Exercise 7 | Electrochemistry

Here is a galvanic cell: Zn(s) | Zn2+ (0.150 M) || Zn2+ (aq) | Zn(s)


Calculate the concentration of Zn2+ (aq) at the cathode if Ecell = 2.50 x 10-2 V at 25°C.

Nernst equation:
 

At 25°C:
 

Ecell = E0cell0.02570νe x ln Q


Ecell = electromotive force emf of the cell (in V) = 2.50 x 10-2 V
E0cell = standard cell voltage (in V)
νe = stoichiometric coefficient of the electrons = 2
Q = reaction quotient = [Zn2+]anode / [Zn2+]cathode

 

In this exercise: the oxidation reaction is the reverse reduction reaction

⇒ E0ox = - E0red

⇒ E0cell = 0

 

Ecell = E0cell0.02570νe x ln Q

Ecell = 0.02570νe x ln Zn2+anodeZn2+cathode


[Zn2+]cathode = [Zn2+]anode x exp 2 × Ecell0.02570

[Zn2+]cathode = 0.150 x exp 2 × 0.0250.02570

[Zn2+]cathode = 1.05 M