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

Hint: 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.