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 E_cell = 2.50 x 10^-2 V at 25°C.

Question

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 E_cell = 2.50 x 10^-2 V at 25°C.

| General Chemistry 3

Accepted Answer

Nernst equation:

At 25°C:

E_cell = E⁰_cell – $\frac{0.02570}{ν_{e}}$ x ln Q

E_cell = electromotive force emf of the cell (in V) = 2.50 x 10^-2 V
E⁰_cell = standard cell voltage (in V)
ν_e = stoichiometric coefficient of the electrons = 2
Q = reaction quotient = [Zn²⁺]_anode / [Zn²⁺]_cathode

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

⇒ E⁰_ox = - E⁰_red

⇒ E⁰_cell = 0

E_cell = E⁰_cell –

\frac{0.02570}{ν_{e}}

x ln Q

E_cell =

\frac{0.02570}{ν_{e}}

x ln

\frac{{[{Zn}^{2 +}]}_{anode}}{{[{Zn}^{2 +}]}_{cathode}}

[Zn²⁺]_cathode = [Zn²⁺]_anode x exp

(\frac{2 \times E_{cell}}{0.02570})

[Zn²⁺]_cathode = 0.150 x exp

(\frac{2 \times 0.025}{0.02570})

[Zn²⁺]_cathode = 1.05 M

Exercise 7 | Electrochemistry

General Chemistry 3 - Exercise 7