2026 ElectrochemistryNernst equation for a hydrogen electrode Hard Question 1 of 74 in Electrochemistry

Calculate emf of the half cell given below: Pt (s) | $H_2$ (g, 2 atm) | HCl (aq, 0.02 M), $E^{\circ}_{H_2/H^+} = 0$ V (Given: $\frac{2.303\,RT}{F} = 0.059$, $\log 2 = 0.3010$)

A 0.109 V
B 0.035 V
C $-0.035$ V
D $-0.109$ V

The answer is A.

Solution

Cell reaction: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$ ; $P_{H_2}$ = 2 atm, $[H^+] = 2 \times 10^{-2}$ M $E_{OP} = 0 - \frac{0.059}{2} \log \frac{(2 \times 10^{-2})^2}{2}$ $= -\frac{0.059}{2}[\log(2 \times 10^{-4})]$ $= -\frac{0.059}{2}[0.3 - 4]$ $= -\frac{0.059}{2}[-3.7]$ $= +0.109$ V

Electrochemistry in past papers

74 questions from this chapter have appeared across 27 exam years.