Electrochemistry: JEE Advanced PYQ, 2007–2026

Answer: (A)

The cell reaction is Ag⁺(1 M) → Ag⁺(0.01 M), with n = 1. E = 0 − 0.0591 log(0.01/1) = 0.0591 × 2 = 0.118 V.

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: (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)

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: (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 the reduction Cl₂ + 2e⁻ → 2Cl⁻, n = 2 and Q = [Cl⁻]²/p(Cl₂). Q = (0.01)²/0.1 = 10⁻³, so log Q = −3. E = E° − (0.059/2) log Q = 1.36 − 0.0295 × (−3) = 1.36 + 0.0885 ≈ 1.45 V. The potential rises because the product concentration is low and the reactant gas pressure is only 0.1 bar.

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)

Conductivity is the sum over ions, κ = Σ c_i λ_i, with c_i in mol cm⁻³ when λ is in S cm² mol⁻¹. Converting, 1.0 × 10⁻⁷ mol L⁻¹ = 1.0 × 10⁻¹⁰ mol cm⁻³. Then κ = (1.0 × 10⁻¹⁰)(350 + 198) = 5.48 × 10⁻⁸ S cm⁻¹. Both H⁺ and OH⁻ carry current and must be included.

Answer: (D)

Anode: Zn → Zn²⁺ + 2e⁻. Moles of Zn = 1.30/65 = 0.020 mol, so electrons released = 0.040 mol. The cathode equation is one-electron: each electron reduces one MnO₂ to MnO(OH) and liberates one NH₃. So 0.040 mol of MnO₂ is reduced, a mass of 0.040 × 87 = 3.48 g, and 0.040 mol of NH₃ is liberated, a mass of 0.040 × 17 = 0.68 g. The key step is that the anode reaction is a two-electron process while the cathode reaction is a one-electron process.

Answer: (C)

Both electrodes are copper, so E°cell = 0 and the emf comes only from the concentration difference. The more concentrated half-cell (1.0 M) is the cathode and the dilute one (0.01 M) is the anode. The cell reaction is Cu(s) + Cu²⁺(1.0 M) → Cu²⁺(0.01 M) + Cu(s), with n = 2 and Q = 0.01/1.0 = 10⁻². Ecell = 0 − (0.059/2) log(10⁻²) = 0.0295 × 2 = 0.059 V. The 0.118 V option comes from using n = 1. The 0 V option comes from treating zero E° as zero emf. The 0.0295 V option comes from dropping the factor of 2 in log 100.