims that corrosion is faster in an alkaline medium than in an acidic medium. However, corrosion rates are generally higher in acidic media because acids provide hydrogen ions ($$H+$$) that facilitate the cathodic reduction reaction. For instance, in acidic conditions, the cathode reaction is: $$2H+ + 2e^- -> H_{2}$$, which accelerates the overall corrosion process. In alkaline media, the low concentration of $$H+$$ ions slows down corrosion. Therefore, Statement II is false.

Conclusion: Statement I is true, but Statement II is false. The correct option is C.

Final Answer: C

We need to identify the strongest reducing agent from the given standard reduction potentials.

$$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33$$ V

$$E^\circ_{Cl_2/Cl^-} = 1.36$$ V

$$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51$$ V

$$E^\circ_{Cr^{3+}/Cr} = -0.74$$ V

Key Concept: A reducing agent is a species that gets oxidized (loses electrons). The strongest reducing agent is the species whose oxidation is most thermodynamically favorable, i.e., it has the most negative (or least positive) standard reduction potential.

Each species listed in the options acts as a reducing agent in the reverse of the given half-reactions:

- $$Cr$$ gets oxidized: $$Cr \rightarrow Cr^{3+} + 3e^-$$; $$E^\circ_{red} = -0.74$$ V (most negative)

- $$Cl^-$$ gets oxidized: $$2Cl^- \rightarrow Cl_2 + 2e^-$$; $$E^\circ_{red} = +1.36$$ V

- $$Mn^{2+}$$ gets oxidized: $$Mn^{2+} \rightarrow MnO_4^-$$; $$E^\circ_{red} = +1.51$$ V

- $$MnO_4^-$$ is already in a high oxidation state and acts as an oxidizing agent, not a reducing agent.

The lower the standard reduction potential of a half-cell, the greater the tendency of that species to get oxidized (act as a reducing agent). $$Cr^{3+}/Cr$$ has $$E^\circ = -0.74$$ V, which is the most negative value among all the half-cells listed.

Therefore, $$Cr$$ is the strongest reducing agent.

The correct answer is Option 1: $$Cr$$.

We need to find which cell gives the most negative $$\Delta G°$$.

Key formula: $$\Delta G° = -nFE°_{cell}$$. Most negative $$\Delta G°$$ requires largest positive $$nE°_{cell}$$ value.

Option 1: Zn|Zn²⁺||Ag⁺|Ag

$$E°_{cell} = E°_{cathode} - E°_{anode} = 0.80 - (-0.76) = 1.56$$ V

n = 2 (Zn → Zn²⁺ + 2e⁻, 2Ag⁺ + 2e⁻ → 2Ag)

$$nE° = 2 \times 1.56 = 3.12$$

Option 2: Zn|Zn²⁺||Mg²⁺|Mg

$$E°_{cell} = -2.37 - (-0.76) = -1.61$$ V (negative, not spontaneous)

Option 3: Ag|Ag⁺||Mg²⁺|Mg

$$E°_{cell} = -2.37 - 0.80 = -3.17$$ V (negative, not spontaneous)

Option 4: Cu|Cu²⁺||Ag⁺|Ag

$$E°_{cell} = 0.80 - 0.34 = 0.46$$ V

n = 2, $$nE° = 0.92$$

The largest $$nE°_{cell}$$ is for Option 1 (3.12), giving the most negative $$\Delta G°$$.

The correct answer is Option 1: Zn|Zn²⁺(1M)||Ag⁺(1M)|Ag.

We need to identify which electrolyte produces $$H_2S_2O_8$$ (peroxodisulphuric acid, also known as Marshall’s acid) by electrolysis.

Peroxodisulphuric acid is formed by the electrolysis of concentrated sulphuric acid. At the anode, sulphate ions ($$HSO_4^-$$) are oxidized as shown below:

$$2HSO_4^- \rightarrow H_2S_2O_8 + 2e^-$$

In dilute sulphuric acid solutions, water is preferentially oxidized at the anode to produce oxygen. Only in a concentrated solution is the concentration of $$HSO_4^-$$ ions high enough for their oxidation to form the peroxo bond (S-O-O-S) in $$H_2S_2O_8$$.

The correct answer is Option 4: Concentrated solution of sulphuric acid.

We need to describe how molar conductivity of a weak electrolyte varies with concentration.

We first recall the behavior of weak electrolytes. For weak electrolytes (like $$CH_3COOH$$, $$NH_4OH$$), the degree of dissociation $$\alpha$$ increases significantly as the solution becomes more dilute, following Ostwald’s dilution law: $$\alpha = \sqrt{K_a/c}$$ for dilute solutions.

Next, we consider the relationship between molar conductivity and concentration. Molar conductivity is given by $$\Lambda_m = \alpha \cdot \Lambda_m^\circ$$. As concentration increases, $$\alpha$$ decreases sharply (since $$\alpha \propto 1/\sqrt{c}$$), causing $$\Lambda_m$$ to decrease sharply.

Finally, when we examine the behavior plotted against $$\sqrt{c}$$, we see that unlike strong electrolytes (which show a linear decrease as per the Debye-Hückel-Onsager equation: $$\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$$), weak electrolytes exhibit a sharp, non-linear decrease in molar conductivity with increasing concentration.

The correct answer is Option B) Molar conductivity decreases sharply with increase in concentration.

The overall fuel‐cell reaction, carried out in acidic medium, is: $$CH_3OH + \tfrac{3}{2}O_2 \rightarrow CO_2 + 2H_2O$$.

In a galvanic cell, the standard cell potential is related to the standard reduction potentials of its two half-cells by the formula $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode(reduction)}}$$ $$-(1)$$.

The problem supplies $$E^\circ_{\text{cell}} = 1.21\;V$$ and the cathodic reduction potential of oxygen, $$E^\circ_{O_2/H_2O} = 1.229\;V$$. Substituting in $$(1)$$, we find

$$E^\circ_{\text{anode(reduction)}} = 1.229\;V - 1.21\;V = 0.019\;V$$.

This value, $$0.019\;V = 19\;mV$$, is the standard reduction potential for the couple $$CO_2 + 6H^+ + 6e^- \rightarrow CH_3OH + H_2O$$, i.e. $$E^\circ_{CO_2/CH_3OH}$$.

Case A: “The standard half-cell reduction potential for the reduction of $$CO_2$$ is 19 mV.” — True, as just calculated.

Case B: “Oxygen is formed at the anode.” — False. Oxygen is consumed at the cathode and reduced to water; it is not evolved at the anode.

Case C: “Reactants are fed at one go to each electrode.” — False. In fuel cells the gaseous (or liquid) reactants are supplied continuously so that the cell can generate a steady current.

Case D: “Reduction of methanol takes place at the cathode.” — False. Methanol undergoes oxidation at the anode; the cathode hosts the reduction of oxygen.

Hence, only Option A is correct.

An oxidising agent gets reduced itself. Hence the easier it is to get reduced, the stronger is its oxidising power.

The ease of reduction is measured by the standard reduction potential $$E^{\ominus}$$.
Larger (more positive) $$E^{\ominus}$$ ⇒ easier reduction ⇒ stronger oxidising capacity.

The given standard reduction potentials are

$$Pb^{4+} + 2e^- \rightarrow Pb^{2+},\; E^{\ominus}=+1.67\text{ V}$$

$$Sn^{4+} + 2e^- \rightarrow Sn^{2+},\; E^{\ominus}=+1.15\text{ V}$$

$$Al^{3+} + 3e^- \rightarrow Al,\; E^{\ominus}=-1.66\text{ V}$$

$$Tl^{3+}/Tl$$ (given), $$E^{\ominus}=+0.26\text{ V}$$ (value supplied in the option)

Comparing the numerical values:

$$+1.67\text{ V} \gt +1.15\text{ V} \gt +0.26\text{ V} \gt -1.66\text{ V}$$

Therefore $$Pb^{4+}$$ (Option A) has the highest positive standard reduction potential and is the most readily reduced species. Hence it possesses the strongest oxidising capacity.

Final answer: Option A

The lead storage battery has two lead-based electrodes dipped in $$\mathrm{H_2SO_4}$$.
• Spongy lead plate  →  traditionally called the anode plate.
• Lead dioxide plate  →  traditionally called the cathode plate.

While discharging, the following galvanic half-reactions take place:

$$\mathrm{At\;anode\;(Pb):}\quad \mathrm{Pb\;(0)} + \mathrm{SO_4^{2-}} \;\rightarrow\; \mathrm{PbSO_4\;(Pb^{2+})} + 2e^-$$
$$\mathrm{At\;cathode\;(PbO_2):}\quad \mathrm{PbO_2\;(Pb^{4+})} + \mathrm{SO_4^{2-}} + 4\mathrm{H^+} + 2e^- \;\rightarrow\; \mathrm{PbSO_4\;(Pb^{2+})} + 2\mathrm{H_2O}$$

When the battery is CHARGED, an external emf forces the reactions to run in the reverse direction (electrolysis).
Therefore:

Reverse of the discharge reaction:

$$\mathrm{PbSO_4\;(Pb^{2+})} + 2e^- \;\rightarrow\; \mathrm{Pb\;(0)} + \mathrm{SO_4^{2-}}$$

Thus, at the anode plate the oxidation state of lead changes
from $$x_1 = +2$$ (in $$\mathrm{PbSO_4}$$) to $$y_1 = 0$$ (in metallic Pb).

Reverse of the discharge reaction:

$$\mathrm{PbSO_4\;(Pb^{2+})} + 2\mathrm{H_2O} \;\rightarrow\; \mathrm{PbO_2\;(Pb^{4+})} + \mathrm{SO_4^{2-}} + 4\mathrm{H^+} + 2e^-$$

Hence, at the cathode plate the oxidation state of lead changes
from $$x_2 = +2$$ (in $$\mathrm{PbSO_4}$$) to $$y_2 = +4$$ (in $$\mathrm{PbO_2}$$).

Collecting the four values:
$$x_1 = +2,\; y_1 = 0,\; x_2 = +2,\; y_2 = +4$$

These correspond to Option B.

Final answer: $$+2,\;0,\;+2,\;+4$$ (Option B).

We need to find the mass of aluminium deposited during electrolysis.

According to Faraday’s law of electrolysis, the mass deposited is given by the relation $$m = \frac{M \times I \times t}{n \times F}$$ where M = molar mass, I = current, t = time, n = number of electrons transferred, and F = Faraday constant.

For the reduction of aluminium ions $$Al^{3+} + 3e^- \rightarrow Al$$ the number of electrons transferred is n = 3. The molar mass M of aluminium is 27 g/mol, the current I is 2 A, the time t is 30 min = 30 × 60 = 1800 s, and the Faraday constant F is 96500 C/mol.

Substituting these values into the formula gives

$$m = \frac{27 \times 2 \times 1800}{3 \times 96500}$$

$$= \frac{97200}{289500}$$

$$= 0.3355 \text{ g}$$

$$\approx 0.336 \text{ g}$$

The correct answer is Option 2: 0.336 g.

For the cell reaction: $$Fe^{2+}(aq) + Ag^+(aq) \to Fe^{3+}(aq) + Ag(s)$$

Given half-cell potentials:
$$Ag^+ + e^- \to Ag, \quad E° = x$$ V
$$Fe^{2+} + 2e^- \to Fe, \quad E° = y$$ V
$$Fe^{3+} + 3e^- \to Fe, \quad E° = z$$ V

We need the standard potential for $$Fe^{3+} + e^- \to Fe^{2+}$$.

Using the relation between Gibbs energies: $$\Delta G° = -nFE°$$.

For $$Fe^{3+} + 3e^- \to Fe$$: $$\Delta G°_1 = -3Fz$$

For $$Fe^{2+} + 2e^- \to Fe$$: $$\Delta G°_2 = -2Fy$$

The half-cell $$Fe^{3+} + e^- \to Fe^{2+}$$ can be obtained by subtracting:
$$\Delta G°_3 = \Delta G°_1 - \Delta G°_2 = -3Fz - (-2Fy) = -3Fz + 2Fy$$

Since $$\Delta G°_3 = -1 \cdot F \cdot E°(Fe^{3+}/Fe^{2+})$$:

$$E°(Fe^{3+}/Fe^{2+}) = 3z - 2y$$

For the given cell reaction, the cathode is $$Ag^+/Ag$$ and the anode is $$Fe^{2+}/Fe^{3+}$$:

$$E°_{cell} = E°_{cathode} - E°_{anode} = x - (3z - 2y) = x + 2y - 3z$$

The correct answer is Option C: $$x + 2y - 3z$$.

Find the correct Nernst equation for the cell $$Mg|Mg^{2+}(aq)||Ag^+(aq)|Ag$$. An anode (oxidation) half-reaction is $$Mg \rightarrow Mg^{2+} + 2e^-$$ and the cathode (reduction) half-reaction is $$2Ag^+ + 2e^- \rightarrow 2Ag$$. Combining these gives the net cell reaction:

$$ Mg + 2Ag^+ \rightarrow Mg^{2+} + 2Ag $$. Here $$n = 2$$ electrons are transferred in the balanced reaction. The reaction quotient $$Q$$ is given by

$$ Q = \frac{[Mg^{2+}]}{[Ag^+]^2} $$. (Pure solids Mg and Ag have activity = 1.)

Hence the Nernst equation is

$$ E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF}\ln Q = E°_{\text{cell}} - \frac{RT}{2F}\ln\frac{[Mg^{2+}]}{[Ag^+]^2} $$. This can be rewritten as

$$ E_{\text{cell}} = E°_{\text{cell}} + \frac{RT}{2F}\ln\frac{[Ag^+]^2}{[Mg^{2+}]} $$ (using $$-\ln(A/B) = \ln(B/A)$$).

The correct answer is Option B: $$E_{\text{cell}} = E°_{\text{cell}} + \frac{RT}{2F}\ln\frac{[Ag^+]^2}{[Mg^{2+}]}$$.

The limiting molar conductivity $$\Lambda_m^{\circ}$$ of an ion in water at $$298 \text{ K}$$ is directly proportional to its ionic mobility $$u^{\circ}$$:
$$\Lambda_m^{\circ} = z \, F \, u^{\circ}$$, where $$z$$ is the ionic charge and $$F$$ is Faraday’s constant.

Ionic mobility depends mainly on (i) ionic size (including hydration), (ii) charge density and (iii) special transport mechanisms, if any.

Case 1: $$H^{+}$$
$$H^{+}$$ is transported via the Grotthuss or proton-hopping mechanism. This “relay race” of protons through the hydrogen-bond network gives it an extremely high mobility, so $$\Lambda_m^{\circ}(H^{+})$$ is the largest for any ion.

Case 2: Monovalent alkali metal ions $$K^{+}$$ and $$Na^{+}$$
Both ions move through ordinary diffusion. Hydration is stronger for the smaller $$Na^{+}$$, which makes its effective (hydrated) radius larger than that of $$K^{+}$$. Therefore
$$u^{\circ}(K^{+}) \gt u^{\circ}(Na^{+})$$ and hence $$\Lambda_m^{\circ}(K^{+}) \gt \Lambda_m^{\circ}(Na^{+})$$.

Case 3: Divalent alkaline-earth ions $$Ca^{2+}$$ and $$Mg^{2+}$$
Both carry double charge, but the smaller $$Mg^{2+}$$ has a much higher charge density, so it is more strongly hydrated and moves more sluggishly than $$Ca^{2+}$$. Hence
$$u^{\circ}(Ca^{2+}) \gt u^{\circ}(Mg^{2+})$$ and $$\Lambda_m^{\circ}(Ca^{2+}) \gt \Lambda_m^{\circ}(Mg^{2+})$$.

Numerical data at $$298 \text{ K}$$ (S cm2 mol−1):
$$\Lambda_m^{\circ}(H^{+}) \approx 349.6$$
$$\Lambda_m^{\circ}(Ca^{2+}) \approx 119.0$$
$$\Lambda_m^{\circ}(Mg^{2+}) \approx 106.1$$
$$\Lambda_m^{\circ}(K^{+}) \approx 73.5$$
$$\Lambda_m^{\circ}(Na^{+}) \approx 50.1$$

Correct descending order:
$$H^{+} \gt Ca^{2+} \gt Mg^{2+} \gt K^{+} \gt Na^{+}$$

This matches Option B.

For electrolysis using inert electrodes we compare the standard reduction potentials $$E^{\circ}$$ of all possible cathodic reactions. The species having the highest (most positive) $$E^{\circ}$$ gets reduced (deposited) first. All solutions are 1 M, so the numerical values of $$E^{\circ}$$ may be used directly without any Nernst correction.

Standard reduction potentials involved:

$$Ag^{+}+e^{-}\rightarrow Ag;\;E^{\circ}=+0.80\;{\rm V}$$
$$Hg_2^{2+}+2e^{-}\rightarrow 2Hg;\;E^{\circ}=+0.79\;{\rm V}$$
$$Cu^{2+}+2e^{-}\rightarrow Cu;\;E^{\circ}=+0.34\;{\rm V}$$

Because $$E^{\circ}_{Ag^{+}/Ag}\gt E^{\circ}_{Hg_2^{2+}/Hg}\gt E^{\circ}_{Cu^{2+}/Cu}$$, the sequence of metal deposition on the cathode as the external voltage is increased will be Ag (first) → Hg (second) → Cu (third).

Therefore Statement I is correct.

For the magnesium salt two possible cathodic reductions are important:

$$Mg^{2+}+2e^{-}\rightarrow Mg;\;E^{\circ}=-2.37\;{\rm V}$$
$$2H_2O+2e^{-}\rightarrow H_2+2OH^{-};\;E^{\circ}=-0.83\;{\rm V}$$

The reduction of water (giving hydrogen gas) requires a far smaller over-potential than the reduction of $$Mg^{2+}$$. Hence, at the cathode hydrogen gas will evolve, and metallic magnesium will not be deposited.

Oxygen gas is produced at the anode (oxidation of water), not at the cathode. Thus Statement II is incorrect.

Conclusion: Statement I is correct, Statement II is incorrect → Option B.

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,

$$\boxed{pK_b=6}$$

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).

• Anode (Oxidation): 2Tl(s) -> 2Tl+(aq) + 2e
• Cathode (Reduction): Cu+2(aq) + 2e -> Cu(s)
• Overall Reaction: 2Tl(s) + Cu+2 ->2Tl+(aq) + Cu(s)

Nersnt Equation :

E_cell = E0_cell - (0.0591 / 2) * log([Tl+]^2 / [Cu2+])

The EMF of this specific cell can be increased by:

1. Increasing the molarity of Cu+2 (e.g., from 0.01M to 0.1M).
2. Decreasing the molarity of Tl+ ions (e.g., from 0.001M to 0.0001M).

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: $$10^{-14}$$.

A weak electrolyte dissociates only to a small extent in solution. The fraction of the electrolyte that actually dissociates is called the degree of dissociation, denoted by $$\alpha$$.

For any weak monobasic acid or monoacidic base the Ostwald-dilution law gives $$K_a = \frac{\alpha^{2}C}{1-\alpha}$$  $$-(1)$$ where $$K_a$$ = acid (or base) dissociation constant, $$C$$ = initial molar concentration of the electrolyte.

The molar conductivity at the given concentration, $$\Lambda_m$$, is related to the molar conductivity at infinite dilution, $$\Lambda_m^{\circ}$$, through the degree of dissociation: $$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$$  $$-(2)$$

Insert $$\alpha = \dfrac{\Lambda_m}{\Lambda_m^{\circ}}$$ from $$(2)$$ into $$(1)$$:

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

Simplify the denominator and then multiply numerator & denominator by $$\Lambda_m^{\circ}$$ to clear the fraction:

$$K_a = \frac{C\Lambda_m^{2}}{\Lambda_m^{\circ 2} - \Lambda_m\Lambda_m^{\circ}}$$

Cross-multiply to obtain an equation that contains only whole-number powers:

$$K_a\bigl(\Lambda_m^{\circ 2} - \Lambda_m\Lambda_m^{\circ}\bigr) = C\Lambda_m^{2}$$

Rearrange every term to the left-hand side:

$$\Lambda_m^{2}C - K_a\Lambda_m^{\circ 2} + K_a\Lambda_m\Lambda_m^{\circ} = 0$$

This is the required relation between molar conductivity and concentration for a weak electrolyte.

Comparison with the given options shows that it matches Option B.

Hence, the correct choice is Option B.

A. Leclanche Cell (Dry Cell)

The Leclanche cell is a primary cell (non-rechargeable). It consists of a zinc container that acts as the anode. The chemical reaction at the anode involves the oxidation of zinc metal into zinc ions.

• Anode Reaction: Zn(s) to Zn^{2+}(aq) + 2e^-

B. Ni - Cd Cell (Nickel-Cadmium

The Ni-Cd cell is a secondary cell, meaning the electrochemical reactions are reversible. When an external electrical current is applied in the opposite direction, the cell "recharges." It has a longer life than lead-acid batteries but is more expensive to manufacture.

C. Fuel Cell 

Unlike conventional batteries that store chemical energy, fuel cells are designed to convert the energy of combustion of fuels (like hydrogen, methane, or methanol) directly into electrical energy.

• Overall Reaction: 2H2(g) + O2(g) to 2H2O(l)

D. Mercury Cell 

The Mercury cell is used in low-current devices like hearing aids and watches. A unique property of this cell is that the overall reaction does not involve any ions in the solution whose concentration can change during life. This allows the cell to provide a constant voltage (1.35 V) throughout its entire lifespan.

• Anode: Zn(Hg) (Zinc amalgam)
• Cathode: HgO (Mercuric oxide)

A calomel electrode consists of mercury (metal) in contact with mercurous chloride (insoluble salt) and a chloride solution (anion). Its representation is: Hg | Hg₂Cl₂ | KCl (solution).

This falls under the category of Metal-Insoluble Salt-Anion electrodes.

The correct answer is Option 4.

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.

Anode reaction: 1/2 H2(g) -> H+(aq) + e-

Cathode reaction: AgCl(s) + e- -> Ag(s) + Cl-(aq)

Cell Notation: Pt | H2(g) | KCl(aq) | AgCl(s) | Ag

This cell is often used to measure the pH of a solution or to determine the standard reduction potential of the silver-silver chloride electrode, which is a very stable and common reference electrode in chemistry

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 $$\boxed{46}$$.

Potassium ferricyanide is $$K_3[Fe(CN)_6]$$, in which iron is in the $$+3$$ oxidation state. When iodide ion $$I^-$$ is added, a redox equilibrium is set up in which iron(III) is reduced to iron(II) and iodide is oxidised to iodine. The iron(II) product is potassium ferrocyanide $$K_4[Fe(CN)_6]$$, designated as the complex P.

Step 1 - Write the half-reactions
Reduction: $$[Fe(CN)_6]^{3-} + e^- \rightarrow [Fe(CN)_6]^{4-}$$
Oxidation: $$I^- \rightarrow \dfrac{1}{2}I_2 + e^-$$

Step 2 - Equalise electrons and add the half-reactions
Each half-reaction involves one electron, so they can be added directly:

$$[Fe(CN)_6]^{3-} + I^- \rightarrow [Fe(CN)_6]^{4-} + \dfrac{1}{2}I_2$$

Step 3 - Remove the fractional coefficient
Multiply the entire equation by 2:

$$2\,[Fe(CN)_6]^{3-} + 2\,I^- \rightarrow 2\,[Fe(CN)_6]^{4-} + I_2$$

Replacing the complex ions with their potassium salts gives:

$$2\,K_3[Fe(CN)_6] + 2\,KI \rightarrow 2\,K_4[Fe(CN)_6]\;(\text{= P}) + I_2$$

Step 4 - Determine the stoichiometry for potassium iodide
From the balanced equation, 1 mol of $$KI$$ is required to form 1 mol of $$K_4[Fe(CN)_6]$$. Therefore, to obtain 2 mol of $$K_4[Fe(CN)_6]$$ (complex P) we need:

$$2\;\text{mol P} \times \dfrac{1\;\text{mol }KI}{1\;\text{mol P}} = 2\;\text{mol }KI$$

Hence, the number of moles of potassium iodide required is 2.

Final answer: 2

In every nuclear decay, both the mass number $$A$$ and the atomic number $$Z$$ must balance on the two sides of the reaction.
  • $$\alpha$$-particle : $$\ ^4_2He \; ( \Delta A = -4,\; \Delta Z = -2)$$
  • $$\beta^-$$-particle : $$\ ^0_{-1}e \; ( \Delta A = 0,\; \Delta Z = +1)$$
  • $$\beta^+$$-particle : $$\ ^0_{+1}e \; ( \Delta A = 0,\; \Delta Z = -1)$$

$$^{238}_{92}U \;\longrightarrow\; {}^{234}_{91}Pa$$
Required changes: $$\Delta A = 238-234 = 4,\; \Delta Z = 92-91 = 1$$ (mass ↓4, atomic ↓1).
One $$\alpha$$ gives $$\Delta A = 4, \Delta Z = 2$$ too much fall in $$Z$$. Adding one $$\beta^-$$ raises $$Z$$ by 1, giving net $$\Delta A = 4, \Delta Z = 1$$ as needed.
Emission: one $$\alpha$$ + one $$\beta^-$$ ⇒ List-II (4).

$$^{214}_{82}Pb \;\longrightarrow\; {}^{210}_{82}Pb$$
Required changes: $$\Delta A = 4,\; \Delta Z = 0$$.
One $$\alpha$$ accounts for $$\Delta A = 4, \Delta Z = 2$$. To restore $$Z$$, two $$\beta^-$$ (each +1) add $$\Delta Z = -2+2 = 0$$.
Emission: one $$\alpha$$ + two $$\beta^-$$ ⇒ List-II (3).

$$^{210}_{81}Tl \;\longrightarrow\; {}^{206}_{82}Pb$$
Required changes: $$\Delta A = 4,\; \Delta Z = -1$$ (mass ↓4, atomic ↑1).
One $$\alpha$$ gives $$\Delta A = 4,\; \Delta Z = 2$$ too low atomic number. Three $$\beta^-$$ (total +3) correct it: $$-2 + 3 = +1$$.
Emission: one $$\alpha$$ + three $$\beta^-$$ ⇒ List-II (2).

$$^{228}_{91}Pa \;\longrightarrow\; {}^{224}_{88}Ra$$
Required changes: $$\Delta A = 4,\; \Delta Z = 3$$ (mass ↓4, atomic ↓3).
One $$\alpha$$ produces $$\Delta Z = 2$$; an extra drop of 1 is still needed. One $$\beta^+$$ (−1) gives the balance: $$-2 -1 = -3$$.
Emission: one $$\alpha$$ + one $$\beta^+$$ ⇒ List-II (1).

Thus the matches are:
P → 4,  Q → 3,  R → 2,  S → 1.

Option A which is: P → 4, Q → 3, R → 2, S → 1

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 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.

In acidic medium, $$\mathrm{KMnO_4}$$ is reduced to $$\mathrm{Mn^{2+}}$$, while $$\mathrm{H_2S}$$ is oxidised to elemental sulphur. Write the two half-reactions and balance them for mass and charge.

Reduction half-reaction
$$\mathrm{MnO_4^- + 8\,H^+ + 5\,e^- \;\rightarrow\; Mn^{2+} + 4\,H_2O}$$ $$-(1)$$

Oxidation half-reaction
Sulphur changes from $$-2$$ in $$\mathrm{H_2S}$$ to $$0$$ in $$\mathrm{S}$$, so the increase is $$2$$ electrons per molecule.
$$\mathrm{H_2S \;\rightarrow\; S + 2\,H^+ + 2\,e^-}$$ $$-(2)$$

To make the electrons equal, take LCM of $$5$$ and $$2,$$ i.e. $$10$$ electrons.
Multiply $$(1)$$ by $$2$$ and $$(2)$$ by $$5$$:

$$\mathrm{2\,MnO_4^- + 16\,H^+ + 10\,e^- \;\rightarrow\; 2\,Mn^{2+} + 8\,H_2O}$$ $$-(3)$$
$$\mathrm{5\,H_2S \;\rightarrow\; 5\,S + 10\,H^+ + 10\,e^-}$$ $$-(4)$$

Add $$(3)$$ and $$(4)$$; the $$10\,e^-$$ cancel:

$$\mathrm{5\,H_2S + 2\,MnO_4^- + 16\,H^+ \;\rightarrow\; 5\,S + 10\,H^+ + 2\,Mn^{2+} + 8\,H_2O}$$

Subtract $$10\,H^+$$ from both sides to obtain the net balanced redox equation in acidic medium:

$$\mathrm{5\,H_2S + 2\,MnO_4^- + 6\,H^+ \;\rightarrow\; 5\,S + 2\,Mn^{2+} + 8\,H_2O}$$ $$-(5)$$

The problem states that exactly $$5$$ moles of $$\mathrm{H_2S}$$ react. Comparing with $$(5)$$:

Number of moles of $$\mathrm{H_2O}$$ produced, $$x = 8$$.

Total electrons transferred per $$(5)$$ are $$10$$; hence for the given scale the number of moles of electrons involved, $$y = 10$$.

Therefore $$x + y = 8 + 10 = 18$$.

Final Answer: 18

The problem gives four phosphorus-based reactions (List-I) and five possible products (List-II). We have to decide which product is formed in each reaction and then select the option that contains the correct matching.

Case P: $$P_2O_3 + 3H_2O \longrightarrow$$ Phosphorus(III) oxide is the anhydride of phosphorous acid. When it is hydrolysed, the only product is phosphorous acid. $$P_2O_3 + 3H_2O \rightarrow 2\,H_3PO_3$$ Hence, for reaction P the required product is $$H_3PO_3$$ (List-II, item 2).

Case Q: $$P_4 + 3NaOH + 3H_2O \longrightarrow$$ White phosphorus in hot alkali undergoes disproportionation. One part is reduced to phosphine, the other is oxidised to hypophosphite. $$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3\,NaH_2PO_2$$ Thus one of the products is phosphine $$PH_3$$ (List-II, item 3).

Case R: $$PCl_5 + CH_3COOH \longrightarrow$$ $$PCl_5$$ is a strong chlorinating agent. With a carboxylic acid it replaces the -OH by -Cl forming an acyl chloride and simultaneously gets converted to phosphoryl chloride. $$PCl_5 + CH_3COOH \rightarrow CH_3COCl + POCl_3 + HCl$$ Therefore the phosphorus-containing product is $$POCl_3$$ (List-II, item 4).

Case S: $$H_3PO_2 + 2H_2O + 4AgNO_3 \longrightarrow$$ Hypophosphorous acid is a good reducing agent. In the presence of $$Ag^+$$ it reduces silver(I) to metallic silver and itself is oxidised to phosphoric acid. $$H_3PO_2 + 2H_2O + 4AgNO_3 \rightarrow H_3PO_4 + 4Ag + 4HNO_3$$ So the phosphorus-containing product is $$H_3PO_4$$ (List-II, item 5).

Collecting the matches:
P → 2, Q → 3, R → 4, S → 5.

The only option containing this set is Option D.

Final Answer: Option D which is: P → 2; Q → 3; R → 4; S → 5

The correct statements are C and E.

Final Answer: 2

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 $$Cr_2O_7^{2-} \to Cr^{3+}$$. 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 $$3$$.

Statement A is incorrect. Molecules present in the bulk of a liquid are surrounded by neighbouring molecules from all directions and experience equal intermolecular attractive forces, resulting in zero net force. Surface tension does not arise due to these balanced forces.

Statement B is correct. Molecules present at the surface experience unbalanced intermolecular forces because there are no molecules above them. This net inward force causes the liquid surface to contract and gives rise to surface tension.

Statement C is incorrect. Molecules in a liquid are in continuous random motion and can move freely between the bulk and the surface. Therefore, a molecule in the bulk can reach the liquid surface.

Statement D is correct. Vapour pressure is produced when surface molecules escape into the vapour phase in a closed system until dynamic equilibrium is established. Hence, the molecules present on the surface are responsible for vapour pressure.

Therefore, Statements B and D are correct, while Statements A and C are incorrect. Hence, the number of correct statements is $$\boxed{2}$$.

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 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 $$\times 10^{-7}$$.

The enthalpy difference between the two phases at any temperature $$T$$ can be obtained by integrating the heat-capacity difference from 0 K to $$T$$.

For the given data, the heat-capacity difference is constant:
$$C_{P,\beta}-C_{P,\alpha}=1\ \text{J mol}^{-1}\text{K}^{-1}$$

Starting from 0 K (where both phases may be taken to have the same enthalpy), the enthalpy difference at temperature $$T$$ is

$$\begin{aligned} H_\beta-H_\alpha &= \int_{0}^{T}\bigl(C_{P,\beta}-C_{P,\alpha}\bigr)\,dT \\ &= (C_{P,\beta}-C_{P,\alpha})\int_{0}^{T} dT \\ &= (C_{P,\beta}-C_{P,\alpha})\,T \end{aligned}$$

Substituting the numerical values for the required temperature $$T=300\ \text{K}$$:

$$H_\beta-H_\alpha = 1\ \text{J mol}^{-1}\text{K}^{-1}\times 300\ \text{K} = 300\ \text{J mol}^{-1}$$

Hence, the enthalpy change at 300 K is 300 J mol$$^{-1}$$.

The combustion taking place inside the bomb calorimeter is
$$Hg(g)+\frac12\,O_2(g)\rightarrow HgO(s)$$

Step 1 : Heat evolved at constant volume (ΔU)
The bomb calorimeter provides the change in internal energy because its volume is fixed.

Temperature rise $$\Delta T = 312.8\,\text{K} - 298.0\,\text{K} = 14.8\,\text{K}$$

Heat capacity of the bomb $$C_{\text{cal}} = 20.00\;\text{kJ K}^{-1}$$

Heat released for 2 mol Hg(g):
$$q_v = -\,C_{\text{cal}}\;\Delta T = -\,(20.00)\,(14.8) = -296\;\text{kJ}$$

Hence for 1 mol Hg(g):
$$\Delta U^{\circ}_{\text{rxn}} = \frac{-296\;\text{kJ}}{2} = -148\;\text{kJ mol}^{-1}$$

Step 2 : Convert ΔU to ΔH for the gaseous-Hg reaction
For any reaction, $$\Delta H = \Delta U + \Delta n_{\text{gas}}\,R\,T$$, where $$\Delta n_{\text{gas}}$$ is the change in moles of gaseous species.

In the reaction, gaseous moles change as
$$\Delta n_{\text{gas}} = n_{\text{products,\,gas}} - n_{\text{reactants,\,gas}} = 0 - \left(1 + \frac12\right) = -1.5$$

Therefore
$$\Delta H^{\circ}_{1} = -148\;\text{kJ mol}^{-1} + (-1.5)(8.3\times10^{-3}\;\text{kJ mol}^{-1}\text{K}^{-1})(298\;\text{K})$$

$$\Delta H^{\circ}_{1} = -148\;\text{kJ mol}^{-1} - 3.71\;\text{kJ mol}^{-1} = -151.71\;\text{kJ mol}^{-1}$$

This is the enthalpy change for
$$Hg(g)+\frac12\,O_2(g)\rightarrow HgO(s)$$

Step 3 : Bring the reactants to their standard states
In the standard enthalpy of formation, mercury must start as liquid Hg(l).
The given data supply the enthalpy of formation of gaseous Hg:

$$Hg(l)\rightarrow Hg(g)\;;\qquad \Delta H^{\circ} = +61.32\;\text{kJ mol}^{-1}$$

Adding this step to the gaseous-Hg reaction yields the desired formation reaction:
$$Hg(l)+\frac12\,O_2(g)\rightarrow HgO(s)$$

Hence
$$\Delta H_f^{\circ}\bigl(HgO,s\bigr)= +61.32\;\text{kJ mol}^{-1} + (-151.71\;\text{kJ mol}^{-1}) = -90.39\;\text{kJ mol}^{-1}$$

Answer
The standard molar enthalpy of formation of $$HgO(s)$$ at 298 K is
$$\boxed{\Delta H_f^{\circ}(HgO,s)= -90.39\;\text{kJ mol}^{-1}}$$
Thus, $$|X| = 90.39$$.

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 $$0.77\ \text{V}$$.

Initially, each solute furnishes $$0.01\text{ mol}$$ in a total volume of $$100\text{ mL}=0.1\text{ L}$$, so their starting concentrations are $$0.1\text{ M}$$ each:

$$H_2CO_3\;(0.1\text{ M}),\;HCO_3^- \;(0.1\text{ M}),\;CO_3^{2-}\;(0.1\text{ M}),\;OH^-\;(0.1\text{ M})$$

Step 1: Immediate neutralisation by the strong base
The strong base $$OH^-$$ will first de-protonate the strongest acid present, $$H_2CO_3$$:

$$H_2CO_3 + OH^- \;\longrightarrow\; HCO_3^- + H_2O$$

Mole balance (in 0.1 L): both $$H_2CO_3$$ and $$OH^-$$ start with $$0.01\text{ mol}$$, so they completely consume each other.

After this reaction:

$$H_2CO_3 : 0$$    $$OH^- : 0$$    additional $$HCO_3^- : +0.01\text{ mol}$$

Step 2: Concentrations after neutralisation
Total moles now present:

$$HCO_3^- : 0.01 + 0.01 = 0.02\text{ mol}$$
$$CO_3^{2-} : 0.01\text{ mol}$$

Convert to concentrations (divide by $$0.1\text{ L}$$):

$$[HCO_3^-] = \frac{0.02}{0.1} = 0.20\text{ M}$$
$$[CO_3^{2-}] = \frac{0.01}{0.1} = 0.10\text{ M}$$

Step 3: Identify the buffer pair
The remaining species constitute the conjugate acid-base pair of the second dissociation of carbonic acid:

$$HCO_3^- \;\rightleftharpoons\; H^+ + CO_3^{2-} \quad\quad (pK_{a2} = 10.32)$$

Step 4: Apply the Henderson-Hasselbalch equation

$$\text{pH} = pK_{a2} + \log\!\left(\frac{[CO_3^{2-}]}{[HCO_3^-]}\right)$$

Substitute the concentrations:

$$\text{pH} = 10.32 + \log\!\left(\frac{0.10}{0.20}\right) = 10.32 + \log(0.50)$$

Using $$\log 2 = 0.30 \;,$$ so $$\log(0.50) = -\log 2 = -0.30$$, we get

$$\text{pH} = 10.32 - 0.30 = 10.02$$

Hence, the pH of the resulting solution is 10.02.

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.

The standard reduction potential, $$E^\circ$$, indicates the tendency of a species to gain electrons and undergo reduction. A more positive value of $$E^\circ$$ corresponds to a greater tendency for reduction.

The given couples are

• A: $$Cl_2/Cl^-$$
• B: $$I_2/I^-$$
• C: $$Ag^+/Ag$$
• D: $$Na^+/Na$$
• E: $$Li^+/Li$$

The standard reduction potentials of these couples are approximately

$$E^\circ(Cl_2/Cl^-)=+1.36\text{ V}$$

$$E^\circ(Ag^+/Ag)=+0.80\text{ V}$$

$$E^\circ(I_2/I^-)=+0.54\text{ V}$$

$$E^\circ(Na^+/Na)=-2.71\text{ V}$$

$$E^\circ(Li^+/Li)=-3.04\text{ V}$$

Among the halogens, chlorine has a higher reduction potential than iodine because chlorine is a stronger oxidizing agent.

$$Cl_2/Cl^- > I_2/I^-$$

Silver is a noble metal and has a positive reduction potential, which lies between those of chlorine and iodine.

$$Cl_2/Cl^- > Ag^+/Ag > I_2/I^-$$

Alkali metals possess highly negative reduction potentials because they readily lose electrons. Lithium has the most negative reduction potential due to the exceptionally high hydration enthalpy of $$Li^+$$.

$$Na^+/Na > Li^+/Li$$

Combining all the values,

$$Cl_2/Cl^- > Ag^+/Ag > I_2/I^- > Na^+/Na > Li^+/Li$$

Hence, the correct order is

$$\boxed{A>C>B>D>E}$$

Therefore, the correct answer is Option D.

Molar mass of $$Cu(NO_3)_2 = 63 + 2(14) + 6(16) = 187 \text{ g mol}^{-1}$$.

Moles of $$Cu(NO_3)_2$$ taken $$n_{Cu^{2+}} = \frac{3.74 \text{ g}}{187 \text{ g mol}^{-1}} = 0.020 \text{ mol}$$.

With excess $$KI$$ the reaction is $$2\,Cu^{2+} + 4\,I^- \;\longrightarrow\; 2\,CuI(s) + I_2(aq)$$.

From the stoichiometry, $$2$$ mol of $$Cu^{2+}$$ give $$1$$ mol of $$I_2$$. Hence moles of $$I_2$$ formed $$n_{I_2} = \frac{0.020}{2} = 0.010 \text{ mol}$$.

The brown colour of the filtrate is due to this $$I_2$$. Passing $$H_2S$$ through it: $$I_2 + H_2S \;\longrightarrow\; 2\,HI + S(s)$$.

Here, $$1$$ mol of $$I_2$$ produces $$1$$ mol of elemental sulfur $$S$$. Therefore, moles of precipitated sulfur $$n_S = 0.010 \text{ mol}$$.

Mass of sulfur obtained $$m_S = n_S \times M_S = 0.010 \times 32 = 0.32 \text{ g}$$.

Hence, the amount of precipitate X is 0.32 g.

White phosphorous is $$P_4$$.  In a boiling, strongly alkaline medium it undergoes disproportionation to give phosphine (the gaseous product Q) and sodium hypophosphite:

$$P_4 + 3\,NaOH + 3\,H_2O \;\longrightarrow\; 3\,NaH_2PO_2 + PH_3 \uparrow$$

Step 1 : Moles of $$P_4$$ taken
Molar mass of $$P_4 = 4 \times 31 = 124\,$$g mol$$^{-1}$$.
$$n(P_4)=\frac{1.24\;\text{g}}{124\;\text{g mol}^{-1}} = 0.01\;\text{mol}$$

Step 2 : Moles of phosphine ($$PH_3$$) formed
From the above equation, 1 mol $$P_4$$ gives 1 mol $$PH_3$$. Therefore
$$n(PH_3)=0.01\;\text{mol}$$

Step 3 : Reaction of $$PH_3$$ with $$CuSO_4$$
In an aqueous $$Cu^{2+}$$ solution, phosphine reduces the cupric ion, giving a black precipitate of copper phosphide $$Cu_3P_2$$. The stoichiometric equation is

$$2\,PH_3 + 3\,CuSO_4 + 3\,H_2O \;\longrightarrow\; Cu_3P_2\!\downarrow + 3\,H_2SO_4 + 3\,H_2$$

Thus, 2 mol $$PH_3$$ consume 3 mol $$CuSO_4$$, i.e. $$\dfrac{3}{2}=1.5$$ mol $$CuSO_4$$ per mol $$PH_3$$.

Step 4 : Moles of $$CuSO_4$$ required
$$n(CuSO_4)=1.5 \times n(PH_3)=1.5 \times 0.01=0.015\;\text{mol}$$

Step 5 : Mass of $$CuSO_4$$
Molar mass of $$CuSO_4=63+32+4\times16=159\,$$g mol$$^{-1}$$.
$$m(CuSO_4)=0.015 \times 159 = 2.385\;\text{g}$$

On rounding to two decimal places, the mass lies between 2.38 g and 2.39 g.

Hence, the amount of $$CuSO_4$$ required is 2.38 - 2.39 g.

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 industrial extraction of aluminium from bauxite involves three main stages.

1. Purification of bauxite by the Bayer process:
  a) Digestion - powdered bauxite is treated with hot concentrated caustic soda. $$Al_2O_3 + 2\,NaOH + 3\,H_2O \rightarrow 2\,Na[Al(OH)_4]$$.   This directly supports Option C because $$Al_2O_3$$ really is dissolved in hot aqueous $$NaOH$$.

  b) Precipitation - the clear sodium aluminate liquor obtained above is cooled and bubbled with $$CO_2$$.     $$2\,Na[Al(OH)_4] + CO_2 \rightarrow Al_2O_3\!\cdot\!3H_2O \downarrow + 2\,NaHCO_3$$.   This hydrated alumina is filtered, washed and ignited to get pure $$Al_2O_3$$.   Thus Option B is also correct.

2. Electrolytic reduction by the Hall-Héroult process:
  Pure $$Al_2O_3$$ has a very high melting point, so it is dissolved in molten cryolite $$Na_3AlF_6$$ (sometimes with $$CaF_2$$) to lower the melting point and increase conductivity.
  Electrolysis is carried out in a carbon-lined steel cell. Molten aluminium is deposited at the cathode, while the carbon anode is oxidised, producing $$CO_2$$:   Cathode: $$Al^{3+} + 3e^- \rightarrow Al(l)$$   Anode: $$C(s) + 2\,O^{2-} \rightarrow CO_2(g) + 4e^-$$.   Hence Option D is correct.

3. Assessment of the remaining option:
  Option A suggests directly reducing $$Al_2O_3$$ with coke above 2500 °C. Such a reaction is thermodynamically difficult because aluminium forms a highly stable oxide; coke cannot reduce it even at those temperatures. Therefore Option A is incorrect.

Therefore, the processes actually used in aluminium extraction correspond to:
Option B (neutralisation with $$CO_2$$), Option C (digestion in hot $$NaOH$$), and Option D (electrolysis of $$Al_2O_3$$ + $$Na_3AlF_6$$).

Final answer: Option B, Option C, Option D.

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 have 1 $$\mu$$g of radioactive element X with a half-life of 30 years, and we need to find the amount remaining after 100 years, expressed as $$\_\_\_ \times 10^{-1}$$ $$\mu$$g.

We use the radioactive decay formula:

$$N = N_0 \times e^{-\lambda t}$$

where $$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{30}$$ per year. After $$t = 100$$ years:

$$N = 1 \times e^{-\frac{0.693}{30} \times 100} = e^{-\frac{69.3}{30}} = e^{-2.31}$$

Now we evaluate $$e^{-2.31}$$. We know that $$\ln 10 = 2.303$$, so $$e^{-2.303} = 10^{-1} = 0.1$$. Since $$2.31$$ is very close to $$2.303$$, we can use the given values more precisely.

Using the alternative form: $$N = N_0 \times \left(\frac{1}{2}\right)^{t/t_{1/2}} = 1 \times \left(\frac{1}{2}\right)^{100/30} = \left(\frac{1}{2}\right)^{10/3}$$

Taking logarithm: $$\log N = -\frac{10}{3} \times \log 2 = -\frac{10}{3} \times 0.30 = -1.0$$

Therefore $$N = 10^{-1.0} = 0.1$$ $$\mu$$g.

Expressing this as $$\_\_\_ \times 10^{-1}$$ $$\mu$$g, we get $$0.1 = 1 \times 10^{-1}$$ $$\mu$$g.

Hence, the correct answer is 1.

We are given that the equilibrium constant for the reaction:

$$Cu(s) + 2Ag^+(aq) \rightleftharpoons Cu^{2+}(aq) + 2Ag(s)$$

is $$K_1 = 2 \times 10^{15}$$ at 298 K.

We need to find the equilibrium constant for:

$$\frac{1}{2}Cu^{2+}(aq) + Ag(s) \rightleftharpoons \frac{1}{2}Cu(s) + Ag^+(aq)$$

The second reaction is the reverse of the first reaction, divided by 2.

When a reaction is reversed, the new equilibrium constant is the reciprocal: $$K_{reverse} = \frac{1}{K_1}$$. When a reaction is divided by 2, the new equilibrium constant is the square root: $$K_2 = \sqrt{K_{reverse}}$$.

Therefore:

$$K_2 = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} = \frac{1}{\sqrt{2 \times 10^{15}}}$$

Calculating the value gives:

$$K_2 = \frac{1}{\sqrt{2 \times 10^{15}}} = \frac{1}{\sqrt{2} \times 10^{7.5}} = \frac{1}{1.414 \times 3.162 \times 10^7} = \frac{1}{4.472 \times 10^7}$$

$$K_2 = 2.236 \times 10^{-8} \approx 2 \times 10^{-8}$$

Since $$K_2 = x \times 10^{-8}$$, we have $$x = 2$$.

Therefore, the correct answer is 2.