The standard reduction potential at `25^@C` of the reaction `2H_2O+2e^(-) leftrightarrow H_2 +2OH^(-)` is `-0.8277` volt. Calculate the equilibrium constant for the reaction `2H_2O=H_3O^+ +OH^(-)` at `25^@C`

Explanation

Let the equilibrium constants for the following equations be as follows:
`H_2O+H_2O leftrightarrow H_3O^+ +OH^(-) K_1` .......(1)
`H_3O^+ +e leftrightarrow 1/2H_2+H_2O K_2`......(2)
Adding them we get
`H_2O+ e leftrightarrow 1/2H_2 +OH^(-) K_3`........(3)
We have now to calculate `K_1`
Now for eqn(2) , `E_(H_3O^+)^@=0`
`therefore K_2=1(because E^@=0.0591/1log K)`......(Eqn 2)
For eqn. (2) `E^@=-0.8277` volt (given)
`therefore E^@=0.0591/1 log K_3=-0.8277`
`logK_3=-0.8277/0.0591, K_3=1 times 10^-14`
Further as eqn (3) is the sum of eqns. (1) and (2) we have
`K_3=K_1.K_2=K_1`
`therefore K_1=1 times 10^-14`