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

1.	A current of 1.608 A 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= 1.609A""t="50 min(or)"50xx60=3000S""V=250mL` `C=0.5M""eta=100%` The number of Faraday.s of ELECTRICITY passed through the `CuSO_(4)` solution `rArr""Q=It rArr Q=1.608xx3000` `Q=4824C` `therefore"Number of Faraday.s of electricity "=(4824C)/(96500C)=0.5F` Electrolysis of `CuSO_(4)` `Cu_((aq))^(2+)+2e^(-)rarrCu_((s))^(-)`. The above equation shows that 2F electricity will DEPOSIT 1 mole of `CU^(2+)` `therefore0.5F" electricity will deposit"(1mol)/(2F)xx0.5F="0.025 mol"` `"Initial number of molar of "Cu^(2+)" in 250 ml of solution"=(0.5)/(1000)xx250mL=0.125mol` `therefore"Number of molar "Cu^(2+)" after electrolysis "=0.125-0.025=0.1mol` `therefore"Concentration of "Cu^(2+)=("0.1 mol")/("250 mL")xx"1000 mL"=0.4M`