Question

A solution of $S{O_2}$ in water reacts with ${H_2}S$ precipitating sulphur. Here, $S{O_2}$ acts as:
A. Oxidizing agent
B. Reducing agent
C. Acid
D. Catalyst

Answer 1

Hint: It is a disproportionation (redox) reaction, in which one atom/molecule/ion is simultaneously undergoing oxidation as well as getting reduced.

Complete step-by-step solution: In order to answer this question, first we need to write the chemical equation for the above mentioned situation. When a solution of sulphur dioxide$S{O_2}$ gas (in water) is mixed with hydrogen sulphide, ${H_2}S$gas, a precipitate of sulphur is formed, along with formation of water. The following chemical reaction can be written as- $2{H_2}S + 2S{O_2} \to 3S + 2{H_2}O$ .
In the above reaction, the oxidation state of sulphur in ${H_2}S$ is -2, and after reaction, its oxidation state changes to 0. So, we can say here sulphur is being oxidized. Therefore, ${H_2}S$ is acting as a reducing agent.
But, if we will consider the oxidation state of sulphur in sulphur dioxide gas, i.e., $S{O_2}$ , we will find that it’s being reduced from +4 to 0. So, in this case $S{O_2}$ is acting as an oxidizing agent.

So, the correct option is A.

Note:
Hydrogen sulphide is a colorless gas, having a characteristic rotten egg smell. It is poisonous, corrosive, and flammable in nature.
$S{O_2}$ is an oxide of sulphur. It is a toxic gas that smells like burnt matches. It is used as an intermediate in the production of sulphuric acid. It is often used as a preservative in dried figs, apricots, because of its antimicrobial activity and its ability to prevent oxidation.