The most commonly used reducing agent among the following 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 D $S n C l_{2}$
+4 oxidation state of Sn is more stable than +2 oxidation state.
Therefore, $S n^{2 +}$ and hence $S n C l_{2}$ acts a reducing agent.
$S n C l_{2} + C l_{2} \to S n C l_{4} + 2 e^{-}$

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