Question

In electrolysis of dilute ${H_2}S{O_4}$ , what is liberated at anode?
A. ${H_2}$
B. $S{O_4}^{2 - }$
C. $S{O_2}$
D. ${O_2}$

Answer 1

Hint: Electrolysis is a process in which electric current is passed through a solution known as electrolyte and there occurs a chemical change. The chemical change here refers to the loss or gains of an electron or in other words oxidation or reduction reaction takes place.

Complete step by step solution:
This process is carried out in an electrolytic cell, an apparatus which consists of positive and negative electrodes or in other words a cathode and an anode which are held apart and dipped into a solution containing positively and negatively charged ions, and also consists of a salt bridge connecting the cathodic and anodic compartments.
Now, according to the question when a dilute solution of sulfuric acid is electrolysed, gases are produced at both the anode and the cathode. The gas produced at the cathodic compartment exhibits a popping sound when a sample is lit in a test tube. This shows that the gas exhibiting a pop-up sound is due to hydrogen gas.
The gas produced in the anodic compartment lights a glowing splint when it is dipped into the sample of the gas. This shows that the oxygen gas is liberated. The reaction at the anode and the cathode is as follows:
At cathodic compartment:
\[2{H^ + } + 2{e^ - } \to {H_2}\]
​At anodic compartment:
\[4O{H^ - } \to 2{H_2}O + {O_2} + 4{e^ - }\]
Hence, the correct answer is Option D.

Note: At the cathodic compartment the hydrogen is produced twice as that of the oxygen produced at the anode this is because the water has two molecules of hydrogen and one oxygen atom.