Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{P}\mathrm{t},\frac{{\mathrm{H}}_{2\left(\mathrm{g}\right)}}{{\mathrm{H}}^{+}}$Hydrogen electrode${\mathrm{E}}_{{\mathrm{H}}_2/{\mathrm{H}}^{+}}^{\mathrm{o}}=0.0\mathrm{ }\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{t}$

${\mathrm{H}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_{\left(\mathrm{a}\mathrm{q}\right)}^{+}+2{\mathrm{e}}^{-}$

$\left[{\mathrm{H}}^{+}\right]={10}^{-7}\mathrm{ }\mathrm{M}$at${25}^{\mathrm{o}}\mathrm{C}$(for Pure water)

$\mathrm{E}=\frac{-0.0591}{2}\log \left(\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}}\right)$

$\mathrm{E}=0=\log \frac{\left[{\mathrm{H}}^{+}\right]}{{\mathrm{P}}_{{\mathrm{H}}_2}}=0$

$\therefore {\left[{\text{H}}^{+}\right]}^2={\text{P}}_{{\text{H}}_2}={\left({10}^{-7}\right)}^2$

$\therefore {\mathrm{P}}_{{\mathrm{H}}_2}={10}^{-14}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$