Electrochemistry JEE questions: formulas, patterns and the mistakes that cost marks

Galvanic and electrolytic cells

A galvanic cell turns a spontaneous reaction into current; an electrolytic cell uses current to force a non-spontaneous one. Oxidation is at the anode in both, but the anode is negative in a galvanic cell and positive in an electrolytic one.

Standard electrode potentials

Tabulated values are reduction potentials against the standard hydrogen electrode. E°_cell = E°_cathode − E°_anode, using both as reduction potentials — never reverse a sign and subtract as well.

The Nernst equation

Concentration changes shift the cell potential: E = E° − (0.0591/n) log Q at 298 K, where Q is the reaction quotient written for the balanced cell reaction. At equilibrium Q = K and E = 0, which is a flat battery.

Linking potential, free energy and equilibrium

ΔG = −nFE and ΔG° = −nFE° = −2.303RT log K. These two lines connect a measured voltage to spontaneity and to the equilibrium constant, and questions routinely ask you to travel between them.

Conductivity and molar conductivity

Conductivity κ falls on dilution because there are fewer ions per unit volume. Molar conductivity Λ_m = 1000κ/c rises, because it counts the conduction per mole of electrolyte, and every ion is freer to move.

Kohlrausch’s law

At infinite dilution each ion contributes independently: Λ°_m = ν₊λ°₊ + ν₋λ°₋. This is how Λ°_m for a weak acid is obtained from strong-electrolyte data, and it leads to α = Λ_m/Λ°_m and Ka = cα²/(1 − α).

Standard cell potential

E°_cell = E°_cathode − E°_anode

Both taken as reduction potentials.

Nernst equation at 298 K

E = E° − (0.0591/n) log Q

n = moles of electrons in the balanced cell reaction.

Free energy from potential

ΔG = −nFE · ΔG° = −nFE°

F = 96500 C mol⁻¹; a positive E means a spontaneous cell.

Equilibrium constant

E°_cell = (0.0591/n) log K

From ΔG° = −2.303RT log K at 298 K.

Conductivity

κ = G × (l/A)

l/A is the cell constant, in cm⁻¹.

Molar conductivity

Λ_m = 1000 κ / c

κ in S cm⁻¹ and c in mol L⁻¹ give Λ_m in S cm² mol⁻¹.

Kohlrausch’s law

Λ°_m = ν₊ λ°₊ + ν₋ λ°₋

Ions contribute independently at infinite dilution.

Degree of dissociation

α = Λ_m / Λ°_m · Ka = cα²/(1 − α)

For weak electrolytes only.

Faraday’s laws

m = (M / nF) × I × t

Charge Q = It; n is the electrons per ion discharged.

Strong electrolyte dilution

Λ_m = Λ°_m − A√c

Debye–Hückel–Onsager: a gentle, linear-in-√c rise on dilution.

Nernst equation · 3 marks · moderate

For the cell Zn(s) | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu(s), with E°_cell = 1.10 V, calculate the cell potential at 298 K.

1. Cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), so n = 2.
2. Reaction quotient: Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10 (solids are omitted).
3. Nernst: E = E° − (0.0591/2) log 10 = 1.10 − 0.02955 × 1.
4. E = 1.10 − 0.03 = 1.07 V.
5. The potential drops slightly because the product ion is more concentrated than the reactant ion, which pushes the reaction towards equilibrium.

Answer: E ≈ 1.07 V.

Faraday’s laws · 3 marks · easy

A current of 5 A is passed through a CuSO₄ solution for 30 minutes. Calculate the mass of copper deposited. (Cu = 63.5 g mol⁻¹)

1. Charge passed: Q = It = 5 × (30 × 60) = 9000 C.
2. Cu²⁺ + 2e⁻ → Cu, so n = 2 and one mole of copper needs 2 × 96500 = 193000 C.
3. Moles of copper = 9000/193000 = 0.0466 mol.
4. Mass = 0.0466 × 63.5 ≈ 2.96 g.

Answer: About 2.96 g of copper is deposited.

Kohlrausch’s law · 4 marks · moderate

The molar conductivity of 0.01 M acetic acid is 16.3 S cm² mol⁻¹, and Λ°_m for acetic acid is 390.5 S cm² mol⁻¹. Find the degree of dissociation and the dissociation constant.

1. Degree of dissociation: α = Λ_m/Λ°_m = 16.3/390.5 = 0.0417.
2. For a weak acid, Ka = cα²/(1 − α).
3. cα² = 0.01 × (0.0417)² = 0.01 × 0.001739 = 1.739 × 10⁻⁵.
4. Divide by (1 − 0.0417) = 0.9583: Ka = 1.739 × 10⁻⁵ / 0.9583 ≈ 1.8 × 10⁻⁵.

Answer: α ≈ 0.042 (4.2% dissociated); Ka ≈ 1.8 × 10⁻⁵.