The standard cell potential for the reaction Zn(s)+Cu²⁺(aq) ⇔ Zn²⁺(aq)+Cu(s) is +1.10 V. The Gibbs free energy change during the reaction is (F=96500. coulomb.mol⁻¹)

(c) -212 kJ mol⁻¹. Explanation: ΔG° = −nFE° = −2 × 96500 × 1.10 = −212,300 J = −212 kJ mol⁻¹.

Q59 · CSIR-NET Chemistry, June 2017

Paper: CSIR-NET June 2017 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

The standard cell potential for the reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \Leftrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ is +1.10 V . The Gibbs free energy change during the reaction is $\left(\mathrm{F}=96500\right.$ coulomb $\left.\mathrm{mol}^{-1}\right)$

(a)$-21.2 \mathrm{kJ} \mathrm{mol}^{-1}$

(b)$+212 \mathrm{kJ} \mathrm{mol}^{-1}$

(c)$-212 \mathrm{kJ} \mathrm{mol}^{-1}$

(d)$-212 \mathrm{J} \mathrm{mol}^{-1}$

Answer

Answer: C ✓ checked by 4AB · confidence high

Explanation

ΔG° = −nFE° = −2 × 96500 × 1.10 = −212,300 J = −212 kJ mol⁻¹.

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