At `25^(@)C, DeltaH_(f)(H_(2)O,l) =- 56700 J//mol` and energy of ionization of `H_(2)O(l) = 19050J//mol`. What will be the reversible `EMF` at `25^(@)C` of the cell,
`Pt|H_(2)(g) (1atm) |H^(+) || OH^(-) |O_(2)(g) (1atm)|Pt`, if at `26^(@)C` the emf increase by `0.001158V`.

Answer

`0.4414V`

Explanation

`H_(2) +(1)/(2)O_(2) rarr H_(2)O DeltaH_(1) =- 56700 J//"mole"` ...(i)
`2H_(2) O rarr 2H^(+) +2OH^(-) DeltaH_(2) = 2 xx 19050` ..(ii)
cell reaction:
at anode: `H_(2) rarr 2H^(+) +2e^(-)`
at cathode: `2e^(-) +H_(2)O + (1)/(2)O_(2) rarr 2OH^(-)`
net reaction: `H_(2) +H_(2)O +(1)/(2)O_(2) rarr 2H^(+) +2OH` ...(iii)
equation (i) +(ii) =(iii)
`DeltaG_(1) +DeltaG_(2) = DeltaG_(3)`
`DeltaH_(1) - T DeltaS_(1) +DeltaH_(2) - T DeltaS_(2)=- nFE`
`(DeltaH_(1) +DeltaH_(2)) - T (DeltaS_(1) +DeltaS_(2)) =- 2FE`
`(DeltaH_(1) +DeltaH_(2)) - T (DeltaS_("net")) =- 2FE`
`DeltaS = nF (dE)/(dT)` so `E = - ((DeltaH_(1)+DeltaH_(2)))/(2F) +(TDE)/(dT)`
`E =- ((-57700+2 xx 19050))/(2 xx 96500) +298 (0.001158)`
`rArr E = 0.09637 +0.345`
`= 0.4414 V`