The reduction potential of a hydrogen electrode at pH 10 at 298 K is: (p = 1 atm)
Answer & explanation
Correct answer: option 3
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, \(E_{red} = \frac{0.0591}{2} × 2 × [−10] = −0.591 V\)