Question

The current of 9.65A is passed for 3 hrs between nickel electrodes in 0.5L of a $2 \times {10^{ - 3}}M$ solution of $Ni{(N{O_3})_2}$, the molarity of the solution after electrolysis would be:

(A) 0.46 M

(B) 0.625 M

(C) $2 \times {10^{ - 3}}M$

(D) 1.25 M

Answer 1

**** In this electrolysis process, as we are giving current to two nickel electrodes dipped in the same solution, one of the nickel strips will behave as an anode and the other will behave as a cathode. Oxidation will occur at anode and reduction will occur at cathode.

Complete step by step solution:

In this process, as we will give current to the nickel electrodes dipped in the nickel solution, the nickel metal at the nickel electrode will undergo oxidation reaction and will turn into nickel ions and will immerse in the solution. The reaction can be written as below.

$Ni \to N{i^{2 + }} + 2{e^ - }$

But as current is given to the electrodes, one of the electrodes will be anode and one will be cathode. The above given reaction will occur at anode because it is an oxidation reaction.

But at cathode, a reduction reaction will also occur. In this reaction, nickel ions will get reduced at the surface of the cathode to its metallic form. The reaction is shown below.

$N{i^{2 + }} + 2{e^ - } \to Ni$

So, we can say that no matter how much amount of current is passed for any time but as both the electrodes are of nickel and are dipped in same solution, the oxidation and reduction reactions will occur at same rate at a time and hence the amount of nickel ions produced at anode will be equal to the amount of nickel atoms getting deposited at cathode.

Hence, we can say that the concentration of the nickel nitrate solution will remain the same as the initial state because when nickel electrodes are used anodic nickel will be dissolved and get deposited at cathode and hence molarity of the solution will remain unchanged.

So, correct answer is (C) $2 \times {10^{ - 3}}M$

Note: Do not calculate directly the number of moles of electrons generated in the process and decide the amount of nickel deposited at the electrode. The first thing you must get is that there will be oxidation and reduction reactions going at the same rate.