A current of 1.70 A is passed through 300.0 mL of 0.160 M solution of $ZnSO_{4}$ for 230 s with a current efficiency of 90 per cent. Find out the molarity of ${{Zn}^{2+}}$ after the deposition of zinc.
Assume the volume of the solution to remain constant during electrolysis.
Quantity of electricity passed = 1.70 X 230 = 391C
As current efficiency = 90%
Effective charge = 90/100 x 391 C = 351.9 C
${{Zn}^{2+}}$ + ${{2e}^{-}}$ ------>Zn
2 X 96500 C deposit ${{Zn}^{2+}}$ =1 mole
.’. 351.9 C will deposit = 1/2 x 96500x 351.9 mol = 0.0018 mol
${{Zn}^{2+}}$ ions present originally in 300 mL of 0.160 M
$ZnSO_{4}$ = 0.160/1000x 300 = 0.048 mol
Amount of ${{Zn}^{2+}}$ in 300 mL of solution after deposition of Zn = 0.048 — 0.0018 = 0.0462 mol
Molarity of ${{Zn}^{2+}}$ after the deposition of Zn = 0.0462/300 x 1000 = 0.154M.