Q95 · CSIR-NET Chemistry, December 2013

Paper: CSIR-NET December 2013 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

Consider the cell: 3 equilibrium constant for the reaction: $\mathrm{Zn}+2 \mathrm{Fe}^{3+} \rightleftharpoons \mathrm{Zn}^{2+}+2 \mathrm{Fe}^{2+}$ at $25^{\circ} \mathrm{C}$ would be close to
(a)$10^{27}$
(b)$10^{54}$
(c)$10^{81}$
(d)$10_{40}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Q = [Zn²⁺][Fe²⁺]²/[Fe³⁺]² = 0.01 × 10⁻⁶/10⁻⁴ = 10⁻⁴; E° = 1.71 + (0.0592/2) log 10⁻⁴ = 1.59 V; log K = 2 × 1.59/0.0592 ≈ 54 ⇒ K ≈ 10⁵⁴.

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