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$SnCI_{2} + 2HgCI_{2} ---> Hg_{2}CI_{2} + SnCI_{4}$ In the given reaction:

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$SnCI_{2} + 2HgCI_{2} ---> Hg_{2}CI_{2} + SnCI_{4}$
In the given reaction:
1). $HgCI_{2}$ is oxidised
2). $SncI_{2}$ is oxidised
3). $Hg_{2}CI_{2}$ is oxidised
4). $HgCI_{2}$ is reduced


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