A current of 4 ampere was passed for 1.5 hours through a solution of copper sulphate when 3.2 g of copper was deposited. Calculate the current efficiency.

Explanation

`Cu^(2+) +2e^(-)toCu`
1 mole of Cu(63.5g) requries charge`=2F=2xx96500`coulombs
`therefore3.2g` Cu will require charge`=(2xx96500)/(63.5)xx3.2`coulombs=9726 coulombs
Quantity of electricity actually passed`=4xx45xx60C=21600C`.
Current efficiency`=(9726)/(21600)xx100=45%`.