Question

A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured 0,422 V. Determine the concentration of silver ion in the cell.
Given : E°Ag+/Ag = + 0.80 V, E° Cu2+/Cu = + 0.34 V.​

Answer 1

The concentraion of silver ions in the cell is 0.0699 M

Explanation:

• E(cell) = 0.042
• Eo Ag+ / Ag > E Cu+2 / Cu
• Ag + / Ag is cathod ena d Cu+ / Cu is anode.

E(cell) = E reduction =- E oxidation

E (cell) = 0.80 - 0.34

E(cell) = 0.46 V

E(cell) = E(cell) - 0.059 / n log [Cu+] (Ag)^2 / [Ag+]^2 (Cu)^2

E(cell) = 0.46  -  - 0.059 / n log (0.1) / (Ag+)^2

0.422 = 0.46 - 0.029 { log (0.1) - log [Ag+]}

After calculations we get

log  [Ag+] = - 1.155

[Ag+] = antilog (-1.155)

[Ag+] = 0.0699 M

Thus the concentraion of silver ions in the cell is 0.0699 M