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The equilibrium constant of the given reaction at 298 K will be
$2 F e^{3 +} (a q .) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q .) + I_{2} (s)$ , $E_{c e l l}^{o} = 0.237 V$

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$10^{12.25}$

$10^{8.02}$

$10^{9.76}$

$10^{6.8}$

answer is B.

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Detailed Solution

$E_{c e l l}^{0} = \frac{2.303 R T}{n F} \log K_{C}$

$E_{c e l l}^{0} = \frac{0.059}{2} \log K_{C}$

$\frac{0.237 \times 2}{0.059} = \log K_{C}$

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