Question

Why does $S{O_2}$ act as oxidizing as well as reducing agent but ${H_2}S$ act as only reducing agent?

Answer 1

Let’s first see what oxidizing and reducing agents are. Oxidizing agents are those which oxidize the other compound and remove hydrogen from the compound. Oxidizing agents are substances that gain electrons. A reducing agent is an element or compound that loses or "donates" an electron to an electron recipient in a redox chemical reaction.

Complete answer:
An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. A reducing agent, or reductant, loses electrons and is oxidized in a chemical reaction. In $S{O_2}$ , Sulphur has an oxidation state $+ 4$ so it can lose its two more electrons to attain a $+ 6$ oxidation state. Therefore it can lose and gain electrons therefore it acts as oxidizing as well as reducing agent.
We know, reducing agents oxidizes itself and reduces others. ${H_2}S$ is a strong reducing agent. The oxidation state of sulfur in ${H_2}S$ and in $H{S^ - }$ is $- 2$ . Thus, ${H_2}S$ can only be oxidized; it cannot act as an oxidant. Oxidation of ${H_2}S$ can lead to various products in which the sulfur can have oxidation numbers up to $+ 6$ .

Note:
Remember ${H_2}S$ Colorless, flammable, poisonous and corrosive, $H_2S$ gas is noticeable by its rotten egg smell and a good reducing agent. And also $S{O_2}$ it is a toxic gas responsible for the smell of burnt matches. Sulphur dioxide is an acidic gas and this can easily be demonstrated by adding water and a few drops of universal indicator to a container of the gas.