Question

Calculate the emf of the following cell at 298 K:

Mg | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu

Given: \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = 0.337 V, \[\ce{E^{\circ}_{{Mg^{2+}/{Mg}}}}\] = −2.37 V, F = 96500 C mol⁻¹.

Answer 1

Given: Mg | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu

\[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = 0.337 V,

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

F = 96500 C mol⁻¹

Temperature (T) = 298 K

We know that

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

= 0.337 V − (−2.37 V)

= 2.707 V

We know that

E_cell = `E_"cell"^circ - 0.0591/n log([[text(Mg)^(2+)]]/[[text(Cu)^(2+)]])`

= `2.707 - 0.0591/2 log(0.001/0.0001)`

= `2.707 - 0.0591/2 log(10)`

= `2.707 - 0.0591/2 xx 1`

= 2.707 − 0.02955

E_cell = 2.677 V

∴ The emf of the given call at 298 K is 2.677 V.