Question

What happens when ${\mathrm{SO}}_2$ is passed in an acidified ${\mathrm{KMnO}}_4$ solution?

Answer 1

Reduction:

1. When ${\mathrm{SO}}_2$ is passed through an acidic potassium permanganate solution, the color change of the solution occurs thus it is a redox reaction. The oxidation number of metal ions changes during redox processes.
2. Potassium sulfate and manganese sulfate are produced when ${\mathrm{SO}}_2$ passed through an acidic potassium permanganate solution. $\underset{\mathrm{potassium}\mathrm{permagnate}}{\stackrel{+7}{2{\mathrm{KMnO}}_4\left(\mathrm{s}\right)}}+\underset{\mathrm{sulfur}\mathrm{dioxide}}{5{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{potassium}\mathrm{sulfate}}{{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{magnessium}\mathrm{sulfate}}{\stackrel{+2}{2{\mathrm{MnSO}}_4\left(\mathrm{s}\right)}}+\underset{\mathrm{sulfuric}\mathrm{acid}}{2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}$
3. The oxidation number of Manganese is in$+7$${\mathrm{KMnO}}_4$. After the reaction, the oxidation number of manganese ${\mathrm{MnSO}}_4$ is$+2$.
4. Reduction, as we know, is the process of reducing an element's oxidation number. The element which undergoes reduction is the oxidizing agent.
5. This shows that Manganese is being reduced as the oxidation number decreases from $+7$to$+2$. This makes the potassium permanganate solution, which is acidic, and colorless. The colorless solution is formed due to the formation of crystals of manganese sulfate ${\mathrm{MnSO}}_4$.

Therefore, when ${\mathrm{SO}}_2$ it is passed in an acidified ${\mathrm{KMnO}}_4$ solution a colorless solution is formed due to reduction.