What will be the potential of a hydrogen electrode at 298 K in contact with a solution whose pH is 2 and the pressure of dihydrogen gas is maintained at 1 bar?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −0.118 V
For a hydrogen electrode the reduction half reaction is:
\( \displaystyle \ce{H^{+}(aq) + e^{-}} \rightarrow \frac{1}{2}\ce{H2(g)} \)
Its standard potential is zero by definition, and n = 1. At 298 K the Nernst equation gives:
\( \displaystyle E=E^{\circ}-\frac{0.059}{n}\log\frac{(p_{\mathrm{H_2}})^{1/2}}{[\mathrm{H^+}]} \)
With p\( \ce{H2} \) = 1 bar the numerator is 1, and since pH = −log[H+], the equation reduces to a simple form:
\( \displaystyle E=-0.059\log\frac{1}{[\mathrm{H^+}]}=-0.059\times\mathrm{pH} \)
Here pH = 2:
\( \displaystyle E=-0.059\times2 \)
\( \displaystyle E=-0.118\ \mathrm{V} \)
Shortcut: The potential of a hydrogen electrode at 1 bar is −0.059 × pH. Any positive pH gives a negative potential, which rules out Options (1) and (4), and 0.059 × 2 = 0.118.
Note: NTA picked this question from an NCERT (2026-27) intext question and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Intext Questions – “… 2.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.”