In the reaction given below, identify the specie undergoing oxidation and reduction
$H_{2} S (g) + C l_{2} \to 2 H C l (g) + S (s)$

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Solution

The Oxidation number of Sulphur changes from $- 2$ to $0$ . So Sulphur is oxidised.

The Oxidation number of Chlorine changes from $0$ to $- 1$ . So Chlorine is reduced

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Similar questions

H_{2} S (g) + C l_{2} (g) \to 2 H C l (g) + S (s)

H_{2} S (g) + {C l}_{2} (g) \to 2 H C l (g) + S (s)

C l O_{3}^{-} + C l^{-} \to C l_{2} + C l O_{2}

Identify from the reaction the reactants that undergo oxidation and reduction.

$F e + S \to F e S$

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