Identify the oxidising and reducing agents in the following reactions:
$H_{2} S + {Br}_{2} \to 2 HBr + S$

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Solution

Oxidation means loss of electron in this oxidation state increases.
Reduction means gain of electron in this oxidation state decreases.
The one who gets oxidised act as a reducing agent and the one who get reduced act as an oxidising agent.
Reaction:
$H_{2} S (g) + {Br}_{2} (g) \to 2 HBr (g) + S (s)$
(Hydrogen sulphide) (Bromine) (Hydrogen bromide) (Sulphur)

⦁ Here oxidation state of $S$ in $H_{2} S$ is $- 2$ and that of $Br$ in ${Br}_{2}$ is $0$ at reactant side but oxidation state of $S$ in $S$ is $0$ and that of $Br$ in $HBr$ is $- 1$ at product side.
⦁ So, we can say that $H_{2} S$ is getting oxidised since its oxidation number is increasing from $- 2$ to $0$ and ${Br}_{2}$ is getting reduced since its oxidation number is decreasing from $0$ to $- 1$ .
⦁ The one who gets oxidised act as a reducing agent and the one who get reduced act as an oxidising agent.
⦁ So here $H_{2} S$ is a reducing agent and ${Br}_{2}$ is an oxidising agent.

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