Q66 · CSIR-NET Chemistry, June 2018

Paper: CSIR-NET June 2018 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

The equilibrium constant of the following reaction:
\[ \mathrm{Sn}(\mathrm{s})+\mathrm{Sn}^{4+}(\mathrm{aq}) \Leftrightarrow 2 \mathrm{Sn}^{2+}(\mathrm{aq}) \]
At 300 K is closest to, (Given : $\mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}}^{\mathrm{o}}=0.15 \mathrm{V}$ and $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\mathrm{o}}=-0.15 \mathrm{V}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ )
(a)$10^{6.08}$
(b)$10^{8.08}$
(c)$10^{10.08}$
(d)$10^{12.08}$
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
E° = 0.15 − (−0.15) = 0.30 V and n = 2. log K = nFE°/(2.303RT) = 2 × 0.30/0.05956 = 10.07, so K ≈ 10^10.08 (c).

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