Hydrogen sulphide $(H_{2} S)$ is a strong reducing agent. Which of the following reactions shows its reducing action?

C d (N O_{3})_{2} + H_{2} S \to C d S + 2 H N O_{3}

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C u S O_{4} + H_{2} S \to C u S + H_{2} S O_{4}

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2 F e C l_{3} + H_{2} S \to 2 F e C l_{2} + 2 H C I + S

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P b (N O_{3})_{2} + H_{2} S \to P b S + 2 C H_{3} C O O H

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Solution

The correct option is C $2 F e C l_{3} + H_{2} S \to 2 F e C l_{2} + 2 H C I + S$
In all the reactions except for option (C), the oxidation state of both Hydrogen and Sulphur remains the same throughout the reaction.

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H_{2} S + 2 F e C l_{3} \to 2 F e C l_{2} + 2 H C l + S

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2 F e C l_{3} + H_{2} S \to 2 F e C l_{2} + 2 H C l + S

2 F e {C l}_{3} + H_{2} S ⟶ 2 F e {C l}_{2} + 2 H C l + S

2 F e {C l}_{3} + H_{2} S ⟶ 2 F e {C l}_{2} + 2 H C l + S

In the reaction, H₂S + Cl₂ → 2HCl + S; hydrogen sulphide is:

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