# Reaction of sulphur dioxide with solution of potassium permanganate

Solution of $\ce{KMnO4}$ is decolorised by $\ce{SO2}$. Is this due to the reducing nature of $\ce{SO2}$ or acidic nature of $\ce{SO2}$ ?

I encountered following two explanations. While first one appears to be reasonable, I could not understand the second one.

Reducing nature of $\ce{SO2}$: $$\ce{2KMnO4 +5SO2 +2H2O -> K2SO4 + 2MnSO4 +2H2SO4}$$

Acidic nature of $\ce{SO2}$:

\begin{align} \ce{SO2 + 2H2O &-> H2SO4 + 2H}\\ \ce{2H + O &-> H2O}\\ \ce{2KMnO4 +3H2SO4 &-> K2SO4 + 2MnSO4 +3H2O + 5[O]}\\ \end{align}

Also, it is said that $\ce{H2SO4}$ is a moderately strong oxidizing agent. How can $\ce{H2SO4}$ reduce $\ce{KMnO4}$ to $\ce{MnSO4}$ (last equation)? Is the second explanation correct?

• Your first equation under "Acidic nature" is incorrect. $\ce{SO2 + H2O -> H2SO3 + H2O}$ Since now there's no nascent hydrogen atoms, the second reaction won't occur, and hence the entire "Acidic nature" reasoning has failed. – Gaurang Tandon Mar 1 '18 at 8:35

Potassium permanganate is a very strong oxidising agent (stronger than sulphuric acid). In acidic medium, $\ce{Mn}$ gets reduced from its +7 oxidation state to +2 oxidation state. (n-factor is 5).
$$\ce{2KMnO4 + 3H2SO4 -> K2SO4 + 2MnSO4 + 3H2O + 5[O]}$$
Since $\ce{SO2}$ is a strong reducing agent, it reduces $\ce{Mn}$ as denoted by the equation $$\ce{5SO2 + 2KMnO4 + 2H2O -> 2H2SO4 + 2MnSO4 + K2SO4}$$
Both the reactions are viable. Because of the formation of $\ce{MnSO4}$ and the disappearance of $\ce{KMnO4}$, decolorisation occurs.
But sulphur dioxide reacts with water to produce sulphurous acid and not sulphuric acid. $$\ce{ SO2 + H2O —> H2SO3}$$ Hence, the acidic nature reasoning doesn’t hold good.
• (1) You can say oxygen is getting oxidised from -2 to 0 because of the formation of nascent oxygen (Although I am not sure about it) (2) $\ce{H2SO4}$is formed by contact process. (en.m.wikipedia.org/wiki/Contact_process) and also by reacting$\ce{SO3}$ with water but the latter is highly exothermic. (3) Read gaurang’s comment on your question. That was why I said it can form only $\ce{H2SO}$ in my answer. I ll add that also. – MollyCooL Mar 1 '18 at 10:34