Q16 · CSIR-NET Chemistry, December 2018

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

The equilibrium constant of the following reaction: $\mathrm{Sn(s)+Sn^{4+}(aq) \rightleftharpoons 2Sn^{2+}(aq)}$ At $300\,K$ is closest to, (Given, $E^{o}_{\mathrm{Sn^{4+}/Sn^{2+}}}=0.15\,V$ and, $E^{o}_{\mathrm{Sn^{2+}/Sn}}=-0.15\,V$, $R=8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$)
(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°cell = E°(Sn⁴⁺/Sn²⁺) − E°(Sn²⁺/Sn) = 0.15 − (−0.15) = 0.30 V, n = 2.
At 300 K, 2.303RT/F = 2.303 × 8.314 × 300 / 96485 = 0.0595 V.
log K = nE°/0.0595 = 2 × 0.30 / 0.0595 ≈ 10.08 ⇒ K ≈ 10^10.08 (c).

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