The standard reduction potential at `25^(@)C` of the reaction
`2H_(2)O+2e^(-)hArrH_(2)+2overset(Θ)(O)H` is `-0.8277V`. Calculate the equilibrium constant for the reaction.
`2H_(2)OhArrH_(3)O^(o+)+overset(Θ)(O)H` at `25^(@)C` .

Explanation

Consider an electrode of H as
`2H^(+)+2e^(-)rarrH_(2),E_(RP_(H))^(@)=0`
Given electrode is
`2H_(2)O+2erarrH_(2)+2OH^(-),E_(RP)^(@)=0.8277V`
`becauseE_(OP)^(@)" for "H_(2)OgtE_(OP)^(@)" for "H`.
`therefore` the cell reactions are:
Anode : `H_(2)+2OH^(-)rarr2H_(2)O+2e, E_(OP)^(@)=+0.8277V`
Cathode: `2H^(+)+2erarrH_(2),E_(RP)^(@)=0`
`therefore`Net reaction is `2H^(+)+2OH^(-)hArr2H_(2)O`
and `K=([H_(2)O]^(2))/([H^(+)][OH^(-)]^(2))`
Thus, for `2H_(2)OhArr[H_(3)O^(+)][OH^(-)]`
`K_(w)=[H_(3)O^(+)][OH^(-)]`
`thereforeK=[(1)/(K_(w))]^(2)`....(1)
Also `E_(cell)=E_(OP_(H_(2)O))+E_(RP_(H))`
`=E_(OP_(H_(2)O))-(0.059)/(2)log_(10).([H_(2)O]^(2))/([P_(H_(2))][OH^(-)])+E_(RP_(H^(+)//H))^(0)+(0.059)/(2)log_(10).([H^(+)]^(2))/(P_(H_(2)))`
`E_(cell)=0.8277+(0.059)/(2)log_(10).([H^(+)].P_(H_(2)).[OH^(-)])/(P_(H_(2)).[H_(2)O])`
`E_(cell)=0.8277+(0.059)/(2)log_(10).([H^(+)]^(2)[OH^(-)]^(2))/([H_(2)O])`
`=0.8277+(0.059)/(2)log_(10).(1)/(K)`
`E_(cell)=0.8277+(0.059)/(2)log_(10)[K_(w)]^(2)" by eq. (1)"`
At equilibrium, `E_(cell)=0`
`therefore-0.8277=0.059log_(10)K_(w)`
or `log_(10)K_(w)=-(0.8277)/(0.059)`
or `K_(w)=9.35xx10^(-15)`