Q30 · CSIR-NET Chemistry, December 2012
Paper: CSIR-NET December 2012 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium
The equilibrium constant for an electrochemical reaction, \[ 2\,\mathrm{Fe}^{3+}+\mathrm{Sn}^{2+} \rightleftharpoons 2\,\mathrm{Fe}^{2+}+\mathrm{Sn}^{4+} \]
is $\left[E^{o}_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.75\,V,\ E^{o}_{\mathrm{Sn^{4+}/Sn^{2+}}}=0.15\,V;\ \dfrac{2.303RT}{F}=0.06\,V\right]$ (a)$10^{10}$
(b)$10^{20}$
(c)$10^{30}$
(d)$10^{40}$
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
E°cell = 0.75 − 0.15 = 0.60 V, n = 2: log K = nE°/0.06 = 2 × 0.60/0.06 = 20 ⇒ K = 10²⁰.
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