Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A constant current of 2mA was passed for 16 min. It was found that after electrolysis the absorbance of the solution was reduced to `50%` of its original value. Calculate the concentration of `CuSO_(4)` in the solution to begin with.

Explanation

Quantity of charge passed `=2xx10^(-3)xx16xx60`
`=1.92C`
`"1.92 C charge "-=(1)/(96500)xx1.92-=1.99xx10^(-5)` mol of electrons
In the reaction, `Cu^(2+)(aq)+2erarrCu(s)`
2 mol of electrons = `"1 mol of "Cu^(2+)` ions discharged
`therefore" "1.99xx10^(-5)` mol of electrons `=9.95xx10^(-6)` of `Cu^(2+)` ions discharged
Absorbance of a solution is directly proportional to the concentration of the solution. `50%` decrease of absorbance of the solution means `50%` of the concentration of the solution is reduced. Hence, initial number of mol of `Cu^(2+)` ions `=2xx9.95xx10^(-6)=1.99xx10^(-5)` mol.
`therefore" Initial concentration of "Cu^(2+)`
i.e., `CuSO_(4)=(1.99xx10^(-5))/(250)xx1000M=7.96xx10^(-5)M`