The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25⁰C. The equilibrium constant of the reaction
would be $(F = 96500 C {mol}^{- 1}; R = 8 .314 {JK}^{- 1} {mol}^{- 1})$

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1.0 × 10¹⁰

2.0 × 10¹¹

4.0 × 10¹²

1.0 × 10²

answer is A.

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Detailed Solution

$\log K e q = \frac{n F E_{c e l l}^{o}}{2.303 R T} = \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298} l o g K_{e q} = 9.97 K_{e q} = 10^{9.97} \approx 1 \times 10^{10}$

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