The $$EMF$$  of a cell corresponding to the reaction : $$Z{n_{\left( s \right)}} + 2H_{\left( {aq} \right)}^ + \to $$     $$Zn_{\left( {aq} \right)}^{2 + }\left( {0.1\,M} \right) + {H_{2\left( g \right)}}\left( {1\,atm} \right)$$      is 0.28 volt at $${15^ \circ }C.$$  The $$pH$$  of the solution at the hydrogen electrode is $$\left( {{\text{Given:}}\,E_{\frac{{Z{n^{2 + }}}}{{Zn}}}^ \circ = - 0.76\,{\text{volt;}}\,E_{\frac{{{H^ + }}}{{{H_2}}}}^ \circ = 0\,{\text{volt}}} \right)$$

Solution :
$$\eqalign{ & {\text{The half - cell reactions are}} \cr & Zn \to Z{n^{2 + }} + 2{e^ - } \cr & 2{H^ + } + 2{e^ - } \to {H_2} \cr & {\text{We know that}} \cr & {E_{\frac{{Zn}}{{Z{n^{2 + }}}}}} = E_{\frac{{Zn}}{{Z{n^{2 + }}}}}^ \circ - \frac{{RT}}{{nF}}\ln \frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {Zn} \right]}} \cr} $$
$$\therefore {E_{\frac{{Zn}}{{Z{n^{2 + }}}}}} = $$    $$0.76 - \frac{{2.303 \times 8.314 \times 298}}{{2 \times 96500}}\log \frac{{0.1}}{1}$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.76 - \left( { - 0.03} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.79\,V \cr & {\text{Also,}}\,\,{E_{\frac{{{H^ + }}}{{{H_2}}}}} = E_{\frac{{{H^ + }}}{{{H_2}}}}^ \circ - \frac{{RT}}{{nF}}\ln \frac{{\left[ {{H_2}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}} \cr & = 0 - \frac{{2.303 \times 8.314 \times 298}}{{2 \times 96500}}\log \frac{1}{{{{\left[ {{H^ + }} \right]}^2}}} \cr & = 0.0591\,\log \,\left[ {{H^ + }} \right] \cr & = - 0.0591\,pH \cr & {\text{Since}}\,\,{E_{{\text{cell}}}} = {E_{\frac{{Zn}}{{Z{n^{2 + }}}}}} + {E_{\frac{{{H^ + }}}{{{H_2}}}}} \cr & {\text{or}}\,\,0.28 = 0.79 - 0.0591\,pH \cr & {\text{or}}\,\,pH = \frac{{0.79 - 0.28}}{{0.0591}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.51}}{{0.0591}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8.62 \cr} $$