Question

What happens when Sulphur dioxide reacts with acidic potassium permanganate? Write balanced chemical equation for this reaction.

Hint:

Always balance redox reactions in acidic medium by adding H$_2$O, H$^+$, and electrons properly. KMnO$_4$ turns colourless as Mn$^{2+}$ is formed.

Answer 1

Step 1: Nature of reaction.
Sulphur dioxide (SO$_2$) acts as a strong reducing agent, while potassium permanganate (KMnO$_4$) in acidic medium acts as a strong oxidising agent. When they react, SO$_2$ is oxidised to sulphuric acid (H$_2$SO$_4$) and KMnO$_4$ is reduced to Mn$^{2+}$. Step 2: Balanced redox reaction.
In acidic medium: \[ 2MnO_4^- + 5SO_2 + 2H_2O \;\longrightarrow\; 2Mn^{2+} + 5SO_4^{2-} + 4H^+ \] Step 3: Reaction with potassium permanganate.
Since KMnO$_4$ provides MnO$_4^-$ ions, the reaction is: \[ 2KMnO_4 + 5SO_2 + 2H_2O \;\longrightarrow\; K_2SO_4 + 2MnSO_4 + 2H_2SO_4 \] Observation.
The purple colour of KMnO$_4$ solution is discharged, showing that reduction of MnO$_4^-$ to Mn$^{2+}$ has taken place. Conclusion:
SO$_2$ reduces KMnO$_4$ in acidic medium, forming MnSO$_4$, K$_2$SO$_4$, and H$_2$SO$_4$, while the pink colour of permanganate disappears.