For a cell involving two electron transfer, $E^{0} = 0 .59 V$ at 298K. Equilibrium constant for the cell reaction will be

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$10^{5}$

$10^{8}$

$10^{20}$

$10^{10}$

answer is C.

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Detailed Solution

$\begin{array}{l} E_{cell}^{o} = \frac{0.0591}{n} \log k_{eq} \\ 0 .59 = \frac{0.0591}{2} \log k_{eq} \\ \log k_{eq} = 20 \\ k_{eq} = 10^{20} \end{array}$

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