Question

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

Mg_(s) | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu_(s)

Answer 1

The electrode reactions are:

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

At cathode: \[\ce{Cu^{2+} (0.0001 M) + 2e- -> Cu_{(s)}}\]

Net reaction: \[\ce{Mg_{(s)} + Cu^{2+} (0.0001 M) -> Mg^{2+} (0.001 M) + Cu_{(s)}}\]

Hence, n = 2,

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

E_cell = \[\ce{E^{\circ}_{cell} - \frac{0.0591}{2} log_10 \frac{[Mg^{2+}]}{[Cu^{2+}]}}\]

∴ \[\ce{E_{cell} = (E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Mg^{2+}/Mg}) - \frac{0.0591}{2} log_10 \frac{0.001}{0.0001}}\]

= \[\ce{(+0.34 - (-2.37)) - \frac{0.0591}{2} log_10 10}\]

= 2.71 − 0.0295

= 2.68 V