Question

In the presence of ${H_2}S$, $S{O_2}$ act as:
(A) An oxidising agent
(B) A reducing agent
(C) A hydrolysis agent
(D) A redox agent

Answer 1

Hint: Try to recall that the maximum oxidation of sulphur is +6 and the minimum oxidation number of sulphur is -2. Now by using this you can easily find the correct option from the given options.

Complete step by step solution:
It is known to you that sulphur can vary its oxidation state -2 to +6. Also, it is known that those species which decrease their oxidation number in a redox reaction are known as oxidising agents and those which increase their oxidation number are known as reducing agents.
Now, coming to the question, sulphur has an oxidation number of (-2) in ${H_2}S$and this is the minimum oxidation number of sulphur and cannot be reduced further. Therefore, sulphur will increase its oxidation number (from -2 to 0) in ${H_2}S$ and changes to S and hence, will act as reducing agent in redox reaction between ${H_2}S$ and $S{O_2}$.
In $S{O_2}$, sulphur has oxidation number of +4 and can either increase or decrease its oxidation number but since, ${H_2}S$undergoes reduction so, sulphur will decrease its oxidation number(from +4 to 0) and hence, will act as an oxidising agent.
The net balanced redox reaction between ${H_2}S$and $S{O_2}$ is as follows:
$2{H_2}S + S{O_2} \to 2{H_2}O + 3S$.

Therefore, from above we can conclude that option A is the correct option to the given question.

Note:It should be remembered to you that ${H_2}S$ is a colorless, flammable gas with a characteristic smell of rotten egg and is a good reducing agent.
Also, you should remember that $S{O_2}$ is a colorless gas which acts as a bleaching agent to remove excess chlorine and also act as a disinfectant.