Q107 · CSIR-NET Chemistry, June 2011

Paper: CSIR-NET June 2011 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

If the ratio of composition of oxidized and reduced species in electrochemical cell, is given as $\frac{[0]}{[R]}=e^{2}$ the correct potential difference will be

(a)$E-E^{o}=\frac{2 R T}{n F}$

(b)$E-E^{O}=-\frac{2 R T}{n F}$

(c)$E-E^{o}=\frac{R T}{n F}$

(d)$E-E^{O}=-\frac{R T}{n F}$

Answer

Answer: A ✓ checked by 4AB · confidence high

Explanation

Nernst: E − E° = (RT/nF) ln([O]/[R]) = (RT/nF) ln e² = 2RT/nF.

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