Question

In the electrolysis of ${\text{NiS}}{{\text{O}}_{\text{4}}}$ using Nickel electrodes the reaction that takes place at anode is:
A. ${\text{N}}{{\text{i}}^{{\text{2 + }}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to {\text{Ni}}$
B. ${\text{2}}{{\text{H}}^{\text{ + }}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to {{\text{H}}_{\text{2}}}$
C. ${\text{N}}{{\text{i}}^{{\text{ + 2}}}}{\text{ }} \to {\text{ N}}{{\text{i}}^{\text{ + }}}{\text{ + }}{{\text{e}}^{\text{ - }}}$
D. ${\text{Ni }} \to {\text{N}}{{\text{i}}^{{\text{ + 2}}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}$

Answer 1

Hint: Electrolysis is a process in which when the electric current is passed through the substance, it results in a chemical change. The chemical change can be oxidation or reduction depending on the loss or gain of electrons. The electrolysis of ${\text{NiS}}{{\text{O}}_{\text{4}}}$ using the Ni electrode is an electrolytic set up in which Ni electrodes are dipped in a solution of nickel sulfate (aqueous).

Complete step by step answer:
We know that on passing electricity, it undergoes electrolysis.
One of the electrodes dipped in the solution is positive and the other is negative.
We know that when the electrolytic cell is charged, it behaves as a primary cell in which the positive electrode is the cathode and the negative electrode is the anode.
One which is connected to the positive terminal of the power supply is an anode and it has a negative charge and that electrode which is connected to the negative terminal of the power supply is a cathode that has a positive charge.
We know that electrons move in the opposite direction of the current. Current flows from positive to the negative terminal. Therefore we can say that the electrons move from anode to cathode.
From the above discussion, we could conclude that oxidation takes place at the anode (negative charge) meaning electrons from the anode move towards the cathode which accepts the electrons. That is, electrons move from negative to positive.
Nickel undergoes oxidation at anode,
${\text{Ni }} \to {\text{ N}}{{\text{i}}^{{\text{2 + }}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}$

So, the correct answer is Option D.

Note:
At the cathode, the reduction occurs. The reduction in the gain of electrons. Since 2 electrodes are present both oxidation and reduction take place at anode and cathode respectively. So, it can be called a redox reaction.