Question

\[{\text{S}}{{\text{O}}_2}\] passed through an acidic potassium permanganate solution (purple in colour) turns it ____.
A) Blue
B) Colourless
C) Remains same
D) Black

Answer 1

Hint: The colour change of a compound occurs when \[{\text{S}}{{\text{O}}_2}\] is passed through acidic potassium permanganate solution thus, it is a redox reaction. In redox reactions, the oxidation number of metal ions is changed. This changes the colour as a result of change in oxidation number.

Complete step by step answer:
We know that the colour of acidified potassium permanganate solution changes when \[{\text{S}}{{\text{O}}_2}\] is passed through it. Initially the acidified potassium permanganate solution is purple in colour.
-When \[{\text{S}}{{\text{O}}_2}\] is passed through acidic potassium permanganate solution, potassium sulphate is formed along with manganese sulphate.
The reaction of acidic potassium permanganate solution with \[{\text{S}}{{\text{O}}_2}\] is as follows:

\[{\text{2KMn}}{{\text{O}}_4} + {\text{5S}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}} \to {{\text{K}}_2}{\text{S}}{{\text{O}}_4} + {\text{2MnS}}{{\text{O}}_4} + 2{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]

In \[{\text{KMn}}{{\text{O}}_4}\], the oxidation number of \[{\text{Mn}}\] is $ + 7$. After the reaction, in \[{\text{MnS}}{{\text{O}}_4}\], the oxidation number of \[{\text{Mn}}\] is $ + 2$. Thus, the oxidation number of \[{\text{Mn}}\] decreases from $ + 7$ to $ + 2$.

-We know that the decrease in oxidation number of the element is known as reduction. The element whose oxidation number decreases gets reduced and thus, it acts as an oxidising agent.
-The decrease in oxidation number suggests that \[{\text{Mn}}\] is getting reduced. The manganate ions get reduced from $ + 7$ to $ + 2$. This makes the acidic potassium permanganate solution colourless.
The formation of manganese sulphate \[\left( {{\text{MnS}}{{\text{O}}_4}} \right)\] makes the solution colourless.
-Thus, \[{\text{S}}{{\text{O}}_2}\] passed through an acidic potassium permanganate solution (purple in colour) turns it colourless.

Thus, the correct option is (B) colourless.

Note: We know that the decrease in oxidation number of the element is known as reduction. The element whose oxidation number decreases gets reduced and thus, it acts as an oxidising agent. The manganate ions get reduced from $ + 7$ to $ + 2$. Thus, potassium permanganate is a strong oxidising agent.