The oxidation potential of a hydrogen electrode at pH = 1 is (T = 298 K):

0.059 volt

The hydrogen electrode can be represented by the given reaction

\(H_2 \rightarrow 2H^+ + 2e^−\)

Given, \(pH  =1\)

also, \([H^+] = 10^{−1}\)

 and for hydrogen electrode, \(E^o = 0\)

∴ \(E = \frac{0.0591}{2}log\frac{[H_2]}{[H^+]^2}\)

or, \(E =\frac{0.0591}{2}log\left[\frac{1}{10^{−1}}\right]^2\)

or, \(E = \frac{0.0591}{2} × 2 = 0.0591 V\)