The most commonly used reducing agent is:

A l C l_{3}

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P b C l_{2}

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S n C l_{4}

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S n C l_{2}

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Solution

The correct option is C $S n C l_{2}$
+4 oxidation state of $S n$ is more stable than +2 oxidation state.
Therefore, ${S n}^{2 +}$ can be easily oxidised to ${S n}^{4 +}$ and hence $S n {C l}_{2}$ acts a reducing agent.
$S n {C l}_{2} + 2 C l ⟶ S n {C l}_{4} + 2 e^{-}$

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Similar questions

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

S n C l_{2} + 2 F e C l_{3} \to S n C l_{4} + 2 F e C l_{2}

P b C l_{4}

S n C l_{4}

P b C l_{2}

S n C l_{2}

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

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