Question

Calculate the EMF of the following cell at 298 K:

Mg | Mg²⁺ (0.1 M) || Ag⁺ (0.1 M) | Ag

\[\ce{E^{\circ}_{{Mg^{2+}/{Mg}}}}\] = −2.37 V, \[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V, R = 8.31 J K⁻¹, F = 96500 C mol⁻¹.

Answer 1

Given: Mg | Mg²⁺ (0.1 M) || Ag⁺ (0.1 M) | Ag

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

\[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V

Temperature = 298 K

R = 8.31 J K⁻¹

F = 96500 C mol⁻¹

Anode (oxidation): \[\ce{Mg -> Mg^2+ + 2e-}\]

Cathode (reduction): \[\ce{Ag+ + e- -> Ag}\]

To balance electrons, multiply Ag half-cell by 2:

\[\ce{Mg + 2Ag+ -> Mg^2+ + 2Ag}\]

Now calculate the standard EMF:

\[\ce{E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}}\]

= 0.80 − (−2.37)

= 3.17 V

The Nernst equation:

E_cell = `E_"cell"^circ - (RT)/(nF) log Q`    ...(i)

Where

n = 2

Q = `0.1/(0.1)^2`

Q = 10

Substituting these values in equation (i), we get

E_cell = `3.17 - (8.31 xx 298)/(2 xx 96500) log(10)`

= `3.17 - 2477.38/193000 xx 2.303`

= 3.17 − 0.02957

= 3.140 V

∴ The emf of the given cell at 298 K is 3.140 V.