Redox and electrochemistry for JEE Main

Electrochemistry is a scoring chapter because its numericals follow a handful of equations: E°cell, Nernst, ΔG° = −nFE°, molar conductivity and Faraday's law. JEE Main sets both multiple-choice and numerical-value questions on it. The marks slip away through sign errors, a wrong n, or a stoichiometric coefficient multiplied into a potential. Redox sits underneath all of it, so start there.

The official 2026 syllabus lists oxidation number and balancing of redox reactions; electrolytic and metallic conduction, molar conductivity and its variation with concentration, Kohlrausch's law and its applications; electrolytic and galvanic cells, types of electrodes, standard electrode potential, EMF, the Nernst equation, the relation between cell potential and Gibbs energy; and the dry cell, lead accumulator and fuel cells.

Oxidation number: rules and tricky cases

Oxidation is loss of electrons (oxidation number rises); reduction is gain (oxidation number falls). The rules in order of priority:

• Free elements are 0. Fluorine is always −1 in compounds.
• Group 1 metals are +1 and group 2 metals +2 in compounds.
• Hydrogen is +1, except −1 in metal hydrides such as NaH.
• Oxygen is −2, except −1 in peroxides, −½ in superoxides (KO₂) and +2 in OF₂.
• The sum equals the charge on the species.

The formula gives an average. When atoms of one element sit in different environments, look at the structure:

Balancing by the ion-electron method

For acidic medium: split into half-reactions, balance atoms other than O and H, add H₂O for O, H⁺ for H, then electrons for charge. Multiply so that electrons cancel. In basic medium, finish in acid and then add OH⁻ to both sides to neutralise H⁺.

Permanganate and iron(II) in acid:

• MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
• Fe²⁺ → Fe³⁺ + e⁻ (multiply by 5)
• Net: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

The electrons transferred per formula unit (5 for MnO₄⁻ here, 6 for Cr₂O₇²⁻ going to 2Cr³⁺) set the mole ratio in redox titrations.

Galvanic cells and electrode potentials

In a galvanic cell, oxidation happens at the anode (negative terminal) and reduction at the cathode (positive terminal). The salt bridge completes the circuit and keeps both solutions neutral. Cell notation puts the anode on the left: Zn | Zn²⁺ || Cu²⁺ | Cu.

Standard electrode potentials are reduction potentials measured against the standard hydrogen electrode (SHE, 0 V). Then:

E°cell = E°cathode − E°anode, and a positive E°cell means a spontaneous cell reaction.

Electrode types you should recognise: metal–metal ion (Cu²⁺/Cu), gas electrodes (the SHE), metal–insoluble salt electrodes (Ag/AgCl, calomel) and redox electrodes (Pt dipping in Fe³⁺/Fe²⁺).

Nernst equation, ΔG and K

For a cell reaction with reaction quotient Q, at 298 K:

• E = E° − (0.0591/n) log Q
• ΔG = −nFE and ΔG° = −nFE°, with F = 96,500 C mol⁻¹
• ΔG° = −2.303RT log K, so log K = nE°/0.0591 at 298 K
• At equilibrium E = 0 and Q = K

Useful special cases: a hydrogen electrode at 1 bar has E = −0.0591 pH. A concentration cell has E° = 0, so its EMF comes entirely from the concentration ratio.

Conductance and Kohlrausch's law

Metallic conduction is by electrons and falls as temperature rises. Electrolytic conduction is by ions and usually rises with temperature.

• Conductivity κ = G × (l/A), where l/A is the cell constant (cm⁻¹)
• Molar conductivity Λm = κ × 1000/C in S cm² mol⁻¹, with C in mol L⁻¹
• Strong electrolytes: Λm rises slowly on dilution, linearly with √C: Λm = Λ°m − A√C. Extrapolate to C = 0 for Λ°m.
• Weak electrolytes: Λm rises steeply on dilution because α increases, so Λ°m cannot be extrapolated.

Kohlrausch's law: at infinite dilution each ion contributes independently, so Λ°m = ν₊λ°₊ + ν₋λ°₋. Its main uses are Λ°m of a weak electrolyte from strong ones, and α = Λm/Λ°m, which then gives Ka.

Electrolysis and cells in use

In an electrolytic cell, electrical energy drives a non-spontaneous reaction. Faraday's first law gives the mass deposited: m = MIt/(nF). By the second law, the same charge deposits masses in the ratio of M/n.

• Aqueous NaCl: H₂ at the cathode, Cl₂ at the anode (overpotential favours Cl₂ over O₂), NaOH left in solution.
• Aqueous CuSO₄ with inert electrodes: Cu at the cathode, O₂ at the anode.

Lead accumulator on discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. The acid is consumed, which is why its density indicates the state of charge.

Worked problems

Problem 1: Nernst equation. Find the EMF of Zn | Zn²⁺ (0.01 M) || Cu²⁺ (1 M) | Cu, given E°cell = 1.10 V.

Q = [Zn²⁺]/[Cu²⁺] = 0.01, so log Q = −2 and n = 2.
E = 1.10 − (0.0591/2)(−2) = 1.10 + 0.059 = 1.159 V. Diluting the anode side raises the EMF.

Problem 2: ΔG° and K. For the same cell, find ΔG° and K at 298 K.

ΔG° = −2 × 96,500 × 1.10 = −212,300 J = −212.3 kJ mol⁻¹.
log K = 2 × 1.10/0.0591 = 37.23, so K ≈ 1.7 × 10³⁷. The reaction goes essentially to completion.

Problem 3: Kohlrausch and Ka. Λ°m values (S cm² mol⁻¹): HCl 425.9, NaCl 126.4, CH₃COONa 91.0. A 0.01 M acetic acid solution has κ = 1.6 × 10⁻⁴ S cm⁻¹. Find α and Ka.

Λ°m(CH₃COOH) = 91.0 + 425.9 − 126.4 = 390.5.
Λm = 1.6 × 10⁻⁴ × 1000/0.01 = 16 S cm² mol⁻¹.
α = 16/390.5 = 0.041. Ka = Cα²/(1 − α) = 0.01 × 0.00168/0.959 ≈ 1.75 × 10⁻⁵.

Problem 4 (numerical answer): Faraday's law. A current of 2 A is passed through aqueous CuSO₄ for 965 s. Find the mass of copper deposited in mg (Cu = 63.5).

Charge = 2 × 965 = 1,930 C = 0.02 F. Cu²⁺ needs 2 F per mole, so 0.01 mol Cu = 0.635 g.
Answer: 635.

Traps to watch

• Multiplying E° by coefficients. Never. Balance electrons, not potentials.
• Wrong n in Nernst. n is the electrons in the balanced overall reaction, not in one half-reaction before multiplying.
• Sign of ΔG. A positive E means negative ΔG and a spontaneous reaction.
• Units of Λm. Use the factor 1000 only when κ is in S cm⁻¹ and C in mol L⁻¹.
• Extrapolating a weak electrolyte to find Λ°m. Use Kohlrausch instead.
• Anode sign. Negative in a galvanic cell, positive in an electrolytic cell. Oxidation happens at the anode in both.

Practice set

1. Find the oxidation state of Cr in CrO₅.
2. How many faradays are needed to reduce 1 mol of MnO₄⁻ to Mn²⁺?
3. Find the average oxidation state of S in Na₂S₂O₃.
4. Find the potential of a hydrogen electrode at pH 2 and 298 K.
5. E°(Ag⁺/Ag) = 0.80 V and E°(Cu²⁺/Cu) = 0.34 V. Find E° for Cu + 2Ag⁺ → Cu²⁺ + 2Ag.
6. A conductivity cell filled with 0.1 M KCl (κ = 0.0129 S cm⁻¹) has a resistance of 100 Ω. Find the cell constant.
7. How many moles of Fe²⁺ does one mole of Cr₂O₇²⁻ oxidise in acid?
8. How long must 0.5 A flow to deposit 1.08 g of silver (Ag = 108)?

Answers

1. +6 (one oxide O at −2, two peroxide groups with O at −1).
2. 5 (Mn goes from +7 to +2).
3. 2 + 2x − 6 = 0, so +2.
4. E = −0.0591 × 2 = −0.118 V.
5. 0.80 − 0.34 = 0.46 V. The 2 in front of Ag⁺ does not change E°.
6. Cell constant = κ × R = 1.29 cm⁻¹.
7. 6 (Cr₂O₇²⁻ gains 6 electrons going to 2Cr³⁺).
8. 0.01 mol Ag needs 965 C, so t = 965/0.5 = 1,930 s.

What to do next

• Write the four core equations (E°cell, Nernst, ΔG°–K, Λm) on one card and solve ten numericals from it.
• Balance five redox equations in acid and five in base by the ion-electron method.
• Revise ionic equilibrium for Ka and pH links, and thermodynamics for ΔG.
• Mix these problems with others in the physical chemistry numericals guide.