A hydrogen gas electrode is made by dipping platinum wire in a solution of $$HCl$$  of $$pH=10$$   and by passing hydrogen gas around the platinum wire at $$1$$ $$atm$$  pressure. The oxidation potential of electrode would be

Solution :
For hydrogen electrode, oxidation half reaction is
$$\eqalign{ & \mathop {{H_2}}\limits_{\left( {1\,\,atm} \right)} \to \mathop {2{H^ + }}\limits_{\left( {{\text{At}}\,pH\,10} \right)} + 2{e^ - } \cr & {\text{If}}\,\,pH = 10 \cr & {H^ + } = 1 \times {10^{ - pH}} = 1 \times {10^{ - 10}} \cr & {\text{From Nemst equation,}} \cr & {E_{cell}} = {E^ \circ }_{cell} - \frac{{0.0591}}{2}{\text{log}}\frac{{{{\left[ {{H^ + }} \right]}^2}}}{{{p_{{H_2}}}}} \cr & {\text{For hydrogen electrode,}}\,{E^ \circ }_{cell} = 0 \cr & {E_{cell}} = - \frac{{0.0591}}{2}{\text{log}}\frac{{{{\left( {{{10}^{ - 10}}} \right)}^2}}}{1} \cr & = + \frac{{0.0591 \times 2}}{2}{\text{log}}\frac{1}{{{{10}^{ - 10}}}} \cr & = 0.0591\,{\text{log}}\,{10^{10}} \cr & = 0.0591 \times 10 \cr & = 0.591\,V \cr} $$