Chemistry - Electrochemistry Question with Solution | TestHub

Reaction $2{\mathrm{H}}^{+}+2\mathrm{e}\to {\mathrm{H}}_2$ $\therefore \mathrm{n}=2,\mathrm{Q}=\frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}$ As ${\mathrm{E}}_{{\mathrm{H}}^{+}/{\mathrm{H}}_2}={\mathrm{E}}_{{\mathrm{H}}^{+}/{\mathrm{H}}_2}^0\frac{-0.059}{\mathrm{n}}\log \mathrm{Q}=-\frac{0.059}{2}\log \mathrm{Q}$ $\therefore$ if $\mathrm{Q}>1{\mathrm{E}}_{{\mathrm{H}}^{+}/{\mathrm{H}}_2}<0$ (A) $Q=\frac{1}{1}=1X$ (B) $Q=\frac{1}{2^2}<1X$ (C) $Q=\frac{2}{12}>1$
(D) $Q=\frac{2}{2^2}<1X$