Question

A current of $1.70$ ampere is passed through $300mL$ of $0.160M$ solution of zinc sulphate for $230$ seconds with a current efficiency of $90$ per cent. Find out the molarity of $Z{n}^{2+}$ ions after the deposition of zinc. Assume the volume of the solution to remain constant during electrolysis.

Answer 1

Amount of charge passed $=1.70\times 230coulomb$
Amount of actual charge passed $=\frac{90}{100}\times 1.70\times 230$
$=351.9coulomb$
No. of moles of $Zn$ deposited by passing $351.9$ coulomb of charge
$=\frac{1}{2\times 96500}\times 351.9=0.000182$
Molarity of $Z{n}^{2+}$ ions after deposition of zinc
$=[0.160-\frac{0.000182\times 1000}{300}]M$
$=0.154M$.