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 Cu^(2+) after electrolysis assuming volume to be constant and the current efficiency is 100%.

A current of 1.608A is passed through 250 mL of 0.5 M solution of copp

1.	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 Cu^(2+) after electrolysis assuming volume to be constant and the current efficiency is 100%.
Answer» Solution :Given : I = 1.608A, t = 50 MIN = `50 times 60` V = 500 mL C = 0.5 M `" "`= 3000 S `eta=100%` Calculate the number of FARADAYS of electricity passed through the `CuSO_(4)` solution `rArr Q=It` `" "Q=1.608 times 3000` `" "Q=4824C` `therefore` number of Faradays of electricity `=(4824C)/(96500C)` `""=0.5F` Electrolysis of `CuSO_(4)` `Cu_((aq))^(2+)+2e^(-) rarr Cu_((s))` The above equation shows that 2F electricity will deposit 1 mole of `Cu^(2+)` to. `therefore` 0.5F electricity will deposit `(1mol)/(2F) times 0.5F=0.025 mol` Initial number of molar of `Cu^(2+)` in 250 ml of solution `=0.5/(1000mL) times 250mL` `""=0.125mol` `therefore` number of `Cu^(2+)` after electrolysis `""=0.125-0.025` `""=0.1` mol `therefore` CONCENTRATION of `Cu^(2+)=(0.1mol)/(250mL) times 1000mL` `""=0.4` M.