Question

Write the Nernst equation and emf of the following cell at 298 K:

\[\ce{Sn_{(s)} | Sn^{2+} (0.050 M) || H^+ (0.020 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]

Answer 1

The cell reaction is as follows:

\[\ce{Sn_{(s)} + 2H^+ (0.020 M) -> Sn^{2+} (0.050 M) + H2 (1 bar)}\]

Hence, n = 2,

According to this, the Nernst equation will be as follows:

∴ \[\ce{E_{cell} = (E{^{\circ}_{H^{+}/\frac{1}{2}H_{2}}} - E{^{\circ}_{Sn^{2+}/Sn}}) - \frac{0.0591}{2} log_10 \frac{[Sn^{2+}] × pH_2}{[H^+]^2}}\]

∴ \[\ce{E_{cell} = [0 - (-0.14)] - \frac{0.0591}{2} log_10 \frac{0.050 × 1}{(0.020)^2}}\]

= \[\ce{0.14 - \frac{0.059}{2} log_10 125}\]

= 0.078 V