Question

Two platinum electrodes were immersed in a solution of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ , and electric current was passed through the solution. After some time, it was found that the colour of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ disappeared with evolution of gas at the electrode. The colorless solution contains:
A.Platinum sulphate
B.Copper hydroxide
C.Copper sulphate
D.Sulphuric acid

Answer 1

To answer this question, you should have an idea of the standard reduction potentials of various elements. That substance is preferably oxidized which has a higher oxidation potential.

Complete step by step answer:
On electrolysis, the cupric ion in ${\text{CuS}}{{\text{O}}_{\text{4}}}$ is reduced to ${\text{Cu}}$ which then deposits on the platinum electrode. The reduction reaction takes place at the anode of the electrolytic cell.
At the anode, the two species present are sulphate ions and water.
The oxidation potential of water is more than the oxidation potential of sulphate ions. As a result, the oxidation of water occurs forming hydrogen ions and water is released as a gas.
Now the anion present in the solution was sulphate and we know that the cation produced is hydrogen ion. Thus, the colourless solution is due to the formation of sulphuric acid $\left( {{H_2}S{O_4}} \right)$

The correct answer is D.

Note:
Electrode potential is the EMF of a galvanic cell which is built from a standard reference electrode and another random electrode. Generally the reference electrode used is standard hydrogen electrode to ease the calculation for the electrode potential as the standard hydrogen electrode has an electrode potential of zero volts. Electrode potential can be either reduction potential or oxidation potential depending upon the reaction taking place. Lower is the standard reduction potential value of the element, higher is its tendency to form positive ions.