Question

Justify the following reaction as a redox reaction.
$2Na\left( s \right) + S\left( s \right)\xrightarrow{{}}N{a_2}S\left( s \right)$
Find out the oxidizing and reducing agents.

Answer 1

Hint: We can define Oxidation and Oxidizing agent as,
In the oxidation process the electrons are lost from an atom. A compound that gains electrons during oxidation is known as oxidizing agent.
Reducing and Reducing agent are,
Reduction is the gain of electrons by an atom and a compound that loses electrons during reduction is called a reducing agent.

Complete step by step answer:
Let us see what a redox reaction is.
Redox reactions:
All the redox reactions have two parts, they are reduced half and an oxidized half which occur at the same time. The reduced half of the reaction accepts electrons and the oxidation number of the species gets decreased, while the oxidized half of the reaction loses electrons and the oxidation number of the species increases. There is no net variation in the number of electrons in a redox reaction.
Now calculate the oxidation state of sodium on the reactant and product side,
We know that the oxidation number of sodium metal is zero. Thus the oxidation number of sodium on the reactant side is zero.
We know that the oxidation number of sulfur is \[ - 2\]
The oxidation number of sodium in $N{a_2}S$ can be calculated as,
$2\left( x \right) - 2 = 0$
$x = \dfrac{2}{2} = 1$
The oxidation number of sodium in $N{a_2}S$ is \[1\]. Sodium loses electrons in the reaction and therefore, it acts as a reducing agent.
We know that the oxidation number of sulfur is zero. Thus the oxidation number of sulfur on the reactant side is zero.
We know that the oxidation number of sulfur is \[ - 2\]
The oxidation number of sulfur in $N{a_2}S$ is \[ - 2\]. Sulfur gains electrons and therefore it acts as an oxidizing agent.

Note:
Let us see few rules for oxidation numbers,
1.A free element will be zero as its oxidation number.
2.Monatomic ions will have an oxidation number equal to charge of the ion.
3.In hydrogen, the oxidation number is ${\text{ + 1}}$, when combined with elements having less electronegativity; the oxidation number of hydrogen is -1.
4.In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.
5.Group 1 elements will have +1 oxidation number.
6.Group 2 elements will have +2 oxidation numbers.
7.Group 17 elements will have -1 oxidation number.
8.Sum of oxidation numbers of all atoms in neutral compounds is zero.