If the standard electrode potential for a cell is $2 \mathrm{~V}$ at 300 $\mathrm{K}$, the equilibrium constant $(\mathrm{K})$ for the reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ at $300 \mathrm{~K}$ is approximately $\left(\mathrm{R}=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~F}=96000 \mathrm{C} \mathrm{mol}^{-1}\right)$
Correct Option: , 4
We know that, $\ln K=\frac{\mathrm{nFE}_{\text {cell }}^{\circ}}{\mathrm{RT}}$
After putting the given values, we get
$\ln K=\frac{2 \times 96000 \times 2}{8 \times 300}=160$
$\therefore \quad K=e^{160}$
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