So2 + kmno4 + h2o = mnso4 + k2so4 + h2so4

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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.

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

Acidic nature of $ce{SO2}$:

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

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?



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).$$ce{2KMnO4 + 3H2SO4 -> K2SO4 + 2MnSO4 + 3H2O + 5}$$

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.


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