A current of 1.70 A is passed through 300.0 ml of 0.160 M solution of ZnSO_(4) for 230 sec with a curent efficiency of 90 percent. Find out the molarity of Zn^(2+) after the deposition of zinc. Assume the volume of the solution to remain constant during electrolysis.

A current of 1.70 A is passed through 300.0 ml of 0.160 M solution of

1.	A current of 1.70 A is passed through 300.0 ml of 0.160 M solution of ZnSO_(4) for 230 sec with a curent efficiency of 90 percent. Find out the molarity of Zn^(2+) after the deposition of zinc. Assume the volume of the solution to remain constant during electrolysis.
Answer» Solution :QUANTITY of electricity PASSED=`1.70xx230C=391C` As current efficiency=90% `therefore` Effective charge`=(90)/(100)xx319C=351.9C` `Zn^(2+)+2e^(-)TOZN` `2xx96500C` deposit `Zn^(2+)=1` mole `therefore351.9C` will deposit `Zn^(2+)=(1)/(2xx96500)xx351.9`mole=0.0018mole `Zn^(2+)` ions present originally in 300 ml of 0.160 M `ZnSO_(4)=(0.160)/(1000)xx300=0.048` mole `therefore`Amount of `Zn^(2+)` in 300 ml of solution after deposition of `Zn=0.048-0.0018=0.0462`mole `therefore`Molarity of `Zn^(2+)` after deposition of `Zn=(0.0462)/(300)xx1000=0.154M`