Chemistry - Electrochemistry Question with Solution | TestHub

Given cell reaction:
${\mathrm{Pt}}_{\left(\mathrm{s}\right)}\left|{\mathrm{H}}_2\left(\mathrm{g},1\mathrm{atm}\right)\right|{\mathrm{H}}^{+}\left(\mathrm{aq},1\mathrm{M}\right)\left|\right|{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right),{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)Pt\left(\mathrm{s}\right)$

at anode(oxidation) ${\mathrm{H}}_2⟶2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

At cathode(reduction) ${\mathrm{Fe}}_{\mathrm{aq}}^{3+}+{\mathrm{e}}^{-}⟶{\mathrm{Fe}}_{\mathrm{aq}}^{2+}$

$\mathrm{E}{°}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{o}} - {\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}}={{\mathrm{E}}^{\mathrm{o}}}_{{\mathrm{H}}_2|{\mathrm{H}}^{+}} +{{\mathrm{E}}^{\mathrm{o}}}_{{\mathrm{Fe}}^{3+}\mid {\mathrm{Fe}}^{2+}}=0.771 \mathrm{V}$

Thus, using Nernst equation,
$\Rightarrow \mathrm{E}=\mathrm{E}°-\frac{0·06}{1}\log \frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}$

$\Rightarrow 0.712=\left(0+0.771\right)-\frac{0.06}{1}\log \frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}$

$\Rightarrow \log \frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}=\frac{0.059}{0.06}\approx 1$

$\boxed{\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}=10}$