Chemistry - Electrochemistry Question with Solution | TestHub

$\text{2Fe}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+\text{4}{\text{H}}^{+}\left(\text{aq}\right)\to \text{2F}{\text{e}}^{2+}\left(\text{aq}\right)+\text{2}{\text{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{l}\right)$

$\mathrm{N}=4$(no. of moles of electrons involved)

From Nernst equation,

${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}={\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \mathrm{Q}$

$=1.67-\frac{0.0591}{4}\mathrm{log log}\frac{{\left(10^{-3}\right)}^2}{0.1\times {\left(10^{-3}\right)}^4}\left[\left({\mathrm{H}}^{+}\right)=10^{-\mathrm{p}\mathrm{H}}\right]$

$=1.\text{67}-0.106=1.\text{57 V}$