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Question
Write the Nernst equation and emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.001 M) || H+ (1 M) | H2(g) (1 bar) | Pt(s)
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Solution
The cell reaction is as follows:
\[\ce{Fe_{(s)} + 2H^+ (1 M) -> Fe^{2+} (0.001 M) + H2 (1 bar)}\]
Hence, n = 2,
The Nernst equation for the emf of this cell will be as follows:
∴ \[\ce{E_{cell} = (E{^{\circ}_{H^{+}/\frac{1}{2}H_{2}}} - E{^{\circ}_{Fe^{2+}/Fe}}) - \frac{0.0591}{2} log_10 \frac{[Fe^{2+}] \times pH_2}{[H^+]^2}}\]
\[\ce{E_{cell} = [0 - (-0.44)] - \frac{0.0591}{2} log_10 \frac{0.001 × 1}{(1)^2}}\]
= \[\ce{0.44 - \frac{0.059}{2} log_10 10^{-3}}\]
= `0.44 - 0.059/2 xx (-3)`
= 0.529 V
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