Question

You might have noticed that when copper powder is heated in a china dish, the surface of copper powder becomes coated with black colour substance.
(i) How has this black-coloured substance formed?
(ii) What is that black substance?
(iii) Write the chemical equation of the reaction that takes place.

Answer 1

Hint: Copper metal is stable in air under normal conditions. But upon heating in air, copper reacts with the oxygen present in the air and forms a black-coloured substance on its surface. Refer to the redox reaction of copper with oxygen upon heating.

Complete step by step answer:
(i) When copper powder is heated in a china dish, the surface of copper powder becomes coated with a black-coloured substance. This black substance formed on the surface is because of the process called surface oxidation. Upon heating, the copper metal melts gradually and turns black on the surface. This is because the copper powder present in the china dish reacts with the atmospheric oxygen and forms copper (II) oxide. The compound formed copper (II) oxide is generally black in colour and thus, the surface of copper powder is coated with this black copper (II) oxide because of the surface oxidation of copper. Oxidation is the loss of electrons or addition of oxygen to a substance.
(ii) It is quite clear from the explanation in (i) part that the black substance formed on the surface of copper powder is copper (II) oxide. Chemical formula of copper (II) oxide is \[CuO\].
(iii) Redox reaction takes place upon heating of copper metal in air. Required chemical equation of the reaction taking place:
$2Cu(s) + {O_2}(g)\xrightarrow{{heat,\Delta }}2CuO(s)$
${\text{copper + oxygen(in air)}}\xrightarrow{{heat}}{\text{copper (II) oxide (black - coloured substance)}}$
In this redox reaction, oxygen is getting reduced and copper is getting oxidised.

Note: When hydrogen gas (${H_2}$) is passed over heated copper oxide (\[CuO\]), then the black-coloured copper (II) oxide is reduced and we get back brown copper metal. Chemical equation of this reaction is:
$CuO + {H_2} \to Cu + {H_2}O$