Question

Name the substance oxidised and reduced in the given reaction
${\text{C}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}} \to {\text{2HCl + S}}$

Answer 1

I must tell you that the oxidation number for the element that is pure is considered as zero. If the hydrogen is attached to the non metal then the oxidation number of that compound is +1. In the neutral molecule we see that the addition of the oxidising number of all the elements is equal to zero.

Complete step by step answer:
So here the chlorine molecule has its oxidation state equal to zero because it is a pure element. If we assume that the oxidation state of the chlorine in hydrochloric acid is x. The hydrogen’s oxidation state in the hydrochloric acid is +1. As we see the hydrogen is being attached to the electronegative element so its oxidation state will be +1. As in the neutral molecule the addition of the oxidation state of all the elements is zero. Now let us calculate the oxidation state of chlorine in hydrochloric acid:
$( + 1) + (x) = 0$
$x = - 1$
Here the oxidation state of chlorine in the hydrochloric acid is -1. So we see that the oxidation state decreases of chlorine here which indicates that it is a reduction reaction. As the oxidation state changes from 0 to -1. So the oxidised substance is hydrochloric acid.
Whereas in the reaction we see that the oxidation state of sulphur in hydrogen sulphide is 2-. But in the product the oxidation number is 0. This shows it has undergone oxidation and the oxidised substance is sulphur as the oxidation state is increased.

Note: We see in the redox reactions where oxidation and reduction occurs simultaneously. This reaction is also an example of redox reaction because chlorine is undergoing the reduction process. The hydrogen sulphide is undergoing the process of oxidation as sulphur is increasing the oxidation state from 2- to 0.