Chemistry - Electrochemistry Question with Solution | TestHub

$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2(\mathrm{g})$ $\mathrm{E}={\mathrm{E}}^{\circ }-0.059\log \left(\frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}\right)$ (here $\mathrm{E}$ is -ve when ${\mathrm{P}}_{{\mathrm{H}}_2}>{\left[{\mathrm{H}}^{+}\right]}^2$ ) $=\frac{-0.0591}{2}{\log }_{10}\left(\frac{2}{1}\right)=\frac{-.0591}{2}\times .3010=$ negative value