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 rarr H_(2), E_(RP_(H))^(@) = 0`
Given electrode is
`2H_(2)O + 2e rarr H_(2) + 2OH^(-), E_(RP)^(@) = -0.8277 V`
`:'` for `H_(2)O gt E_(OP)^(@)` for `H. :.` The cell reactions are:
Anode: `H_(2) + 2OH^(-) rarr 2H_(2)O + 2e , E_(OP)^(@) = +0.8277 V`
Cathode: `2H^(+) + 2e rarr H_(2) , E_(RP)^(@) = 0`
`:.` Net reaction is `2H^(+) + 2OH^(-) hArr 2H_(2)O`
and `K = ([H_(2)O]^(2))/([H^(+)]^(2)[OH^(-)]^(2))`
Thus, for `2H_(2)O hArr [H_(3)O^(+)][OH^(-)]`
`K_(w) = [H_(3)O^(+)][OH^(-)]`
`:. K = [(1)/(K_(w))]^(2)`
Also `E_(cell) = E_(cell) = E_(OP_(H_(2)O)) + E_(RP_(H))`
`E_(cell) = E_(OP_(H_(2)O))^(@) - (0.059)/(2)log_(10).[H_(2)O]^(2)/([P_(H_(2))][OH^(-)]^(2)]+`
`E_(RP_(H^(+)//H))^(@)+(0.059)/(2)log_(10).[H^(+)]^(2)/(P_(H_(2)))`
`E_(cell) = 0.8277 + (0.059)/(2)log_(10).([H^(+)]^(2).P_(H_(2)).[OH^(-)]^(2))/(P_(H_(2)).[H_(2)O]^(2))`
`E_(cell) = 0.8277 + (0.059)/(2)log_(10).([H^(+)]^(2).[OH^(-)]^(2))/([H_(2)O]^(2))`
` = 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`
`:. -0.8277 = 0.059log_(10) K_(w)`
or `log_(10)K_(w) = -(0.8277)/(0.059)`
or `K_(w) = 9.35 xx 10^(-15)`