Question

A current of 1.608A is passed through 250 mL of 0.5 M 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 = 1.608A

t = 50 min = 50 × 60 = 3000 S

V = 250 mL

C = 0.5M

η = 100%

Calculate the number of Faradays of electricity passed through the CuSO₄ solution

Q = It

Q = 1.608 × 3000

Q = 4824 C

∴ Number of Faradays of electicity = `(4824  "C")/(96500  "C")` = 0.05F

Electrolysis of CuSO₄

\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]

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

∴ 0.05F electricity will deposit

`(1  "mol")/(2"F") xx 0.05 "F"` = 0.025 mol

Initial number of molar of Cu²⁺ in 250 ml of solution

= `0.5/(1000  "mL") xx 250  "mL"`

= 0.125 mol

∴ Number of moles of Cu²⁺ after electrolysis

= 0.125 − 0.025

= 0.1 mol

∴ Concentration of Cu²⁺

= `(0.1  "mol")/(250  "mL") xx 1000  "mL"`

= 0.4 M