Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{Z}\mathrm{n}+\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{Z}{\mathrm{n}}^{2+}+\mathrm{F}\mathrm{e}(\mathrm{n}=2)$
$\text{E}={\text{E}}^{\text{o}}-\frac{0.0591}{\text{n}}\text{log}\text{Q}$
$0.02905={\mathrm{E}}^{\mathrm{o}}-\frac{0.0591}{2}\mathrm{l}\mathrm{o}\mathrm{g}\frac{0.01}{0.001}$
${\text{E}}^{\text{o}}=0.2905+0.0295=0.32\text{volt}$
${\text{E}}^{\text{o}}=\frac{0.0591}{\text{n}}\text{log}{\text{K}}_{\text{eq}}$
$0.32=\frac{0.0591}{2}\text{log}{\text{K}}_{\text{eq}}=0.02945\text{log}{\text{K}}_{\text{eq}}$
${\text{K}}_{\text{eq}}={10}^{0.32/0.0295}$