Question

A current of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5 litre of a 2M solution of $Ni{{(N{{O}_{3}})}_{2}}$. What will be the molarity of the solution at the end of electrolysis? What will be the molarity of the solution if nickel electrodes are used? (1F=96500 coulomb; Ni=58.7)

Answer 1

Hint: Attempt this question by calculating total no. of electrons participating in the reaction and the total quantity of charge used in the reaction. Using them, we can further calculate no. of moles of $Ni{{(N{{O}_{3}})}_{2}}$ decomposed and then molarity of the solution.

Formula used:
We need to use the following formulas:-
$q=I\times t$
$\begin{align}
  & n=\dfrac{q}{(n-factor)\times F} \\
 & M(molarity)=\dfrac{n}{V(\text{in L})} \\
\end{align}$

Complete step-by-step answer:
-The reaction of electrolysis of $Ni{{(N{{O}_{3}})}_{2}}$ is as follows:-
$N{{i}^{2+}}+2{{e}^{-}}\to Ni$
From this reaction we can observe that 2 electrons were used for the reduction of $N{{i}^{2+}}$ to Ni.
-The values provided in the question are as follows:-
I= 3.7 A
t= 6hours =$6\times 60\times 60=21600s$
-The total quantity of charge passed in the reaction:-
$q=I\times t$
Where
 q=charge passed during electrolysis
I= current used during electrolysis
t= time period
Therefore,$q=3.7A\times 21600s=79920C$
-Calculation of no. of moles of $N{{i}^{2+}}$ deposited or no. of moles of $Ni{{(N{{O}_{3}})}_{2}}$decomposed:-
$n=\dfrac{q}{(n-factor)\times F}$
In this reaction, n-factor will be no. of electrons i.e.,2 and F= 96500C/mol (Faraday’s Constant)
\[{{n}_{decomposed}}=\dfrac{79920C}{2\times 96500C/mole}=0.414moles\]
-Calculation of no. of moles of$Ni{{(N{{O}_{3}})}_{2}}$ present in the solution after electrolysis:-
Total no. of moles of $Ni{{(N{{O}_{3}})}_{2}}$present before electrolysis= $M\times V(in\text{ L)= }n$
Molarity (M) of solution before electrolysis = 2M
Volume (V) of solution = 0.5L
${{n}_{before}}=(0.5L\times 2M)=(0.5L\times 2mol/L)=1mole$
Total no. of moles of $Ni{{(N{{O}_{3}})}_{2}}$present after electrolysis= ${{n}_{before}}-{{n}_{decomposed}}=(1-0.414)=0.586moles$

-Molarity of solution after electrolysis is as follows:-
$M(molarity)=\dfrac{n}{V(\text{in L})}=\dfrac{0.586moles}{0.5L}=1.172M$

Therefore, the molarity of the solution is =1.172M
-When nickel is used as electrodes, the anodic nickel tends to dissolve in the solution and simultaneously get deposited at the cathode due to which there is no change in the concentration of the solution i.e., the molarity remains unaffected.

Note: -It is to be noted that during electrolysis, oxidation takes place at anode and reduction takes place at cathode. Therefore, at the end, nickel gets deposited at cathode as it gains electrons and gets reduced.
-Kindly prefer to note down all the values and convert them into required units so as to obtain the desired result accurately.