Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{\Delta G}=\mathrm{\Delta G}°-\frac{0.591}{\mathrm{n}}\mathrm{logQ}$

at equilibrium

$\mathrm{\Delta G}=0,{\mathrm{E}}_{\mathrm{cell}}=0,\mathrm{Q}={\mathrm{K}}_{\mathrm{C}},\mathrm{\Delta G}°=-\frac{0.591}{1}\mathrm{logQ}$

${\text{E}}_{\text{ cell}}\text{=}{\text{E}}_{\text{ cell}}^{\text{o}}-\frac{\text{0}\text{.0591}}{\text{n}}{\text{log K}}_{\text{c}}$

$0 = \text{0}\text{.591}-\frac{\text{0}\text{.0591}}{1} {\text{log K}}_{\text{c}}$

$\Rightarrow -\text{0}\text{.591} = -\text{0}\text{.0591} {\text{log K}}_{\text{c}}$

$\Rightarrow \; {\text{log K}}_{\text{c}} = \frac{\text{0.591}}{\text{0.0591}} = 10$

$\therefore \; {\text{K}}_{\text{c}} = \text{antilog 10} = 1\times {\text{10}}^{\text{10}}$.