Question

A current of $\,0.965\,ampere\,$ is passed through $\,500ml\,$ of $\,0.2M\,$ solution of $\,ZnS{O_4}\,$​ for $\,10\,$ minutes. The molarity of $\,Z{n^{2 + }}\,$ after deposition of zinc is:
A.$\,0.1M\,$
B.$\,0.5M\,$
C.$\,0.8M\,$
D.$\,0.194M\,$

Answer 1

Hint: The solution to this question can be found out by first finding the charge deposited and then the number of moles deposited and from that the molarity can be found out using the respective formula. The Faraday constant in electrolysis measurements is the single most significant piece of knowledge.
$\,96500C\,$ is called $\,1faraday\,$.
Formula used: $\,Q = I \times t\,$
$\,Molarity = \dfrac{{Number\,of\,moles}}{{Volume\,in\,litres}}\,\,$

Complete step by step answer:
Let us first find out the charge supplied in this reaction;
For that we have to use the formula mentioned above;
$\,Q = I \times t\,$
Where,
$\,Q = $ charge supplied in coulombs
$\,I = \,$ current in amperes \[ = {\text{ }}0.965A\,\]
$\,t = \,$ time in seconds $\,10 \times 60 = 600s\,$
Therefore, $\,Q = 0.965 \times 600\,$
$\, = 579C\,$
In the reaction of zinc;
$\,Z{n^{2 + }} + 2{e^ - } \to Zn\,$
So, for one mole of $\,Zn\,$ charge required $\, = 2F\,$ as two electrons are gained in this reaction
$\, \Rightarrow 2 \times 96500\,$$\, = 193000C\,$ since, $\,F = 96500C\,$
$\therefore Number\,of\,moles\,deposited\,here = \dfrac{{579}}{{193000}}\,$
$\, = 0.003\,moles\,$
Then, to calculate the number of moles in $\,500ml\,$;
$\,Molarity = \dfrac{n}{V}\,$
Where, $\,n = number\,of\,moles\,of\,solute\,,\,V = volume\,in\,litres\,$
$\therefore \,Number\,of\,moles = Molarity \times Volume\,in\,litres\,$
$ \Rightarrow 0.5 \times 0.2\,$$\, = 0.1\,moles\,$since,$\,500ml = 0.5L\,$
Then, the remaining number of moles in the solution is;
$\,0.1 - 0.003\,$
$\, = 0.097moles\,$
$\,\therefore Molarity = \dfrac{{0.097}}{{0.5}}\, = \,0.194M\,$

Hence, for this question option D is the correct answer.

Additional information: Whenever an electrolyte is dissolved in water, such as metal sulphate, the molecules break into positive and negative ions. The positive ions travel to the electrodes attached to the battery's negative terminal, where they take up electrons and thereby become a pure metal atom and placed on the electrode. The negative ions travel to the electrode linked to the battery's positive terminal, giving up their extra electrons and they also become pure metal atoms

Note:
The number of moles atoms deposited on electrodes also depends on the amount of valency they have. If the value is greater, the number of deposited atoms will be lower for the same amount of electricity, and if the value is lower, the number of deposited atoms will be higher for the same amount of electricity.