# When zinc is added to $CuS{O_4}$ solution, copper is precipitated because:A.$C{u^{2 + }}$ is reduced B.$C{u^{2 + }}$ is oxidised C.$CuS{O_4}$ is ionised D.$CuS{O_4}$ is hydrolysed

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Hint: It is a displacement reaction when zinc reacts with copper sulphate and displaces copper from copper sulphate and forms zinc sulphate. It is a redox reaction as well because the oxidation number of copper changes from +2 to 0 and the oxidation number of zinc changes from 0 to +2.

$Zn(s) + CuS{O_4}(aq) \to ZnS{O_4}(aq) + Cu(s)$
We can see from the equation that when zinc forms zinc sulphate, its oxidation state changes from zero to +2 and when copper sulphate changes from +2 to zero. The substance which is reduced in a reaction is called the oxidizing agent as it gains electrons. The substance which is oxidized in a reaction is called a reducing agent as it loses electrons. Here, copper gains electron, so we can say that $C{u^{2 + }}$ is reduced whereas zinc loses two electrons to become $Z{n^{2 + }}$ , then we say it is oxidized.
$C{u^{2 + }} + 2{e^ - } \to Cu$
$Zn \to Z{n^{2 + }} + 2{e^ - }$