Question

A current of 4 amperes was passed for 2 hours through a solution of copper sulphate when 5.0 g of copper was deposited. Calculate the current efficiency.

Answer 1

Given: I = 4 A

t = 2 hours = 2 × 3600 = 7200 s

Q = I × t

= 4 × 7200

= 28800 C

\[\ce{Cu^2+ + 2e− -> Cu}\]

2 moles of electrons deposit 1 mole of Cu.

Molar mass of Cu = 63.5 g/mol

Moles of e⁻ = `Q/F`

= `28800/96500`

= 0.298 mol e⁻

Moles of Cu = `0.298/2`

= 0.149 mol

Mass = 0.149 × 63.5

Mass = 9.47 g

Actual mass = 5.0 g

Efficiency = `"Actual mass"/"Theoretical mass" xx 100`

= `5.0/9.47 xx 100`

= 52.8%

∴ The current efficiency is 52.8%.