Chemistry - Electrochemistry Question with Solution | TestHub

${\text{Fe}}^{2+}+{\text{Ce}}^{4+}\rightleftharpoons {\text{Fe}}^{3+}+{\text{Ce}}^{3+}$

${{\text{E}}^{\text{o}}=\text{E}}^{\text{o}}\left({\text{Ce}}^{4+}/{\text{Ce}}^{3+}\right){ -\; \text{E}}^{\text{o}}\left({\text{Fe}}^{3+}/{\text{Fe}}^{2+}\right)$

$= \text{1.44}-\text{0.68} = \text{0.76 V}$

$\because {\text{E}}^{\text{o}} = \text{0.059}\mathit{\text{ log }}\text{K}$

$\Rightarrow \mathit{\text{ log }}\text{K} = \frac{{\text{E}}^{\text{o}}}{\text{0.0592}}$

$= \frac{\text{0.76}}{\text{0.059}}=\text{12.85}$

$\Rightarrow \; \text{K} = \text{7.07}\times {\text{10}}^{12}$