2 Cu_2O_{(s)} + Cu_2S_{(s)} longrightarrow 6 | vedclass

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(C) In the reaction $2 Cu_2O_{(s)} + Cu_2S_{(s)} \longrightarrow 6 Cu_{(s)} + SO_{2(g)}$:$1$. The oxidation state of $Cu$ in $Cu_2O$ and $Cu_2S$ is $+

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$Cu(I)$ of $Cu_2O$ and $Cu_2S$,and sulphide of $Cu_2S$

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(C) In the reaction $2 Cu_2O_{(s)} + Cu_2S_{(s)} \longrightarrow 6 Cu_{(s)} + SO_{2(g)}$:$1$. The oxidation state of $Cu$ in $Cu_2O$ and $Cu_2S$ is $+1$. It decreases to $0$ in $Cu_{(s)}$,so $Cu(I)$ undergoes reduction and acts as an oxidizing agent (oxidant).$2$. The oxidation state of $S$ in $Cu_2S$ is $-2$. It increases to $+4$ in $SO_{2(g)}$,so the sulphide ion undergoes oxidation and acts as a reducing agent (reductant).Therefore,$Cu(I)$ of $Cu_2O$ and $Cu_2S$ acts as the oxidant,and the sulphide of $Cu_2S$ acts as the reductant.

$2 Cu_2O_{(s)} + Cu_2S_{(s)} \longrightarrow 6 Cu_{(s)} + SO_{2(g)}$
In the above reaction,the oxidant and reductant respectively are:

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$2 Cu_2O_{(s)} + Cu_2S_{(s)} \longrightarrow 6 Cu_{(s)} + SO_{2(g)}$In the above reaction,the oxidant and reductant respectively are: