Copper sulphate is prepared by blowing a curr | vedclass

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(B) Copper is a metal with a positive standard reduction potential $(E^0_{Cu^{2+}/Cu} = +0.34 \ V)$,meaning it does not react with dilute $H_2SO_4$ to

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to oxidise $Fe^{2+}$ to iron $(III)$ sulphate,which remains in solution after crystallisation of $CuSO_4$

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(B) Copper is a metal with a positive standard reduction potential $(E^0_{Cu^{2+}/Cu} = +0.34 \ V)$,meaning it does not react with dilute $H_2SO_4$ to release hydrogen gas.To prepare $CuSO_4$,copper scrap is treated with dilute $H_2SO_4$ in the presence of air (oxygen) or a small amount of $HNO_3$.$HNO_3$ acts as an oxidizing agent to oxidize $Cu$ to $Cu^{2+}$ ions,which then react with $SO_4^{2-}$ ions from $H_2SO_4$ to form $CuSO_4$.Additionally,copper scrap often contains iron impurities. $HNO_3$ oxidizes $Fe^{2+}$ to $Fe^{3+}$,which forms iron $(III)$ sulphate. This is beneficial because iron $(III)$ sulphate remains in the solution during the crystallization of $CuSO_4$,allowing for the separation of pure copper sulphate.

List-$I$ (Method)	List-$II$ (Metal refined)
$(A)$ Zone refining	$(I)$ $Sn$
$(B)$ Liquation	$(II)$ $In$
$(C)$ Vapour phase refining	$(III)$ $Zn$
$(D)$ Distillation	$(IV)$ $Zr$
-	$(V)$ $Al$

Copper sulphate is prepared by blowing a current of air through copper scrap and dilute $H_2SO_4$. Dilute $HNO_3$ is also added:

Similar Questions

Match the following List-$I$ (Method) with List-$II$ (Metal refined):
List-$I$ (Method) List-$II$ (Metal refined)
$(A)$ Zone refining $(I)$ $Sn$
$(B)$ Liquation $(II)$ $In$
$(C)$ Vapour phase refining $(III)$ $Zn$
$(D)$ Distillation $(IV)$ $Zr$
- $(V)$ $Al$

The correct answer is:

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