Calculate the emf of the following cell at `25^@C`
`H_2(g) (1atm)//H^+ (aq)|| OH^(-) (aq)//O_2(g) (1atm)` from the following data:
`H_2+1/2O_2=H_2O,DeltaG^@=-226.8kJ//mol e`

Explanation

The electrodic process are,
`1/2 O_2+H_2O +2e=2OH^-` ……….at cathode (RHS)
`H_2=2H^+ +2e`…………at anode (LHS)
The cell reaction is
`1/2 O_2+ H_2O+ H_2=2H^+ + 2OH^(-)`
for which,
`E_(cell)=E_(cell)^@-.0591/2log[H^+]^2[OH^-]^2`
`E_(cell)=E_(cell)^@-.05921/2 log (K_w)^2`
`E_(cell)=E_(cell)^@+0.8274 (K_w=10^-14)`
Now multiply the given eqn. (ii) by 2 and add eqn. (i) we get the cell reaction
`H_2+ 1/2O_2+H_2O=2H^+ +2OH^(-), Delta G^@=-226.8+2 times 76.8`
`=-74.4 kJ`
`=-74400J`
Again we have,
`Delta G^@=-2.303RT logk`
`-74400=-2.303 times 8.314 times 298 times log k`
`log k=13.04`
`therefore E^@=(2.303 RT)/(n F) log k`
`=.591/2times 13.04`
`=0.3853`
Substituting the `E^@` values in eqn. (1)
`E_(cell)= 0.3853+0.8274`
`=1.2128V`