At $$298K$$  the standard free energy of formation of $${H_2}O\left( l \right)$$  is $$ - 237.20\,kJ/mol$$     while that of its ionisation into $${H^ + }\,ion$$   and hydroxyl ions is $$80\,kJ/mol,$$   then the emf of the following cell at $$298\,K$$  will be
[ Take Faraday constant $$F = 96500\,C$$   ]
$${H_2}\left( {g,1\,bar} \right)\left| {{H^ + }\left( {1M} \right)} \right|\left| {O{H^ - }\left( {1M} \right)} \right|{O_2}\left( {g,1\,bar} \right)$$

Solution :
$$\eqalign{ & {\text{Cell reaction}} \cr & {\text{cathode :}}{H_2}O\left( l \right) + \frac{1}{2}{O_2}\left( g \right) + 2{e^ - } \to 2O{H^ - }\left( {aq} \right) \cr & \underline {{\text{anode :}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_2}\left( g \right) \to 2{H^ + }\left( {aq} \right) + 2{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr & {H_2}O\left( l \right) + \frac{1}{2}{O_2}\left( g \right) + {H_2}\left( g \right) \to 2{H^ + }\left( {aq} \right) + 2O{H^ - }\left( {aq} \right) \cr & {\text{Also we have}} \cr & {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right) \cr & \Delta G_f^ \circ = - 237.2\,kJ/mol \cr & {H_2}O\left( l \right) \to {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right); \cr & \Delta {G^ \circ } = 80\,kJ/mol \cr & {\text{Hence for cell reaction}} \cr & \Delta {G^ \circ } = - 237.2 + \left( {2 \times 80} \right) \cr & = - 77.20\,kJ/mol \cr & \therefore \,\,{E^ \circ } = - \frac{{\Delta {G^ \circ }}}{{nF}} = - \frac{{77200}}{{2 \times 96500}} = - 0.40\,V \cr} $$