$Z n (s) + C u {C l}_{2} (a q) \to Z n {C l}_{2} (a q) + C u (s)$
In the reaction shown above, which substance is reduced?

C u

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C u {C l}_{2}

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Z n

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Z n {C l}_{2}

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Solution

The correct option is D $C u {C l}_{2}$
Here ,
$C u C l_{2}$ is reduced.
Oxidation state of $C u$ in $C u C l_{2}$ is $⟹ x + 2 (- 1) = 0, x = + 2.$
But in $C u$ oxidation state is $0$ .
So, it reduced from $+ 2$ to $0$ .

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