Question

When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction is as shown. Here, hydrogen sulphide is acting as:
$S{{O}_{2}}+2{{H}_{2}}S\to 2{{H}_{2}}O+3S.$
A. An oxidizing agent
B. A reducing agent
C. A dehydrating agent
D. A catalyst

Answer 1

A chemical which donates electrons to other then it is called a reducing agent because the chemical undergoes oxidation and reduces other chemicals. A chemical which accepts electrons from others is called an oxidizing agent because the chemical is going to make others undergo oxidation.

Complete answer:
- In the question is asked to find the role of hydrogen sulphide in the given reaction.
- The given reaction is as follows.
$S{{O}_{2}}+2{{H}_{2}}S\to 2{{H}_{2}}O+3S$
- In the above reaction one mole of sulphur dioxide reacts with two moles of hydrogen sulphide and forms two moles of water and three moles of sulphur as the products.
- The oxidation state of sulphur in sulphur dioxide is
$x + 2(-2) = 0$
$x = 4$
Here x = oxidation state of the sulphur on sulphur dioxide.
- The oxidation state of the pure sulphur formed as the product is 0.
- Means the oxidation state of sulphur in sulphur dioxide is 4 and it decreases to ‘0’ after completion of the reaction.
- So, the sulphur dioxide got reduced by accepting electrons from hydrogen sulphide.
- Means sulphur dioxide is a oxidizing agent and hydrogen sulphide is a reducing agent.
- Therefore the role of hydrogen sulphide in the given reaction is a reducing agent.

So, the correct option is B.

Note:
The oxidation state of sulphur in hydrogen sulphide is
$x +2 (1) = 0$
$x = -2$
Here x =oxidation state of sulphur in hydrogen sulphide.
The oxidation state of sulphur in hydrogen sulphide is -2 it is changed to ‘0’ (oxidation state of sulphur produced in reaction as the product). So, hydrogen sulphide is oxidized and acts as a reducing agent in the given reaction.