At `25^(@)C` , the free energy of formation of `H_(2)O(l)` is `-56,700 Cal mol^(-1)`. The free energy of ionization of water to `H^(o+)` and `overset(c-)(O)H` is `19050 cal mol^(-1)`. What is the reversible `EMF` of the following cell at `25^(@)C:`
`H_(2)(g)(1 atm)|H^(o+)||overset(c-)(O)H,O_(2)(g),1atm`

Explanation

Given `:`
`H_(2)(g)+(1)/(2)O_(2)(g)hArrH_(2)O(l)" "(i)`
`H_(2)OhArrH^(o+)+overset(c-)(O)H" "(ii)`
`DeltaG_(1)=-56,700 cal mol^(-1),DeltaG_(2)=19050 cal mol^(-1)`
Cell reaction
`H_(2) hArr 2h^(o+)+cancel(2e^(-))`
`(1)/(2)O_(2)+H_(2)O+2e^(-) hArr 2overset(c-)(O)H`
`ulbar(H_(2)+(1)/(2) O_(2)+H_(2)OhArr2H^(o+)+2overset(c-)(O)H)`
Multiply Eq. `(ii)` by 2 and adding Eq. `(i)`, we get
`H_(2)+(1)/(2)O_(2)+H_(2)O hArr 2H^(o+)+2overset(c-)(O)H`
`:. Delta G=Delta G_(2)+2DeltaG_(2)`
`=-56700+2xx19050`
`=-18600 cal mol ^(-1)`
`=-18600xx4.2 J mol^(-1)(1cal =4.184J~~4.2J)`
`=-78120J mol^(-1)`
`DeltaG=-nFE_(cell)`
` -78120 J mol^(-1)=-2xx96500CxxE_(cell)`
`:. E_(cell)=0.405V`