Question

A current of 9.65A is passed for 3 hours between nickel electrodes immersed in $0.5{\text{ d}}{{\text{m}}^3}$ solution of concentration of \[2mold{m^{ - 3}}{\text{ Ni(N}}{{\text{O}}_2})\] The molarity of the solution after electrolysis would be:
A. 0.46M
B. 0.92M
C. 1.25M
D. 0.625M

Answer 1

Hint: Quantity of charged passed (Q) is equal to \[Q = It\] where I=current and t is time for which it is passed. Electrode reaction of Nickel is \[N{i^{2 + }} + 2{e^ - } \to Ni\]$$

Complete step by step answer:
> We know that the Given quantities are Current =9.65 A, Time(t)=3 hours, Solution concentration is \[2{\text{ }}mold{m^{ - 3}}\]Now we will use the formula to find the Quantity of charged passed.
> We know \[Q = It\]
Therefore, \[Q = 9.65 \times 3 \times 60 \times 60\]
\[Q = 104220{\text{ C}}\]
 Here, We change the Value of t from hour to second to get our result in Coulomb. The SI unit of time is Seconds.
Second step is to find the amount of Ni deposited, which is equal to=
\[{\text{Amount of Nickel deposited = }}\frac{{104220{\text{ C}}}}{{2 \times 96500{\text{ Cmo}}{{\text{l}}^{ - 1}}}}\] = 0.54 mol
> Now our next step is to find out the moles left. In starting we had 1 mol of nickel, now 0.54 mol of nickel has been deposited,
So, Now the mol of the nickel left behind are 1mol-0.54mol = 0.46 mol.
\[{\text{Molarity of solution = }}\frac{{{\text{Number of moles}}}}{{{\text{Volume in litres}}}}\]
\[{\text{Molarity of solution = }}\frac{{0.46}}{{0.5}}\]
\[{\text{Molarity of solution = 0}}{\text{.92mol}}{{\text{L}}^{ - 1}}\]
Therefore option B is Correct.

So, the correct answer is “Option B”.

Additional Information:
Nickel or any derivative of nickel is the most commonly used electrodes materials in commercial alkaline electrolysers. Reasons why it is used as electrodes:
- Its High Electrical Conductivity.
- Reduced specific energy consumption for hydrogen production.
- Exhibits good resistance to corrosive solutions.
- Low cost as compared to platinum group metals.

Note:
Don't forget to convert all the given Units into SI units. Students may sometimes confuse that the Given electrolysis is between the active electrodes, So the concentration of electrolyte will not change.
Yes concentration will not change But after attaining the molarity of \[{\text{0}}{\text{.92mol}}{{\text{L}}^{ - 1}}\].