Calculate the standard reduction potential for the reaction `H_2O+ e to 1/2 H_2+OH^-` using the Nernst equation and the fact that the standard reduction potential for the reaction `H^+ + e to 1/2 H_2` is by definition equal to 0.00V at `25^@C`

Explanation

Given that
`H_2O + e to 1/2 H_2 +OH^- , E^@=xV` (say)
`therefore` for the reaction,
`H_2O to H^+ +OH^- , E^@=xV`
The Nernst equation is
`E_(cell)=E^@- 0.0591/1 log[H^+][OH^-]`
At equilibrium `E_(cell)=0 and [H^+][OH^-]=K_w=1 times 10^-14`
`therefore 0=x-0.0591/1 log 10^-14`
or `x=-0.8274 V`