Question

Calculate the emf of the cell:

Mg_(s) | Mg²⁺ (0.1 M) || Cu²⁺ (1 × 10⁻³ M) | Cu_(s)

Given \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = +0.34 V, \[\ce{E^{\circ}_{{Mg^{2+}/{Mg}}}}\] = −2.37 V

Answer 1

The given cell is:

Mg_(s) | Mg²⁺ (0.1 M) || Cu²⁺ (1 × 10⁻³ M) | Cu_(s)

The half-cell reactions are:

Anode (oxidation): \[\ce{Mg -> Mg^{2+} + 2e-}\], E° = −2.37 V

Cathode (reduction): \[\ce{Cu^{2+} + 2e- -> Cu}\], E° = +0.34 V

We know that,

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]

= 0.34 − (−2.37)

= 2.71 V

Using the Nernst equation:

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

= `2.71 - 0.0591/2 xx log(0.1/(1 xx 10^-3))`

= `2.71 - 0.0591/2 xx log(100)`

= `2.71 - 0.0591/2 xx 2`

= 2.71 − 0.0591

= 2.6509 V