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

Mg(s) / Mg2+ (0.001 M) // Cu2+ (0.0001 M) / Cu(s)

How does Ecell vary with the concentration of both Mg²⁺ and Cu²⁺ ions? (Given E₀ cell= 2.71 V).

• The net reation for the given cell :

Mg(s) +Cu²⁺(aq)↔ Mg²⁺(aq) + Cu(s)

• Nerst equation =
E_cell = E⁰_cell – log k_c

where , E_cell=cell emf

E⁰_cell = standard eletrode potentialof the cell

k_c = rate constant for the reaction and n = no. Of electrons transferred.

• E_cell = 2.71 - 0.059 log [Mg²⁺(aq)] [Cu(s)] ( n=2 beacause there are 2

2 [Cu²⁺(aq)] [Mg(s)] electrons transferred)

=2.71 - 0.059 log (.001)

2 (.0001)

= 2.71 - 0.059 log 10

2

=2.71 – 0.0295 V ( log10 = 1 and .059/2 = .0295)

= 2.6805 V .

• As the reaction proceeds[ Mg²⁺]in the anode compartment increases whereas [Cu²⁺] in the cathode compartment decreases as Cu is being deposited as metallic Cu in the electrode .

• During this process, the ratio [Mg²⁺]/[Cu²⁺] steadily increases, and the cell voltage therefore steadily decreases and Eventually,

[Mg²⁺] = [Cu²⁺], so the ratio becomes = 1 and

E_cell = E°_cell .

• Beyond this point, [Mg²⁺] will continue to increase in the anode compartment, and [Cu²⁺] will continue to decrease in the cathode compartment. Thus the value of [Mg²⁺]/[Cu²⁺] will increase further, leading to a further decrease in E_cell .