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For a cell involving one electron transfer $E_{c e l l}^{0} =' X'$ at 298K. The equilibrium constant for the cell reaction is 10¹⁰. ‘X’ is

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0.0295 V

0.295 V

0.0591 V

0.591 V

answer is D.

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Detailed Solution

$\begin{array}{l} E_{c e l l}^{0} = \frac{0.059}{n} \log K c \\ 0.59 = 0.059 \log K c \\ \log K c = 10 \\ K c = 10^{10} \end{array}$

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