Question

Verify the change in oxidation number \[2C{{u}_{2}}{{O}_{\left( s \right)}}\text{ }+\text{ }C{{u}_{2}}{{S}_{\left( s \right)}}\text{ }\to \text{ }6C{{u}_{\left( s \right)}}\text{ }+\text{ }S{{O}_{2}}_{\left( s \right)}\]?

Answer 1

Hint: The change in the oxidation number of an element refers to the alteration in the oxidation number of the same element between the reactant side and the product side. An increase in the oxidation state of an element denotes its oxidation and a decrease in the oxidation state of an element denotes its reduction.

Complete answer:
The given equation is a balanced chemical equation where Cuprous oxide reacts with Copper sulphide to form Copper and sulphur-di-oxide. On both sides the number of atoms of copper is 6, oxygen is 2 and sulphur is 1.
We know that the oxidation state of a compound as a whole is zero. To understand the change in the oxidation let’s find out the oxidation state of the element on the reactant side and then on the product side.
On the reactant side:
The oxidation number of oxygen is -2. Hence the oxidation number of copper will be,
$2\text{ }\times \text{ }\left( Oxidation\text{ No}\text{. of Cu} \right)\text{ + }\left( -2 \right)\text{ = 0}$
$2\text{ }\times \text{ }\left( Oxidation\text{ No}\text{. of Cu} \right)\text{ = +2}$
$\left( Oxidation\text{ No}\text{. of Cu} \right)\text{ = +}\dfrac{2}{2}$
$\left( Oxidation\text{ No}\text{. of Cu} \right)\text{ = +1}$
Let’s find the oxidation number of Sulphur in Copper sulphide,
$2\text{ }\times \text{ }\left( +1\text{ } \right)\text{ + }\left( Oxidation\text{ No}\text{. of S} \right)\text{ = 0}$
$\left( +2\text{ } \right)\text{ + }\left( Oxidation\text{ No}\text{. of S} \right)\text{ = 0}$
$\left( Oxidation\text{ No}\text{. of S} \right)\text{ = -2}$
The oxidation state of Copper in Cuprous oxide (\[C{{u}_{2}}O\]) and Copper sulphide \[(C{{u}_{2}}S)\] is +1, Sulphur and Oxygen is -2.
On the product side:
The oxidation number of elemental copper is 0 and that of Oxygen is -2.
The oxidation number of Sulphur in sulphur-di-oxide is,
$\left( Oxidation\text{ No}\text{. of S} \right)\text{ + 2 }\times \text{ }\left( -2 \right)\text{ = 0}$
$\left( Oxidation\text{ No}\text{. of S} \right)\text{ + -4 = 0}$
$\left( Oxidation\text{ No}\text{. of S} \right)\text{ = -4 }$
The oxidation state of Copper is 0, Sulphur is +4 and Oxygen is -2.
So, in the above reaction,
The oxidation number of Copper decreased from +1 to 0. Thus Copper is reduced. Therefore, the change is by -1.
The oxidation number of Sulphur increased from -2 to +4. Thus Sulphur is oxidized. Therefore, the change is by +6.

Note:
The oxidation number of Oxygen can be used to find out the oxidation number of other elements in a compound. This is because the oxidation number of Oxygen is fixed to -2. However, the oxidation state of transition elements is variable.