A current of 1.70 A is passed through 300.0 mL of 0.160 M solution of a ZnSO 4 for 230 sec. with a current efficiency of 90%. Find out the molarity of Zn 2+ after the deposition Zn. Assume the volume of the solution to remain constant during the electrolysis.
Text Solution
Verified by Experts96
Sol. i = 
ampere
∴ equivalent of Zn 2+ lost = 
= 
= 3.646 × 10 –3 milliequivalent of Zn
2+ lost = 3.646
Initial milliequivalent of Zn 2+ = 300 × 0.160 × 2 = 96
( ∴ M × 2 = N for Zn 2+ ∴ meq. = N × V(mL)
∴ milliequivalent of Zn 2+ left in solution
= 96 – 3.646
= 92.354
∴ [ZnSO4] = 
= 0.154 M
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