Question

A current of 1 .608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu²⁺ after electrolysis assuming volume to be constant and the current efficiency is 100%

Answer 1

Given, I = I .608A 

t = 50 min (or) 50 x 60 = 3000 S 

V = 250 mL 

C = 0.5M 

η = 100% 

The number of Faraday’s of electricity passed through the CuSO₄ solution 

Q = It 

= Q = 1.608 x 3000 

Q = 4824C

Number of Faraday’s of electricity = \(\frac{4824C}{96500C}\) = 0.5F

Electrolysis of CuSO₄ 

Cu²⁺_(aq) + 2e^- → Cu_(s) 

The above equation shows that 2F electricity will deposit 1 mole of Cu²⁺ 

0.5F electicity will deposit  \(\frac{1mol}{2F}\) x 0.5F = 0.025 mol 

Initial number of molar of Cu²⁺ in 250 ml of solution  = \(\frac{1mol}{250mL}\)  x 250mL = 0.125 mol 

Number of molar of Cu²⁺ after electrolysis 0.125 – 0.025 = 0.1 mol

Concentration of Cu²⁺ = \(\frac{0.1mol}{250mL}\) X 1000 mL = 0.4 M