JEE Chemistry Electrochemistry MCQs Set 14

Answer: (a) 0.917 V

Explanation:
1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, n = 6.
2. [Cr₂O₇²⁻] = 0.01 M, [Cr³⁺] = 0.1 M, [H⁺] = 10⁻³ M.
3. \( E=1.330-\dfrac{0.059}{6}\log\dfrac{(0.1)^2}{(0.01)(10^{-3})^{14}}=1.330-\dfrac{0.059}{6}\times42=0.917 \) V.

Teacher's Note:
1. Write the balanced half-reaction first to get n = 6 and the power 14 on [H⁺].
2. H⁺ is raised to the 14th power, so pH has a huge effect on the dichromate potential.

Answer: (b) Conductance rises slowly up to the equivalence point and then rises sharply

Explanation:
1. Before the equivalence point the weak acid is replaced by its sodium salt, which is fully ionised, so the conductance rises slowly.
2. After the equivalence point excess OH⁻ (very high conductivity) is added, so the conductance rises sharply.
3. A V-shaped curve (fall then rise) is for a strong acid against a strong base.

Teacher's Note:
1. Weak acid vs strong base: two rising lines, the second steeper.
2. Strong acid vs strong base: fall to a minimum, then rise.

Answer: (d) 3

Explanation:
1. Cathode reaction: M³⁺ + 2e⁻ → M⁺ (n = 2); anode: H₂ → 2H⁺ + 2e⁻ with E° = 0.
2. E_cell = 0.2 − (0.059/2) log([M⁺]/[H⁺]... = 0.2 − 0.0295 log([M⁺]/[M³⁺]) since [H⁺] = 1 M and p_H₂ = 1 bar.
3. 0.1115 = 0.2 − 0.0295a, so a = 0.0885/0.0295 = 3.

Teacher's Note:
1. Q = [M⁺]/[M³⁺] for the cathode half-reaction when the hydrogen electrode is standard.
2. Use n = 2 because M³⁺ gains two electrons to become M⁺.

Answer: (c) 10

Explanation:
1. Cathode: Fe³⁺ + e⁻ → Fe²⁺ (n = 1); E = 0.771 − 0.059 log([Fe²⁺]/[Fe³⁺]).
2. 0.712 = 0.771 − 0.059 log r, so log r = 0.059/0.059 = 1.
3. r = 10.

Teacher's Note:
1. Cell potential is below the standard value when the product concentration exceeds the reactant concentration.
2. Nernst equation for n = 1: E = E° − 0.059 log Q.

Answer: (b) 2

Explanation:
1. (A) is incorrect: for a weak electrolyte the curve rises too steeply, so Λ°_m cannot be extrapolated; Kohlrausch's law of independent migration is used instead.
2. (B) is correct (Debye–Hückel–Onsager equation) and (D) is correct (Kohlrausch's law).
3. (C) is incorrect: at infinite dilution the degree of dissociation approaches one. Incorrect statements: A and C = 2.

Teacher's Note:
1. Strong electrolytes: Λ_m = Λ°_m − A√c, a straight line.
2. Every electrolyte is completely dissociated at infinite dilution.

Answer: (a) 0.17 V

Explanation:
1. E°_cell = (0.059/n) log K = (0.059/2) × 20 = 0.59 V.
2. E°_cell = E°(Sn²⁺/Sn) − E°(Zn²⁺/Zn), so E°(Sn²⁺/Sn) = 0.59 − 0.76 = −0.17 V.
3. The magnitude is 0.17 V (17 × 10⁻² V).

Teacher's Note:
1. Zinc is oxidised and tin(II) is reduced in this reaction, so tin is the cathode.
2. Use E°_cell = (0.059/n) log K at 298 K.

Answer: (d) 2.75 V

Explanation:
1. E°_cell = 0.36 − (−2.36) = 2.72 V; n = 2.
2. Q = [X²⁺]/[Y²⁺] = 0.001/0.01 = 0.1.
3. E = 2.72 − (0.06/2) log 0.1 = 2.72 + 0.03 = 2.75 V (275 × 10⁻² V).

Teacher's Note:
1. The cell with the product concentration lower than the reactant concentration has a potential above E°.
2. Anode on the left, cathode on the right.

Answer: (d) OH⁻ is formed at the cathode

Explanation:
1. At the cathode water is reduced: 2H₂O + 2e⁻ → H₂ + 2OH⁻, so H₂ gas and OH⁻ are produced.
2. At the anode chloride is oxidised: 2Cl⁻ → Cl₂ + 2e⁻.
3. So OH⁻ is formed at the cathode.

Teacher's Note:
1. Brine electrolysis gives Cl₂ (anode), H₂ and NaOH (cathode region).
2. Reduction always takes place at the cathode in an electrolytic cell as well.

Answer: (c) 6

Explanation:
1. Required reaction = first half-reaction − second half-reaction.
2. E°_cell = 0.83 − 0.65 = 0.18 V, with n = 2.
3. log K = nE°/0.06 = 2 × 0.18/0.06 = 6.

Teacher's Note:
1. Subtract the half-reaction with the lower E° (reverse it) to get the overall reaction.
2. At 298 K, log K = nE°/0.059.

Answer: (b) 2.5 × 10⁵

Explanation:
1. Conductivity κ = 1/ρ = 1/(5 × 10⁻³) = 200 Ω⁻¹ cm⁻¹.
2. Λ_m = κ × 1000/M = 200 × 1000/0.8 = 2.5 × 10⁵ Ω⁻¹ cm² mol⁻¹ (25 × 10⁴).

Teacher's Note:
1. With κ in S cm⁻¹ and concentration in mol L⁻¹, Λ_m = 1000κ/M.
2. Take the reciprocal of resistivity first.

Answer: (a) 3

Explanation:
1. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, n = 5.
2. \( 1.282=1.51-\dfrac{0.059}{5}\log\dfrac{0.001}{0.1\,[H^+]^8} \), so \( \log\dfrac{0.001}{0.1[H^+]^8}=19.3 \).
3. \( -2+8\,\mathrm{pH}=19.3 \), so pH ≈ 2.67, i.e. 3 to the nearest integer.

Teacher's Note:
1. The potential of permanganate depends on [H⁺]⁸, so it falls sharply as pH rises.
2. Q = [Mn²⁺]/([MnO₄⁻][H⁺]⁸).