The oxidation potential of a hydrogen electrode at pH = 1 is (T = 298 K):
Answer & explanation
Correct answer: option 1
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, \(E = \frac{0.0591}{2} × 2 = 0.0591 V\)