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` .

Answer

`9.35xx10^(-15)M

Explanation

Consider an electrode of hydrogen `(H_(2))` as
`2H^(o+)+2e^(-) rarr H_(2), " "E^(c-)=0`
Given electrode is
`2H_(2)O+2e^(-) rarr H_(2)+2overset(c-)(O)H," "e^(C-)._(red)=-0.82227V`
`:' E^(c-)._(o x i d)` for `H_(2)OgtE^(c-)._( o x i d)` for `H_(2)`
Therefore, cell reactions are `:`
Anode reaction `:`
`H_(2)+2overset(c-)(O)H rarr 2H_(2)O+2e^(-)," "E^(c-)._(o x i d) =-0.8277V.`
Cathode reaction `:`
`2H^(o+)+2e^(-) rarr H_(2),`
Net reaction
Thus, `K=([H_(2)O]^(2))/([H^(o+)]^(2)[overset(c-)(O)H]^(2))`
For the reaction,

`K_(w)=[H_(3)O^(o+)][overset(c-)(O)H]`
`K=[(1)/(K_(w))] .......(i) `
Also, `E_(cell)=E_(o x i d(H_(2)O ))+E_(red(H))`
`=E^(c-)._(o x i d(H_(2)O))-(0.059)/(2)log.([H_(2)O]^(2))/([H_(2)][overset(c-)(O)H]^(2))+E^(c-)._(red(H^(o+)//H_(2)))+(0.059)/(2)log.([H^(o+)]^(2))/([H_(2)])`
`=0.8277+(0.059)/(2)log.([H^(o+)]^(2).[H_(2)].[overset(c-)(O)H]^(2))/([H_(2)].[H_(2)O]^(2))`
`=0.8277+(0.059)/(2)log.([H^(o+)]^(2)[overset(c-)(O)H]^(2))/([H_(2)O]^(2))`
`=0.8277+(0.059)/(2)log.(1)/(K) ....(ii)`
From `Eqs. (i) ` and `(ii)`,
`E_(cell)=0.8277+(0.059)/(2)log[K_(w)]^(2)`
At equilibrium, `E_(cell)=0`
`:. -0.8277=0.059logK_(w)`
or `log K_(w)=(0.8277)/(0.059)`
or `K_(w)=9.35xx10^(-15)`