Question

The pH of the solution _________ during the electrolysis of dilute aqueous solution of $CuS{{O}_{4}}$:
[A] First increases then decreases
[B] Decreases
[C] Remains constant
[D] Increases

Answer 1

Hint: Copper sulphate is an electrolyte. It will completely dissolve in water and dissociate into respective cations and anions. When this solution is subjected to electrolysis, the cathode will have a stable neutral metal but the anode species will be unstable and will react with the water molecules in the solution.

Complete Step by Step Solution: We define electrolysis as a process in which we pass electric current through a substance for it to react chemically. Electrolysis involves breaking of electrolytes into ions.
In the process of electrolysis, there is a cathode and an anode. Anode is the negative end and oxidation takes place at the anode and cathode is the positive end and reduction takes place here. An electrolyte is a substance which dissociates into cations and anions in a solute like water even if its atoms are tightly held through ionic bonds.
Now, let us discuss the electrolysis of copper sulphate.
Copper sulphate is an electrolyte. It dissolves completely in water and dissociates into $C{{u}^{2+}} and \text{ S}{{\text{O}}_{4}}^{2-}$ ions. Now, we immerse the two copper electrodes in the solution where one is anode and the other is cathode.
The negative ions will be attracted towards the anode and the cations are attracted towards the cathode. On reaching the cathode and the anode $C{{u}^{2+}} and \text{ S}{{\text{O}}_{4}}^{2-}$respectively, copper gains two electrons and become neutral and $\text{S}{{\text{O}}_{4}}^{2-}$ gives up 2 electrons and becomes a radical.
However, the sulphate radical formed is not stable and thus it will attack copper anode and will form copper sulphate again and finally end up reacting with the water and form sulphuric acid.
$2S{{O}_{4}}+2{{H}_{2}}O\to 2{{H}_{2}}S{{O}_{4}}+{{O}_{2}}$
As we can see an acid is produced here which will decrease the pH of the solution.

Therefore, the correct answer is option [B] Decreases.

Note: If we use carbon electrodes instead of the copper electrode, the electrolysis takes place differently. $S{{O}_{4}}$ present in the solution cannot react with carbon so instead of attacking at the cathode, it will react with water in the solution and it will form sulphuric acid and oxygen. This is the case even if we use any other metal electrode too.
$2S{{O}_{4}}+2{{H}_{2}}O\to 2{{H}_{2}}S{{O}_{4}}+{{O}_{2}}$