Question

A volume of 200 cm3 of a 2-normal sodium chloride solution (ρ = 1.10 g cm-3) was electrolysed at permanent stirring in an electrolytic cell with copper electrodes. Electrolysis was stopped when 22.4 dm3 (at STP) of a gas were liberated at the cathode.

Calculate the mass percentage of NaCl in the solution after electrolysis.

Relative atomic masses: Ar(H) = 1; Ar(O) = 16; Ar(Na) = 23; Ar(Cl) = 35.5; Ar(Cu) = 64.

Answer 1

Calculations are made on the assumption that the following reactions take place:

2 NaCl → 2 Na+ + 2 Cl

cathode: 2 Na+ + 2 e– → 2 Na

anode: 2 Cl- – 2 e– → Cl–

Cl2 + Cu → CuCl2

Because the electrolyte solution is permanently being stirred the following reaction comes into consideration:

CuCl2 + 2 NaOH → Cu(OH)2 + 2 NaCl

On the assumption that all chlorine reacts with copper, the mass of NaCl in the electrolyte solution *** unchanged during the electrolysis

m(NaCl) = n M = c V M = 2 mol dm-3 × 0.2 dm3 × 58.5 g mol-1 = 23.4

V(H2) = 22.4 dm3 , i. e. n(H2) = 1 mol

The amount of water is decreased in the solution by:

n(H2O) = 2 mol

m(H2O) = 36 g

Before electrolysis:

m(solution NaCl) = V ρ = 200 cm3 × 1.10 g cm-3 = 220 g

% NaCl = 23.4 g/220 g x 100 = 10.64

After electrolysis:

m(solution NaCl) = 220 g – 36 g = 184 g

% NaCl = (23.4 g/184 g x 100) = 12.72