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Which of the following is correct for
$H_{2} S + {Cl}_{2} \to 2 HCl + S$

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$H_{2} S and {Cl}_{2} both are oxidised$

$H_{2} S and {Cl}_{2} both are reduced$

$H_{2} S is oxidised to S, {Cl}_{2} is reduced to HCl$

$H_{2} S is reduced to S, {Cl}_{2} is oxidised to HCl$

answer is C.

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Detailed Solution

$S^{2 -}$ changes to S by loss of two $e^{-} .$ ${Cl}_{2}$ gains two electrons to produce $2 {Cl}^{-}$ .Gain of $e^{-} is reduction, so, {Cl}_{2} is reduced .$

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