The reduction potential of a hydrogen electrode at pH 10 at 298 K is: (p = 1 atm)

–0.591 volt

The correct answer is option 3. –0.591 volt.

The  hydrogen electrode can be represented as

\(2H^+  +  2e− \rightarrow H_2\)

Given,

\(pH = 10\)

\([H^+] = 10^{−10}\)

For hydrogen electrode, \(E^o = 0\)

∴ \(E_{red} = \frac{0.0591}{2}log\frac{[H^+]^2}{P_{H_2}(g)}\)

or, \(E_{red} = \frac{0.0591}{2} × 2 log[10^{−10}]\)

or, \(E_{red} = \frac{0.0591}{2} × 2 × [−10] = −0.591 V\)