The black color of copper oxide turns blue on treating with ${\text{HCl}}$ because of:
(A) Formation of cupric chloride
(B) Formation of water
(C) Release of hydrogen
(D) Release of carbon dioxide
Answer
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Hint: Copper (II) oxide or cupric oxide is an inorganic compound having the chemical formula ${\text{CuO}}$ . It is a solid with blackish color. It is one of the two stable oxides of copper, the other is cuprous oxide. It is produced by heating copper in air at a temperature range of around $300 - {800^\circ }{\text{C}}$ .
$2{\text{Cu}} + {{\text{O}}_2} \to 2{\text{CuO}}$
Complete step by step answer:
* Copper is a transition metal with atomic number 29. It belongs to the 3d series of transition elements. Although copper does not react with hydrochloric acid, its oxide copper oxide does react with hydrochloric acid. A metal - acid reaction always happens to be a redox reaction. Since the reduction potential of copper is higher than that of hydrogen, it does not react with non-oxidising acids like hydrochloric acid.
* But metal oxides are basic in nature and hence they react with acids producing the corresponding salts and water. These type of acid – base reactions or neutralization reactions are non-redox in nature.
* Thus, copper oxide being an oxide of the copper metal is basic and hence it reacts with hydrochloric acid to form the corresponding copper (II) chloride or cupric chloride salt along with the release of water.
${\text{CuO + 2HCl}} \to {\text{CuC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
Now, this cupric chloride ${\text{CuC}}{{\text{l}}_{\text{2}}}$ salt is blue in color and hence the whole resulting solution appears to be blue in color.
Thus, the black color of copper chloride turns blue on treating with hydrochloric acid due to the formation of the corresponding cupric chloride salt.
So, the correct option is A.
Note: (i) Cupric chloride can also be prepared commercially by the chlorination of copper. In this process, copper at red heat reacts with chlorine gas to give molten cupric chloride. ${\text{Cu(s) + C}}{{\text{l}}_{\text{2}}}{\text{(g)}} \to {\text{CuC}}{{\text{l}}_{\text{2}}}({\text{l}})$
(ii) Another method is to combine cupric oxide with ammonium chloride to give cupric chloride, ammonia and water.
${\text{CuO + 2N}}{{\text{H}}_{\text{4}}}{\text{Cl}} \to {\text{CuC}}{{\text{l}}_{\text{2}}}{\text{ + 2N}}{{\text{H}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
$2{\text{Cu}} + {{\text{O}}_2} \to 2{\text{CuO}}$
Complete step by step answer:
* Copper is a transition metal with atomic number 29. It belongs to the 3d series of transition elements. Although copper does not react with hydrochloric acid, its oxide copper oxide does react with hydrochloric acid. A metal - acid reaction always happens to be a redox reaction. Since the reduction potential of copper is higher than that of hydrogen, it does not react with non-oxidising acids like hydrochloric acid.
* But metal oxides are basic in nature and hence they react with acids producing the corresponding salts and water. These type of acid – base reactions or neutralization reactions are non-redox in nature.
* Thus, copper oxide being an oxide of the copper metal is basic and hence it reacts with hydrochloric acid to form the corresponding copper (II) chloride or cupric chloride salt along with the release of water.
${\text{CuO + 2HCl}} \to {\text{CuC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
Now, this cupric chloride ${\text{CuC}}{{\text{l}}_{\text{2}}}$ salt is blue in color and hence the whole resulting solution appears to be blue in color.
Thus, the black color of copper chloride turns blue on treating with hydrochloric acid due to the formation of the corresponding cupric chloride salt.
So, the correct option is A.
Note: (i) Cupric chloride can also be prepared commercially by the chlorination of copper. In this process, copper at red heat reacts with chlorine gas to give molten cupric chloride. ${\text{Cu(s) + C}}{{\text{l}}_{\text{2}}}{\text{(g)}} \to {\text{CuC}}{{\text{l}}_{\text{2}}}({\text{l}})$
(ii) Another method is to combine cupric oxide with ammonium chloride to give cupric chloride, ammonia and water.
${\text{CuO + 2N}}{{\text{H}}_{\text{4}}}{\text{Cl}} \to {\text{CuC}}{{\text{l}}_{\text{2}}}{\text{ + 2N}}{{\text{H}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
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