A current of 1.70 A is passed through 300 ML of a 0.16 M solution of ZnSO_(4) for 230 s with a current efficienty of 90%. Find out molarity of Zn^(2+) after the deposition of Zn. Assume the volume of the solution to remain constant during the electrolysis.

A current of 1.70 A is passed through 300 ML of a 0.16 M solution of Z

1.	A current of 1.70 A is passed through 300 ML of a 0.16 M solution of ZnSO_(4) for 230 s with a current efficienty of 90%. Find out molarity of Zn^(2+) after the deposition of Zn. Assume the volume of the solution to remain constant during the electrolysis.
Answer» Solution :Mole of electric CHARGE `= (1.70 xx 230)/(96500) = 0.004052 F`. `therefore` eq. of Zn to be deposited for 100% CURRENT efficiency = 0.004052 or mole of Zn to be deposited for 100% current efficiency = 0.004052/2 = 0.002026 or mole of Zn to be deposited for 90% current efficiency `= 0.9 xx 0.002026 = 0.00118234` Initial mole of Zn (or `ZnSO_(4)`) `= 0.16 xx 0.3 = 0.48`. Mole of Zn remained undeposited = 0.048 - 0.0018234 `= 0.0461766`. MOLARITY after electrolysis `= (0.0461766)/(0.3) = 0.154 M`