The reaction $H_{2} S + H_{2} O_{2} ⟶ 2 H_{2} O + S$ shows

Oxidizing action of $H_{2} S$

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Reducing action of $H_{2} S$

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Alkaline nature of $H_{2} S$

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Acidic nature of $H_{2} S$

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Solution

The correct option is B
Reducing action of $H_{2} S$

In this reaction $H_{2} O_{2}$ acts as an oxidising agent and $H_{2} S$ acts as a reducing agent as sulphur gains electron and changes its oxidation state from -2 to 0. As its oxidation state increases, it reduces other reactants.

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Similar questions

H_{2} S + H_{2} O_{2} \to 2 H_{2} O + S

H_{2} S + H_{2} O_{2} \to 2 H_{2} O + S

H_{2} S + H_{2} O_{2} \to S + 2 H_{2} O

H_{2} S + H_{2} O_{2} \to S + 2 H_{2} O

Equation $H_{2} S + H_{2} O_{2} \to S + 2 H_{2} O$ represents

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