Question

A voltaic cell consists of two Ag/Ag+half-cells, A and B. The electrolyte in A is 0.10 M AgNO3. Theelectrolyte in B is 0.90 M AgNO3. Which half-cell houses the cathode? What is the voltage of the cell?

Answer 1

The electrolyte in A is 0.10 M AgNO₃.

The electrolyte in B is 0.90 M AgNO₃.

Concentration of the electrolyte in B is more (0.90 M), so reduction occurs here. This half-cell is the cathode

Concentration of the electrolyte in A is less (0.10 M), so oxidation occurs here. This half-cell is the anode. Electrons flow from high concentration to low concentration.

Voltage of cell -

E = Eº - (0.0592)/n * log ( [Ag+]dil / [Ag+]con )

[Ag+]dil = concentration of the dilute solution(0.10M)

[Ag+]con = concentration of the concentrated solution 0.90(M)

n = 1

Since both half cells involve the same ion, Ag+, Eº = 0

E = Eº - (0.0592) * log ( [Ag+]dil / [Ag+]con )

E = Eº - (0.0592) * log (0.10/ 0.90)

E = 0 - 0.0592 * -0.954

E = 0.0565V