Electrochemistry (JEE Advanced)

Question 1

Use the standard reduction potentials E° values at 298 K (V): Ag⁺/Ag = +0.80, Fe³⁺/Fe²⁺ = +0.77, I₂/I⁻ = +0.54, Cu²⁺/Cu = +0.34, Sn⁴⁺/Sn²⁺ = +0.15, Fe²⁺/Fe = −0.44. Which of the following reactions is

Accepted Answer

B. For each reaction, E°_(cell) = E°(species reduced) − E°(species oxidised); spontaneous if E°_(cell) > 0.
(A) Ag⁺ reduced, Cu oxidised: 0.80 − 0.34 = +0.46 V, spontaneous.
(B) I₂ reduced, Fe²⁺ oxidised: 0.54 − 0.77 = −0.23 V, NOT spontaneous (the reverse, Fe³⁺ oxidising I⁻, is spontaneous).
(C) Fe³⁺ reduced, Sn²⁺ oxidised: 0.77 − 0.15 = +0.62 V, spontaneous.
(D) Sn⁴⁺ reduced, Fe oxidised: 0.15 − (−

Question 2

Acidified permanganate oxidises chloride: 2MnO₄⁻ + 16H⁺ + 10Cl⁻ arrow 2Mn²⁺ + 5Cl₂ + 8H₂O. At 298 K, E°(MnO₄⁻/Mn²⁺) = +1.51 V and E°(Cl₂/Cl⁻) = +1.36 V. Take 2.303RT/F = 0.059 V at 298 K. Keep [MnO₄⁻]

Accepted Answer

D. For MnO₄⁻ + 8H⁺ + 5e⁻ arrow Mn²⁺ + 4H₂O: E = 1.51 − 0.059/5log([Mn²⁺])/([MnO₄⁻][H⁺]⁸) = 1.51 − (8(0.059))/5 pH = 1.51 − 0.0944 pH.
The Cl₂/Cl⁻ potential is pH-independent (1.36 V). Oxidation stops when 1.51 − 0.0944 pH = 1.36, i.e. pH = 0.15/0.0944 = 1.59.
Distractors: 12.71 drops the factor 8; 4.07 inverts 8/5; 0.32 uses n = 1.
Answer: (D).
Difficulty: Hard

Question 3

A galvanic cell is built so that the spontaneous cell reaction is MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) arrow Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l). Given E°(MnO₄⁻/Mn²⁺) = +1.51 V, E°(Fe³⁺/Fe²⁺) = +0.77 V, F = 965

Accepted Answer

C. Oxidation (Fe²⁺ arrow Fe³⁺) occurs at the anode, written on the left; both couples are ionic, so inert Pt electrodes are used. Notation: Pt | Fe²⁺, Fe³⁺ ∥ MnO₄⁻, Mn²⁺, H⁺ | Pt.
E°_(cell) = 1.51 − 0.77 = 0.74 V. Electrons transferred for the reaction as written: n = 5.
Δ G° = −nFE° = −5(96500)(0.74) = −3.57×10⁵ J = −357 kJ.
(A) reverses the electrodes; (B) uses n = 1; (D) uses n = 10.
Answer: (C).

Electrochemistry: questions with solutions

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