When SO_2 is passed through cupric chloride ( | vedclass

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(C) When $SO_2$ is passed through a solution of cupric chloride $(CuCl_2)$,it acts as a reducing agent. The reaction is: $2CuCl_2 + SO_2 + 2H_2O \righ

Vedclass · Accepted Answer

$C$. The solution becomes colourless and a white precipitate of $Cu_2Cl_2$ is obtained

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(C) When $SO_2$ is passed through a solution of cupric chloride $(CuCl_2)$,it acts as a reducing agent. The reaction is: $2CuCl_2 + SO_2 + 2H_2O \rightarrow Cu_2Cl_2 + H_2SO_4 + 2HCl$. In this reaction,$Cu^{2+}$ is reduced to $Cu^+$,resulting in the formation of a white precipitate of cuprous chloride $(Cu_2Cl_2)$. As the blue $Cu^{2+}$ ions are converted to $Cu^+$,the solution loses its blue colour and becomes colourless.

List-$I$	List-$II$
$(a) \ [S_2 O_3]^{2-}$	$(I)$ Paramagnetic
$(b) \ H_2 O_2$	$(II)$ Toxic gas
$(c) \ H_2 S$	$(III)$ Strong reducing agent
$(d) \ S_2$	$(IV)$ Disproportionate in presence of $Mn^{2+}$

When $SO_2$ is passed through cupric chloride $(CuCl_2)$ solution,what is the observation?

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Match the following items of List-$I$ with those in List-$II$ and choose the correct option.
List-$I$ List-$II$
$(a) \ [S_2 O_3]^{2-}$ $(I)$ Paramagnetic
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$(c) \ H_2 S$ $(III)$ Strong reducing agent
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