Electrochemistry: JEE Main PYQ, 2020–2025

Answer: (B)

Conductance G = κA/l, where κ is the conductivity, A the electrode area and l the separation. κ depends only on the solution, so it stays the same. Doubling A multiplies G by 2. Halving l multiplies G by 2. The total factor is 4.

Answer: (A)

A metal ion attacks copper only if its couple has a higher reduction potential than Cu²⁺/Cu (+0.34 V). Zn²⁺/Zn is at −0.76 V, so for Cu + Zn²⁺ → Cu²⁺ + Zn the emf would be −0.76 − 0.34 = −1.10 V and the reaction does not occur. The other three ions all oxidise copper.

Answer: (A)

Both electrodes are silver, so E°cell = 0 and the cell is a concentration cell. The Nernst equation with n = 1 gives E = 0.0591 log([Ag⁺]cathode/[Ag⁺]anode). The cathode has [Ag⁺] = 0.1 M. Then 0.8274 = 0.0591 log(0.1/[Ag⁺]anode), so log(0.1/[Ag⁺]anode) = 14 and [Ag⁺]anode = 10⁻¹⁵ M. In the anode compartment AgI(s) is in equilibrium with 0.1 M I⁻, so Ksp = [Ag⁺][I⁻] = 10⁻¹⁵ × 0.1 = 1.0 × 10⁻¹⁶.

Answer: (A)

Principle: a heat engine converts chemical energy to heat and then to work, so its efficiency is limited by the Carnot relation that depends on the temperatures of the source and sink. A fuel cell converts chemical energy to electrical energy directly through redox reactions, so this limit does not apply. Its efficiency is therefore higher, though still less than 100 % because of real losses.

Answer: (D)

A salt-bridge electrolyte must meet two conditions. Its cation and anion should have nearly equal mobilities, and neither ion may react with the species in either half-cell. In this cell the right-hand compartment contains Ag⁺, which precipitates with Cl⁻ as AgCl, so KCl and NaCl are ruled out. The left-hand compartment contains SO₄²⁻, which precipitates with Ba²⁺ as BaSO₄, so Ba(NO₃)₂ is ruled out. Only KNO₃ satisfies both conditions.

Answer: (C)

Molar mass of HgO = 200 + 16 = 216 g mol⁻¹, so the cell holds 2.16/216 = 0.0100 mol. Zn amalgam is in excess, so HgO is the limiting reagent. HgO is reduced to Hg at the cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻. That is 2 mol of electrons per mole of HgO, so Q = 2 × 0.0100 × 96 500 = 1930 C. Time = Q/I = 1930 C / (1.00 × 10⁻³ A) = 1.93 × 10⁶ s. Dividing by 3600 s h⁻¹ gives about 536 h.

Answer: (A)

For Al³⁺ + 3e⁻ → Al(s), the solid aluminium has activity one and drops out of Q, leaving Q = 1/[Al³⁺]. With [Al³⁺] = 0.01 M, Q = 100 and log Q = 2. The Nernst equation is E = E° − (0.059/n) log Q, and with n = 3 the correction is (0.059/3) × 2 ≈ 0.039 V. Subtracting gives E ≈ −1.66 − 0.039 = −1.70 V.

Answer: (A)

The Debye–Hückel–Onsager (Kohlrausch) relation is linear in √c. The two data points are √c = 0.200 with Λm = 120 and √c = 0.100 with Λm = 130. The slope is A = (130 − 120)/(0.200 − 0.100) = 100. The intercept at √c = 0 is Λ° = 130 + 100 × 0.100 = 140 S cm² mol⁻¹. As a check, 140 − 100 × 0.2 = 120.

Answer: (A)

At equilibrium E_cell = 0, so the Nernst equation gives E°cell = (0.059/n) log K_c, i.e. log K_c = nE°cell/0.059. First, E°cell = E°(cathode) − E°(anode) = −0.25 − (−1.66) = 1.41 V. Next, n is the total number of electrons transferred in the balanced equation as written. Al is oxidised to Al³⁺ (3e⁻ each) and 2 Al atoms react, so 6 electrons are lost. Three Ni²⁺ ions gain 2e⁻ each, so 6 electrons are gained. Hence n = 6, and log K_c = (6 × 1.41)/0.059 = 8.46/0.059 ≈ 143.

Answer: (A)

Use E = E° − (2.303RT/nF) log Q for the reduction Sn⁴⁺ + 2e⁻ → Sn²⁺. The number of electrons, n = 2, appears in the denominator of the coefficient, so the coefficient is 0.059/2 = 0.0295 V. Q = [Sn²⁺]/[Sn⁴⁺] = 0.1/0.001 = 100 and log 100 = 2. The correction is 0.0295 × 2 = 0.059 V, so E = 0.15 − 0.059 = 0.091 V.