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Question

The most commonly used reducing agent is:


A
AlCl3
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B
PbCl2
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C
SnCl4
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D
SnCl2
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Solution

The correct option is C $$SnCl_2$$
+4 oxidation state of $$Sn$$ is more stable than +2 oxidation state. 
Therefore, $${Sn}^{2+}$$ can be easily oxidised to $${Sn}^{4+}$$ and hence $$Sn{Cl}_{2}$$ acts a reducing agent.
$$Sn{Cl}_{2} + 2Cl \longrightarrow Sn{Cl}_{4} + 2{e}^{-}$$

Chemistry

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