Molecular weight of Cu_2S is M. In the follow | vedclass

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(C) To find the equivalent weight,we need to determine the change in oxidation state of the species in $Cu_2S$.The oxidation states are: $Cu$ in $Cu_2

Vedclass · Accepted Answer

$8$

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(C) To find the equivalent weight,we need to determine the change in oxidation state of the species in $Cu_2S$.The oxidation states are: $Cu$ in $Cu_2S$ is $+1$,$S$ in $Cu_2S$ is $-2$.In the products: $Cu$ is $+2$ and $S$ in $SO_2$ is $+4$.The change in oxidation state for $Cu_2S$ is calculated as:$2 	imes (Cu^{+1} ightarrow Cu^{+2}) = 2 	imes 1 = 2$$1 	imes (S^{-2} ightarrow S^{+4}) = 1 	imes 6 = 6$Total change in oxidation state = $2 + 6 = 8$.Since the equivalent weight is $M / n$-factor,and the $n$-factor is $8$,the value of $x$ is $8$.

Molecular weight of $Cu_2S$ is $M$. In the following reaction its equivalent weight is $M / x$. What is $x$?
$Cu_2S + MnO_4^{-} \rightarrow Cu^{2+} + Mn^{2+} + SO_2$

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Molecular weight of $Cu_2S$ is $M$. In the following reaction its equivalent weight is $M / x$. What is $x$?$Cu_2S + MnO_4^{-} \rightarrow Cu^{2+} + Mn^{2+} + SO_2$