Question

A silver voltmeter,copper voltameter and oxygen voltameter are connected in series. Given that 4.32g of silver is deposited during the electrolytic process, determine(I) the mass of copper deposited(ii) the volume of oxygen liberated (Ag=108, Cu=64, O=16, G.M.V=22.4dm³

Answer 1

Ag⁺ + Cu⁰ = Ag⁰ + Cu²⁺;

Ag⁺ +1e = Ag⁰ * 2

Cu⁰ -2e = Cu²⁺ * 1

Then,

2Ag⁺ +1e = 2Ag⁰;

Cu⁰ -2e = Cu²⁺;

And: 2Ag⁺ + Cu⁰ = 2Ag⁰ + Cu²⁺;

n(Ag) = m(Ag)/M(Ag) = 4.32/108 = 0.04 mol.

By the ionic reaction: n(Cu) = 1/2 * n(Ag) = 1/2 * 0.04 = 0.02 mol;

m(Cu) = n(Cu) * M(Cu) = 0.02 * 64 = 1.28 g.

Ag⁺ + O^2- = Ag⁰ + O₂⁰;

Ag⁺ +1e = Ag⁰ * 4

2O^2- -4e = O₂⁰ * 1

Then,

4Ag⁺ +1e = 4Ag⁰;

2O^2- -4e = O₂⁰;

And: 4Ag⁺ + 2O^2- = 4Ag⁰ + O₂⁰;

n(Ag) = m(Ag)/M(Ag) = 4.32/108 = 0.04 mol.

By the ionic reaction: n(O₂) = 1/4 * n(Ag) = 1/4 * 0.04 = 0.01 mol;

V(O₂) = n(O₂) * V_m = 0.01 * 22.4 = 0.224 L.

Answer: m(Cu) = 1.28 g;

V(O₂) = 0.224 L.