Question

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

Sn_(s) | Sn²⁺ (0.050 M) || H⁺ (0.020 M) | H_2(g) (1 bar) | Pt_(s)

Answer 1

The electrode reactions are:

At anode: \[\ce{Sn_{(s)} -> Sn^{2+} (0.05 M) + 2e-}\]

At cathode: \[\ce{2H+ (0.02 M) + 2e- -> H2 (1 bar)}\]

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

Hence, n = 2,

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

∴ E_cell = \[\ce{(E{^{\circ}_{H^{+}/\frac{1}{2}H_{2}}} - E{^{\circ}_{Sn^{2+}/Sn}}) - \frac{0.0591}{2} log_10 \frac{[Sn^{2+}] × P_{H_2}}{[H^+]^2}}\]

= \[\ce{(0 - (-0.14)) - 0.0296  log_10 \frac{0.050 × 1}{(0.020)^2}}\]

= 0.14 − 0.0296 × log₁₀ 125

= 0.14 − 0.0296 × 2.0969

= 0.14 − 0.062

= 0.078 V