Question

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

Fe_(s) | Fe²⁺ (0.001 M) || H⁺ (1 M) | H_2(g) (1 bar) | Pt_(s)

Answer 1

The electrode reactions are:

At anode: \[\ce{Fe_{(s)} -> Fe^{2+} (0.001 M) + 2e-}\]

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

Net reaction: \[\ce{Fe_{(s)} + 2H^+ (1 M) -> Fe^{2+} (0.001 M) + H2 (1 bar)}\]

Hence, n = 2,

The Nernst equation for the emf of this cell will be as follows:

∴ \[\ce{E_{cell} = (E{^{\circ}_{H^{+}/\frac{1}{2}H_{2}}} - E{^{\circ}_{Fe^{2+}/Fe}}) - \frac{0.0591}{2} log_10 \frac{[Fe^{2+}] \times P_{H_2}}{[H^+]^2}}\]

\[\ce{E_{cell} = [0 - (-0.44)] - \frac{0.0591}{2} log_10 \frac{0.001 × 1}{(1)^2}}\]

= \[\ce{0.44 - \frac{0.059}{2} log_10 10^{-3}}\]

= 0.44 − 0.0296 × (−3)

= 0.529 V