Question

$2H_{2}S+S{O}_2\to 3S+2H_{2}O$

In the above reaction:
(a) $H_{2}S$ gets oxidized.
(b) $S{O}_2$ gets reduced.
(c) $H_{2}S$ gets reduced.
(d) $S{O}_2$ gets oxidized.

• A. (a) and (c)
• B. (b) and (c)
• C. (a) and (b)
• D. (b) and (d)

Answer 1

Answer: (C)

(a) and (b)

Explanation:

The given reaction is a type of redox reaction.

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

From the above given reaction:

$H_{2}S\to S+2H^{+}+2e^{-}$

Oxidation state of Sulphur changes from -2 to 0

Sulfide has been oxidized, and sulfur dioxide is the oxidizing agent.

from $S{O}_2\to 2S$

oxidation state changes from +4 to 0

So $H_{2}S$ gets oxidized and $S{O}_2$ gets reduced

Option C is correct.