# 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 = E0cell$\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ 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 = E0cell$\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ x ln Q

Ecell = $\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ x ln $\frac{{\left[{\mathrm{Zn}}^{2+}\right]}_{\mathrm{anode}}}{{\left[{\mathrm{Zn}}^{2+}\right]}_{\mathrm{cathode}}}$

[Zn2+]cathode = [Zn2+]anode x exp

[Zn2+]cathode = 0.150 x exp

[Zn2+]cathode = 1.05 M