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.

Column $-I$	Column $-II$
$(i)$ Iron and copper	$(P)$ Poling
$(ii)$ Zirconium and Titanium	$(Q)$ Bessemerisation
$(iii)$ Lead and Tin	$(R)$ Van Arkel
$(iv)$ Copper and Tin	$(S)$ Liquation

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

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Match Column $-I$ with Column $-II$ and select the correct answer using the codes given below.

Column $-I$ Column $-II$

$(i)$ Iron and copper $(P)$ Poling

$(ii)$ Zirconium and Titanium $(Q)$ Bessemerisation

$(iii)$ Lead and Tin $(R)$ Van Arkel

$(iv)$ Copper and Tin $(S)$ Liquation

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