Question

There is also formation of $CuS$ (black), when copper reacts with concentrated ${H_2}S{O_4}$ . What is the change in oxidation number of sulphur?

Answer 1

Hint : The number of electrons lost or gained by an element during the formation of a chemical bond, is known as its oxidation number or oxidation state. After the formation of a chemical bond, if the oxidation number of an element increases then it is termed as oxidation whereas if the oxidation number of an element decreases, then it is termed as reduction.

Complete Step By Step Answer:
When copper reacts with concentrated sulphuric acid, then the formation of copper sulphide and copper sulphate takes place along with the removal of water. The reaction proceeds as follows:
$4{H_2}S{O_4} + 4Cu \to CuS + 3CuS{O_4} + 4{H_2}O$
It is a redox reaction i.e.; the reduction and oxidation of elements is taking place simultaneously in the reaction.
Let the oxidation number of sulphur in ${H_2}S{O_4}$ be x.
Therefore,
$x + 2 + 4 \times ( - 2) = 0$
$\Rightarrow x - 6 = 0$
$\Rightarrow x = + 6$
Let us consider the oxidation state of sulphur in $CuS$ be y.
As we know, that copper generally exists in its $+ 2$ oxidation state, therefore the oxidation state of sulphur will be as follows:
$y + 2 = 0$
$\Rightarrow y = - 2$
Change in oxidation number $= x - y$
$\Rightarrow 6 - ( - 2)$
$\Rightarrow 8$
Hence, the change in the oxidation number of sulphur observed during the reaction is $8$ .

Note :
In a redox reaction, when a compound itself gets reduced to oxidized another molecule, then it is known as oxidizing agent. For the given reaction, as the oxidation number of sulphur in concentrated ${H_2}S{O_4}$ reduces from $+ 6$ to $- 2$ , so during the reaction, sulphuric acid itself is reducing but acts as an oxidizing agent and oxidized copper from $0$ to $+ 2$ oxidation state.