Chemistry - Electrochemistry Question with Solution | TestHub

At anode :${\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{H}}^{+}\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{e}}^{-}$
At cathode :${\mathrm{M}}^{4+}\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{M}}^{2+}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }$
Net cell reaction :${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{M}}^{4+}\left(\mathrm{a}\mathrm{q}\right)\rightleftharpoons 2{\mathrm{H}}^{+}\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{M}}^{2+}\left(\mathrm{a}\mathrm{q}\right)$
$E_{cell}=E_{cell}^0-\frac{0.059}{2}\log \frac{[M^{2+}]{[H^{+}]}^2}{[M^{4+}](P_{H_2})}$
Now,${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}=\left({\mathrm{E}}_{{\mathrm{M}}^{4+}/{\mathrm{M}}^{2+}}^0-{\mathrm{E}}_{{\mathrm{H}}^{+}/{\mathrm{H}}_2}^0\right)-\frac{0.059}{\mathrm{n}}.\log \frac{{\left[{\mathrm{H}}^{+}\right]}^2\mathrm{ }\left[{\mathrm{M}}^{2+}\right]}{{\mathrm{P}}_{{\mathrm{H}}_2}.\left[{\mathrm{M}}^{4+}\right]}$
$0.092=\left(0.151-0\right)-\frac{0.059}{2}.\log \frac{1^2\times \left[{\mathrm{M}}^{2+}\right]}{1\times \left[{\mathrm{M}}^{4+}\right]}$
$\therefore \frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}={10}^2\Rightarrow \mathrm{x}=2$