Calculate the emf of the half cell given below: \ce{Pt(s) | H2(g, 2 atm) | HCl(aq, 0.02 M)} $E^\circ_{\ce{H+/H2}} = 0\ \text{V}$ (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

Solution & Explanation

\textbf{Answer:} (D) $0.109\ \text{V}$ \textbf{Solution:} For the hydrogen electrode the Nernst equation (NEET official key convention) is written with reaction quotient $Q = \dfrac{[\ce{H+}]^2}{P_{\ce{H2}}}$ and $n = 2$: $E = E^\circ - \dfrac{0.059}{2}\log\dfrac{[\ce{H+}]^2}{P_{\ce{H2}}}$ With $[\ce{H+}] = 0.02\ \text{M}$ (from \ce{HCl}) and $P_{\ce{H2}} = 2\ \text{atm}$: $E = 0 - 0.0295\log\dfrac{(0.02)^2}{2} = -0.0295\log\dfrac{4\times10^{-4}}{2} = -0.0295\log(2\times10^{-4})$ $E = -0.0295\,(\log 2 - 4) = -0.0295\,(0.301 - 4) = -0.0295\times(-3.699)$ $E = +0.109\ \text{V}$, which is option (D).