Question

A current of 3.7 A passed for 6 h between nickel electrodes in 0.5 L of a 2 M solution of \[Ni{{(N{{O}_{3}})}_{2}}\]. The molarity (M) of the solution at the end of electrolysis will be: (write your answer to the nearest integer)

Answer 1

Hint: Electrolysis is a process of decomposition of ionic compounds into their elements by passing some amount of electric current through the substance in a fluid form. During the electrolysis some of the ions deposits on one the electrodes and few ions will remain in the solution itself.
Formula used:
Formula to calculate electrical charge is
Q = i \[\times \] t
Q = Charge
i = current,
t = time

Complete step by step answer:
In the question it is given that current i = 3.7 A
Time t = 6 hours = 21600 seconds.
We know the formula to calculate electrical charge (Q) from the above data.
Q = i \[\times \] t
    = 3.7 \[\times \] 21600 C
    = 79,920 C.
We know that 96,500 C of electricity deposits 1 gm equivalents of nickel.
We have to calculate 79,920 C of electricity deposits and how much amount of nickel.
\[\begin{align}
 & =\frac{1\times 79,920}{96,500} \\
 & =0.828g\text{ }eq \\
\end{align}\]
79,920 C of electricity deposits 0.828 gm equivalents of nickel.
Now we have to calculate number of moles of nickel by using the formula
Number of moles of Nickel \[\begin{align}
 & \text{=}\frac{\text{gm equivalents of nickel}}{\text{valency of nickel}} \\
 & =\frac{0.828}{2} \\
 & =0.414mol \\
\end{align}\]
(Valency of nickel in \[Ni{{(N{{O}_{3}})}_{2}}\] = 2)
In the question it is given that volume (V) is 0.5 L and the molarity of the solution (M) is 2 M
Now from the data of volume and molarity we can calculate the total number of moles of nickel in the original solution by using the following formula.
\[\begin{align}
 & \text{Molarity=}\frac{\text{number of moles}}{\text{Volume (in liter)}} \\
 & \text{2=}\frac{\text{number of moles}}{\text{0}\text{.5}} \\
 & \text{number of moles =1}\text{.0 mol} \\
\end{align}\]
Therefore nickel left in 0.5 L solution = number of moles of nickel in original solution- number
of moles of nickel deposited
         = 1.0 – 0.414
         = 0.586 mol
Therefore molarity of the solution at the end of the electrolysis
\[\]
\[\begin{align}
 & \text{Molarity =}\frac{\text{number of moles}}{\text{Volume (in liter)}} \\
 & \text{ = }\frac{0.586}{0.5} \\
 & \text{ = 1}\text{.172 M} \\
\end{align}\]
So, the molarity (M) of the solution at the end of electrolysis is 1.172 M.
The nearest integer of 1.172 is 1 M.

Note:
We have to calculate first nickel deposited at electrodes during electrolysis and later nickel left in the solution after electrolysis to find the molarity of the solution at the end of the electrolysis. Also remember that we have to subtract the number of moles of Nickel deposited from the number of moles of Nickel in the original solution. If you don’t subtract this value and calculate the molarity based on the number of moles of Nickel deposited, you will end up getting the wrong answer.