Question

A current of 1.07 ampere is passed through 300 ml of 0.160 M solution of zinc sulphate for 30 seconds with a current efficiency of 90%.Find out the molarity of zn^+2 ions after the deposition of zinc. Assume the volume of the remaining solution is constant during electrolysis.

Answer 1

This is  how  I  did  please  do send the correct answer if this one is correct then please give your views

Answer 2

Answer: The Molarity of $Zn^{2+}$ will be 0.1595M.

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 1.07 A

t= time in seconds =$30s$

$Q=1.07A\times 30s=32.1C$

As current efficiency is 90%, the current actually passed is=$\frac{90}{100}\times 32.1C=28.89C$

$Zn^{2+}+2e^{-1}\rightarrow Zn$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of  Zn.

28.89 C of electricity deposits =$\frac{1}{193000}\times 28.89=0.00015mole$ of  Zn.

$\text{no of moles of} Zn^{2+}={\text{Molarity}\times {\text{Volume in L}}$

$\text{no of moles of} Zn^{2+}={0.160M}\times {0.3L}=0.048mole$

Thus $Zn^{2+}$ remaining in solution = (0.048-0.00015)=0.04785moles

Thus $Molarity=\frac{\text {no of moles}}{\text {Volume of solution in L}}$

$Molarity=\frac{0.04785moles}{0.3L}=0.1595M$