The voltage generated by the zinc concentration cell described by, zn (s) |zn2 (aq, 0.100 m) ||zn2 (aq,? m) |zn (s), is 16.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode.

The concentration cell is:

Zn (s) / Zn²⁺ (aq, 0.100 M) / / Zn²⁺ (aq, x M) / Zn (s)

voltage = 16 mV x (1V / 10³ mV) = 16 x 10⁻³ V

- In the cell notation, the concentration on the left is that of the anode and that on the right is that of the cathode.

- Oxidation takes place at the anode and reduction takes place at the cathode.

so [Zn²⁺]oxidation = 0.100 M

[Zn²⁺] reduction = x M

From Nernst equation:

Ecell = - 0.0592 / n log [Zn²⁺] oxidation / [Zn²⁺]reduction

Number of electrons, n = 2. Substitute and solve for x:

16 x 10⁻³ V = - 0.0592 / 2 log (0.100 / x)

log 0.100 / x = - 0.54

0.100 / x = 0.288

x = 0.347

So the concentration of Zn²⁺ at the cathode = 0.406