Electrochemistry Formulas For JEE 2026
Redox Reactions and Electrochemistry is an important chapter in JEE Chemistry and is known for being quite scoring in the exam. This topic explains how oxidation and reduction reactions occur and how electrons are transferred between substances during chemical reactions. Students also learn concepts like balancing redox equations, identifying oxidation states, and understanding electrochemical cells, which are essential for many chemical processes.
The chapter further includes important topics such as the Nernst equation and Faraday’s laws of electrolysis, which are often used in numerical questions. These concepts also explain real-life applications like batteries, electrolysis, and corrosion. When students clearly understand these formulas, solving exam questions becomes easier. For quick revision during preparation, they can also refer to a well-structured JEE Mains Chemistry Formula PDF to review key formulas and concepts efficiently.
Oxidation and Reduction
Definition: Oxidation
Oxidation is the loss of electrons by a substance. When a species is oxidized, its oxidation number increases.
Definition: Reduction
Reduction is the gain of electrons by a substance. When a species is reduced, its oxidation number decreases.
Definition: Oxidizing Agent (Oxidant)
The substance that gets reduced (gains electrons) and causes oxidation of the other substance.
Definition: Reducing Agent (Reductant)
The substance that gets oxidized (loses electrons) and causes reduction of the other substance.
Tip: Remember "LEO the lion says GER" — Loss of Electrons is Oxidation; Gain of Electrons is Reduction. Alternatively: "OIL RIG" — Oxidation Is Loss, Reduction Is Gain.
Oxidation Numbers
Definition: Oxidation Number
A hypothetical charge assigned to an atom in a compound, calculated by assuming all bonds are completely ionic. It represents the number of electrons an atom has gained or lost relative to its elemental state.
Rules for Assigning Oxidation Numbers
- Free elements have oxidation number = 0 (e.g., Na, O$$_2$$, P$$_4$$, S$$_8$$)
- For monoatomic ions, oxidation number = charge on the ion (e.g., Na$$^+$$ = $$+1$$, Cl$$^-$$ = $$-1$$)
- Hydrogen is $$+1$$ in most compounds; $$-1$$ in metal hydrides (NaH, CaH$$_2$$)
- Oxygen is $$-2$$ in most compounds; $$-1$$ in peroxides (H$$_2$$O$$_2$$, Na$$_2$$O$$_2$$); $$-\frac{1}{2}$$ in superoxides (KO$$_2$$); $$+2$$ in OF$$_2$$
- Fluorine is always $$-1$$ (most electronegative element)
- Alkali metals are always $$+1$$; alkaline earth metals are always $$+2$$
- Sum of oxidation numbers = 0 for a neutral compound; = charge for a polyatomic ion
Worked Example: Oxidation Number of Mn in KMnO$$_4$$
Let the oxidation number of Mn be $$x$$.
K = $$+1$$, O = $$-2$$ (four oxygens)
Sum = 0: $$(+1) + x + 4(-2) = 0$$
$$1 + x - 8 = 0$$
$$x = +7$$
Mn has an oxidation number of $$+7$$ in KMnO$$_4$$.
Worked Example: Oxidation Number of S in Na$$_2$$S$$_2$$O$$_3$$
Let the oxidation number of S be $$x$$.
$$2(+1) + 2x + 3(-2) = 0$$
$$2 + 2x - 6 = 0$$
$$2x = 4$$, so $$x = +2$$
Each S atom has an oxidation number of $$+2$$.
Balancing Redox Reactions: Half-Reaction Method
Half-Reaction Method Steps
- Step 1: Identify oxidation and reduction half-reactions
- Step 2: Balance atoms other than O and H in each half-reaction
- Step 3: Balance oxygen by adding H$$_2$$O
- Step 4: Balance hydrogen by adding H$$^+$$ (acidic medium) or OH$$^-$$ (basic medium)
- Step 5: Balance charge by adding electrons (e$$^-$$)
- Step 6: Multiply half-reactions so that electrons cancel
- Step 7: Add the two half-reactions together
Worked Example: Balancing in Acidic Medium
Balance: $$\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$$
Reduction half: $$\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$$
Balance O with H$$_2$$O: $$\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
Balance H with H$$^+$$: $$\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
Balance charge with e$$^-$$: $$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
Oxidation half: $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$$
Multiply oxidation half by 5 and add:
$$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$$$
Types of Redox Reactions
| Type | Description and Example |
|---|---|
| Combination | Two substances combine: $$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$ |
| Decomposition | One substance breaks down: $$2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$$ |
| Displacement | One element displaces another: $$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$$ |
| Disproportionation | Same element is both oxidized and reduced: $$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$$ |
| Comproportionation | Two different oxidation states of same element combine into one |
Important Note
In a disproportionation reaction, the same element is simultaneously oxidized and reduced. Example: In $$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$$, oxygen goes from $$-1$$ in H$$_2$$O$$_2$$ to $$-2$$ in H$$_2$$O (reduced) and 0 in O$$_2$$ (oxidized).
Electrochemical Cells
An electrochemical cell that produces electrical energy from a spontaneous redox reaction. Examples: batteries, fuel cells.
Components of a Galvanic Cell
- Anode: The electrode where oxidation occurs (negative terminal in galvanic cell)
- Cathode: The electrode where reduction occurs (positive terminal in galvanic cell)
- Salt bridge: Provides ionic contact between half-cells (usually KCl or KNO$$_3$$ in agar)
- Electron flow: From anode to cathode through the external wire
- Ion flow: Anions move toward anode, cations move toward cathode through salt bridge
Tip: Remember "An Ox" and "Red Cat" — Anode = Oxidation, Reduction = Cathode. This works for both galvanic and electrolytic cells.
Cell Notation (Cell Representation)
Cell Notation Convention
$$$\text{Anode} \,|\, \text{Anode solution} \,||\, \text{Cathode solution} \,|\, \text{Cathode}$$$
Example — Daniell Cell:
$$$\text{Zn}(s) \,|\, \text{Zn}^{2+}(aq) \,||\, \text{Cu}^{2+}(aq) \,|\, \text{Cu}(s)$$$
Anode (left): $$\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$$ (oxidation)
Cathode (right): $$\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$$ (reduction)
Overall: $$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$$
EMF and Standard Electrode Potential
Definition: Standard Electrode Potential ($$E^\circ$$)
The potential of a half-cell measured against the Standard Hydrogen Electrode (SHE, defined as $$E^\circ = 0.00$$ V) under standard conditions: 1 M concentration, 1 bar pressure, 298 K.
Cell EMF
$$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$$
For a spontaneous reaction: $$E^\circ_{\text{cell}} > 0$$
Note: Standard reduction potentials are used. The more positive the $$E^\circ$$, the stronger the oxidizing agent (greater tendency to be reduced).
Worked Example: Cell EMF for the Daniell Cell
Given: $$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$$ V, $$E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76$$ V
Cathode = Cu (higher $$E^\circ$$), Anode = Zn (lower $$E^\circ$$)
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (+0.34) - (-0.76) = +1.10$$ V
Since $$E^\circ_{\text{cell}} > 0$$, the reaction is spontaneous.
Electrochemical Series
Electrochemical Series (Selected Values)
| Half-Reaction (Reduction) | $$E^\circ$$ (V) |
|---|---|
| Li$$^+$$ + e$$^-$$ $$\rightarrow$$ Li | $$-3.05$$ |
| K$$^+$$ + e$$^-$$ $$\rightarrow$$ K | $$-2.93$$ |
| Ca$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Ca | $$-2.87$$ |
| Na$$^+$$ + e$$^-$$ $$\rightarrow$$ Na | $$-2.71$$ |
| Mg$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Mg | $$-2.37$$ |
| Al$$^{3+}$$ + 3e$$^-$$ $$\rightarrow$$ Al | $$-1.66$$ |
| Zn$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Zn | $$-0.76$$ |
| Fe$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Fe | $$-0.44$$ |
| Ni$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Ni | $$-0.26$$ |
| Sn$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Sn | $$-0.14$$ |
| 2H$$^+$$ + 2e$$^-$$ $$\rightarrow$$ H$$_2$$ | $$0.00$$ |
| Cu$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Cu | $$+0.34$$ |
| Ag$$^+$$ + e$$^-$$ $$\rightarrow$$ Ag | $$+0.80$$ |
| Au$$^{3+}$$ + 3e$$^-$$ $$\rightarrow$$ Au | $$+1.50$$ |
| F$$_2$$ + 2e$$^-$$ $$\rightarrow$$ 2F$$^-$$ | $$+2.87$$ |
Uses of the Electrochemical Series
- Predicting spontaneity: A species higher in the series (more negative $$E^\circ$$) can reduce a species lower in the series. Metal higher up displaces metal lower down from its salt solution.
- Strength of oxidizing/reducing agents: Higher $$E^\circ$$ = stronger oxidizing agent; lower $$E^\circ$$ = stronger reducing agent.
- Reactivity with water/acids: Metals above hydrogen can displace H$$_2$$ from dilute acids.
Worked Example: Displacement Reaction Prediction
Will iron displace copper from CuSO$$_4$$ solution?
$$E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44$$ V is more negative than $$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$$ V.
So Fe is a stronger reducing agent and will displace Cu:
$$\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$$
$$E^\circ_{\text{cell}} = 0.34 - (-0.44) = +0.78$$ V $$> 0$$ (spontaneous)
Nernst Equation
$$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$$$
At 298 K (converting to log base 10):
$$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$$
where:
- $$n$$ = number of electrons transferred in the balanced equation
- $$F$$ = Faraday constant = 96485 C/mol $$\approx$$ 96500 C/mol
- $$Q$$ = reaction quotient
- $$R$$ = 8.314 J/(mol·K)
Worked Example: Nernst Equation
Calculate the EMF of the cell: $$\text{Zn} | \text{Zn}^{2+}(0.1\text{ M}) || \text{Cu}^{2+}(0.01\text{ M}) | \text{Cu}$$
$$E^\circ_{\text{cell}} = 1.10$$ V, $$n = 2$$
For $$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$$:
$$Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.1}{0.01} = 10$$
$$E_{\text{cell}} = 1.10 - \dfrac{0.0591}{2} \log 10 = 1.10 - \dfrac{0.0591}{2} \times 1 = 1.10 - 0.0296 = 1.07$$ V
Important: Nernst Equation at Equilibrium
At equilibrium, $$E_{\text{cell}} = 0$$ and $$Q = K$$. Substituting into the Nernst equation:
$$0 = E^\circ_{\text{cell}} - \dfrac{0.0591}{n}\log K$$
$$$\log K = \frac{nE^\circ_{\text{cell}}}{0.0591}$$$
Gibbs Energy and Cell Potential
Gibbs Energy and EMF
$$$\Delta G = -nFE_{\text{cell}}$$$
Under standard conditions:
$$$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$$
- If $$E^\circ_{\text{cell}} > 0$$: $$\Delta G^\circ < 0$$ (spontaneous — galvanic cell)
- If $$E^\circ_{\text{cell}} < 0$$: $$\Delta G^\circ > 0$$ (non-spontaneous — electrolytic cell needed)
Worked Example: Gibbs Energy for the Daniell Cell
Calculate $$\Delta G^\circ$$ for the Daniell cell ($$E^\circ_{\text{cell}} = 1.10$$ V, $$n = 2$$).
$$\Delta G^\circ = -nFE^\circ_{\text{cell}} = -(2)(96500)(1.10) = -212300$$ J $$= -212.3$$ kJ
The large negative value confirms the reaction is highly spontaneous.
Electrolysis and Faraday's Laws
Definition: Electrolysis
The process of using direct current (DC) to drive a non-spontaneous chemical reaction. It takes place in an electrolytic cell. In an electrolytic cell, the anode is positive and the cathode is negative (opposite of galvanic cell), but oxidation still occurs at the anode and reduction at the cathode.
Faraday's First Law of Electrolysis
The mass of substance deposited or dissolved at an electrode is directly proportional to the quantity of electricity (charge) passed:
$$$m = \frac{M \times I \times t}{n \times F} = \frac{M \times Q}{n \times F}$$$
where:
- $$m$$ = mass deposited (g)
- $$M$$ = molar mass (g/mol)
- $$I$$ = current (A), $$t$$ = time (s), $$Q = It$$ = charge (C)
- $$n$$ = number of electrons transferred per ion
- $$F$$ = 96500 C/mol (charge of 1 mol of electrons)
Faraday's Second Law of Electrolysis
When the same quantity of electricity passes through different electrolytes in series, the masses deposited are proportional to their equivalent weights:
$$$\frac{m_1}{m_2} = \frac{E_1}{E_2}$$$
where $$E = M/n$$ is the equivalent weight (molar mass divided by number of electrons exchanged per atom/ion).
Worked Example: Faraday's First Law
How long must a current of 3 A be passed through a CuSO$$_4$$ solution to deposit 1.27 g of copper? ($$M_{\text{Cu}} = 63.5$$ g/mol)
Cu$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Cu, so $$n = 2$$
$$m = \dfrac{M \times I \times t}{n \times F}$$
$$1.27 = \dfrac{63.5 \times 3 \times t}{2 \times 96500}$$
$$t = \dfrac{1.27 \times 2 \times 96500}{63.5 \times 3} = \dfrac{245110}{190.5} = 1287$$ s $$\approx 21.4$$ min
Worked Example: Faraday's Second Law
A current of 5 A is passed for 1930 s through two cells in series: one with AgNO$$_3$$ and one with CuSO$$_4$$. Find the mass deposited in each.
Charge $$Q = 5 \times 1930 = 9650$$ C $$= 0.1 F$$
Silver: Ag$$^+$$ + e$$^-$$ $$\rightarrow$$ Ag ($$n = 1$$, $$M = 108$$)
$$m_{\text{Ag}} = \dfrac{108 \times 9650}{1 \times 96500} = 10.8$$ g
Copper: Cu$$^{2+}$$ + 2e$$^-$$ $$\rightarrow$$ Cu ($$n = 2$$, $$M = 63.5$$)
$$m_{\text{Cu}} = \dfrac{63.5 \times 9650}{2 \times 96500} = 3.175$$ g
Conductance and Molar Conductivity
Definition: Specific Conductance ($$\kappa$$)
The conductance of a solution contained between two electrodes of unit area separated by unit distance. Unit: S/cm or S/m (S = siemens = ohm$$^{-1}$$).
Definition: Molar Conductivity ($$\Lambda_m$$)
The conductance of a solution containing 1 mole of electrolyte placed between electrodes 1 cm apart. It accounts for the effect of dilution.
Conductance Formulas
Conductance (G) = reciprocal of resistance:
$$$G = \frac{1}{R} = \kappa \times \frac{A}{l}$$$
Cell constant: $$G^* = l/A$$ (where $$l$$ = distance between electrodes, $$A$$ = area)
$$$\kappa = G \times G^* = \frac{G^*}{R}$$$
Molar conductivity:
$$$\Lambda_m = \frac{\kappa \times 1000}{c}$$$
where $$c$$ = concentration in mol/L, $$\kappa$$ in S/cm, and $$\Lambda_m$$ in S·cm$$^2$$/mol.
Variation of $$\Lambda_m$$ with Dilution
For strong electrolytes (e.g., NaCl, KCl):
$$\Lambda_m$$ increases slightly with dilution and approaches a limiting value $$\Lambda_m^\circ$$ (molar conductivity at infinite dilution):
$$$\Lambda_m = \Lambda_m^\circ - A\sqrt{c} \quad \text{(Debye–Hückel–Onsager equation)}$$$
For weak electrolytes (e.g., CH$$_3$$COOH):
$$\Lambda_m$$ increases steeply with dilution because ionization increases. $$\Lambda_m^\circ$$ cannot be obtained by extrapolation — we use Kohlrausch's law instead.
Kohlrausch's Law
At infinite dilution, the molar conductivity of an electrolyte is the sum of the individual contributions of the cation and anion, each multiplied by the number of ions produced per formula unit.
Kohlrausch's Law Formula
$$$\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ$$$
where $$\nu_+$$ and $$\nu_-$$ are the number of cations and anions per formula unit, and $$\lambda_+^\circ$$ and $$\lambda_-^\circ$$ are the limiting ionic conductivities.
Applications:
- Finding $$\Lambda_m^\circ$$ of weak electrolytes (which cannot be measured directly)
- Calculating degree of dissociation: $$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$$
- Finding $$K_a$$ of a weak acid: $$K_a = \dfrac{c\alpha^2}{1 - \alpha} = \dfrac{c(\Lambda_m)^2}{\Lambda_m^\circ(\Lambda_m^\circ - \Lambda_m)}$$
Worked Example: Finding $$\Lambda_m^\circ$$ of a Weak Electrolyte
Find $$\Lambda_m^\circ$$ of CH$$_3$$COOH given:
$$\Lambda_m^\circ(\text{CH}_3\text{COONa}) = 91.0$$ S·cm$$^2$$/mol
$$\Lambda_m^\circ(\text{HCl}) = 426.2$$ S·cm$$^2$$/mol
$$\Lambda_m^\circ(\text{NaCl}) = 126.5$$ S·cm$$^2$$/mol
$$\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl})$$
$$= 91.0 + 426.2 - 126.5 = 390.7$$ S·cm$$^2$$/mol
(We added Na$$^+$$ + CH$$_3$$COO$$^-$$ and H$$^+$$ + Cl$$^-$$, then subtracted Na$$^+$$ + Cl$$^-$$, leaving H$$^+$$ + CH$$_3$$COO$$^-$$.)
Worked Example: Degree of Dissociation and $$K_a$$
The molar conductivity of 0.001 M acetic acid is 48.15 S·cm$$^2$$/mol. If $$\Lambda_m^\circ = 390.7$$ S·cm$$^2$$/mol, find the degree of dissociation and $$K_a$$.
$$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{48.15}{390.7} = 0.1232$$ (12.32%)
$$K_a = \dfrac{c\alpha^2}{1 - \alpha} = \dfrac{0.001 \times (0.1232)^2}{1 - 0.1232} = \dfrac{0.001 \times 0.01518}{0.8768} = 1.73 \times 10^{-5}$$
Batteries
Types of Batteries
| Type | Description |
|---|---|
| Primary (non-rechargeable) | Reaction is irreversible; once used, cannot be recharged |
| Dry cell (Leclanche) | Zn anode, MnO$$_2$$/C cathode, NH$$_4$$Cl paste; $$E \approx 1.5$$ V |
| Mercury cell | Zn–HgO cell; constant voltage ($$\approx 1.35$$ V); used in hearing aids |
| Secondary (rechargeable) | Reaction is reversible; can be recharged by passing current |
| Lead-acid battery | Pb anode, PbO$$_2$$ cathode, H$$_2$$SO$$_4$$ electrolyte; $$E \approx 2$$ V per cell |
| Ni–Cd battery | Cd anode, NiO(OH) cathode; rechargeable, portable devices |
| Fuel cell | Reactants supplied continuously from outside |
| H$$_2$$–O$$_2$$ fuel cell | 2H$$_2$$ + O$$_2$$ $$\rightarrow$$ 2H$$_2$$O; $$E \approx 1.23$$ V; used in spacecraft |
Lead-Acid Battery Reactions
At anode (during discharge):
$$\text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2\text{e}^-$$
At cathode (during discharge):
$$\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}$$
Overall:
$$$\text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O}$$$
During charging, the reactions are reversed.
Corrosion
The deterioration of a metal due to an electrochemical reaction with its surroundings. It is essentially a galvanic cell where the metal acts as the anode and gets oxidized.
Mechanism of Rusting
Iron rusting occurs in the presence of both water and oxygen:
At anodic region (Fe surface):
$$\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2\text{e}^-$$ (oxidation)
At cathodic region (near water droplet):
$$\text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}(l)$$ (reduction)
The Fe$$^{2+}$$ ions are further oxidized to Fe$$^{3+}$$, forming rust: $$\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$$ (hydrated iron(III) oxide).
Prevention methods:
- Galvanization (coating with Zn — Zn is more reactive, acts as sacrificial anode)
- Painting, oiling, or greasing the surface
- Electroplating with a less reactive metal (Cr, Ni)
- Cathodic protection (connecting to a more reactive metal like Mg)
- Alloying (e.g., stainless steel contains Cr which forms a protective oxide layer)
Tip: In galvanization, zinc corrodes instead of iron because Zn has a more negative $$E^\circ$$ ($$-0.76$$ V vs $$-0.44$$ V for Fe). This is called sacrificial protection. Even if the zinc coating is scratched, the iron is still protected as long as zinc is present.
Summary of Key Formulas: Quick Reference
- $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
- Nernst: $$E = E^\circ - \dfrac{0.0591}{n}\log Q$$ (at 298 K)
- Gibbs–EMF: $$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$
- Equilibrium: $$\log K = \dfrac{nE^\circ}{0.0591}$$
- Faraday's 1st law: $$m = \dfrac{MIt}{nF}$$
- Molar conductivity: $$\Lambda_m = \dfrac{\kappa \times 1000}{c}$$
- Kohlrausch: $$\Lambda_m^\circ = \nu_+\lambda_+^\circ + \nu_-\lambda_-^\circ$$
- Degree of dissociation: $$\alpha = \Lambda_m / \Lambda_m^\circ$$
Important Note
The Nernst equation and Faraday's laws are the two most frequently tested topics from electrochemistry in JEE Mains. For Nernst equation problems, always identify $$n$$ (electrons transferred) and write the correct expression for $$Q$$. For Faraday's law problems, ensure consistent units — current in amperes, time in seconds, and $$F = 96500$$ C/mol.
Electrochemistry Formulas For JEE 2026: Conclusion
Electrochemistry is a crucial chapter in JEE Chemistry because it explains how chemical reactions involve electron transfer and how chemical energy can be converted into electrical energy. Concepts such as oxidation–reduction reactions, electrochemical cells, and electrode potentials help students understand how different substances behave in redox reactions and electrical systems.
A clear understanding of topics like the Nernst equation, Faraday’s laws of electrolysis, and conductance helps students solve numerical problems more efficiently. When these formulas and concepts are revised regularly, students can improve their problem-solving speed and accuracy in electrochemistry questions during competitive exams like JEE.
