Chemistry - Electrochemistry Question with Solution | TestHub

The cell reaction is
${\mathrm{H}}_{2\left(\mathrm{g}\right)}+{\mathrm{I}}_{2\left(\mathrm{s}\right)}\rightleftharpoons 2{\mathrm{H}}_{\left(\mathrm{a}\mathrm{q}\right)}^{+}+2{\mathrm{I}}_{\left(\mathrm{a}\mathrm{q}\right)}^{-}$
Here$\mathrm{n}=2$(n = number of electons involved )
$\mathrm{E}=\mathrm{E}°-\frac{0.0591}{\mathrm{n}}\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[{\mathrm{H}}^{+}{]}^2\right[{\mathrm{I}}^{-}{]}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}}$
$0.7714=0.535-\frac{0.0591}{2}\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[{\mathrm{H}}^{+}{]}^2\right[0.1{]}^2}{1}$
On solving$\left({\mathrm{H}}^{+}\right)=10^{-3}$
$\mathrm{p}\mathrm{H}=-\mathrm{l}\mathrm{o}\mathrm{g}\left({\mathrm{H}}^{+}\right)=-\mathrm{l}\mathrm{o}\mathrm{g}\left(10^{-3}\right)$
So$\mathrm{p}\mathrm{H}=3$