Question

Calculate emf of the following cell at 298 K:

\[\ce{Mg_{(s)} | Mg^{2+} (0.1 M) || Cu^{2+} (0.01) | Cu_{(s)}}\]

[Given \[\ce{E^{\circ}_{cell}}\] = +2.71 V, 1 F = 96500 C mol^–1]

Answer 1

Given: The given cell reaction is:

\[\ce{Mg_{(s)} | Mg^{2+} (0.1 M) || Cu^{2+} (0.01) | Cu_{(s)}}\]

Using the Nernst equation:

\[\ce{E_{cell} = E{^{\circ}_{cell}} - \frac{0.0591}{n} log \frac{[Mg^{2+}]}{[Cu^{2+}]}}\]

Where:

\[\ce{E^{\circ}_{cell}}\] = +2.71 V

n = 2 (number of electrons transferred)

Mg²⁺ = 0.1 M

Cu²⁺ = 0.01 M

\[\ce{E_{cell} = 2.71 - \frac{0.0591}{2} log \frac{0.1}{0.01}}\]

= 2.71 − 0.02955 log (10)

= 2.71 − 0.02955 × 1

= 2.71 − 0.02955

= 2.680 V