Question

Calculate the emf of the following cell at 298 K:

\(Fe_{(s)} |Fe^{2+}(0.01 M)||H^{+} \ _{(1M)}|H_{2(g)}(1 \ bar), Pt_{(s)}\)

Given  \(E^{\circ} _{cell} = 0.44 \ V\).

Answer 1

\(Fe_{(s)} |Fe^{2+} (0.01 M)| |H^+ \ _{(1M)}| H_{2(g)} \text{(1 bar)}, Pt _{(s)}\) 

\(E^\circ_{cell} = 0.44 \ V\)

\(Fe(s) + 2H^+(aq)\longrightarrow Fe^{2+} (aq) + H_2(g)\) 

\(E_{cell = E^\circ _{cell}} - \frac{RT}{nF} ln \frac{[Fe^{2+}]P_{H_2}}{[H^{+}]^2}\) 

\(= 0.44 - \frac{2.303 \times 8.314 \times 298}{2 \times 96485} \log \frac{(10^{-2})(1)}{1^2}\)

= 0.44 – 0.03(–2) log10

= 0.44 + 0.06

= 0.50 V