Question

At what $pH$, oxidation potential of water is $-0.81V$?

Answer 1

The half reaction for the oxidation of water is $2H_{2}O\to O_2+4H^{+}+4e^{-}$.

The standard oxidation potential is $-1.23$ V.

The Nernst equation for the electrode potential is, $E_{cell}=E_{cell}^0-\frac{0.059}{n}logQ$.

Substitute values in the above expression:

$-0.81=-1.23-\frac{0.059}{4}log[H^{+}{]}^4$.

Hence, $\left[H^{+}\right]=1\times {10}^{-7}M$.

The corresponding pH is $-log\left[H^{+}\right]=-log(1\times {10}^{-7})=7$.