# Reaction between manganese(IV) oxide and sulfuric acid

$$\ce{2MnO2 + 2H2SO4 -> 2MnSO4 + O2 ^ + 2H2O}$$

How does this reaction occur? We get the oxygen moiety of the manganese dioxide suddenly oxidized to gaseous oxygen, while the manganese gets reduced from +4 to +2.

The sulfuric acid is apparently unperturbed: its proton $\ce{H+}$ and its $\ce{SO4^{2-}}$ parts remain the same in terms of their oxidation states. What role does the sulfuric acid play here at all?

• Mechanisms in inorganic chemistry are often not well-studied and not well-characterised. There are thousands of possible compounds that you can form with varying amounts of oxygen and varying charges and you can bet that the reaction produces a lot of them. They do not involve "arrow pushing" like in organic chemistry and often you don't get much chemical insight from them. This is an example of what I am talking about: i.stack.imgur.com/euf6M.png I don't think there is a definitive answer to this question. Jun 2, 2016 at 19:45
• I edited your question, as it isn't really about exact mechanism Jun 3, 2016 at 0:47

I am afraid I cannot provide the exact mechanism. Nevertheless, here is what I know.

The reaction is most certainly more complex than the equation

$$\ce{2 MnO2_{(s)} + 2 H2SO4_{(aq)} -> 2 MnSO4_{(aq?)} + 3 O2_{(g)} + H2O_{(l,g)}}$$

suggests. One might say it is even misleading, but it is definitely not the whole story.

• A purple smoke

After adding concentrated sulfuric acid to manganese(IV) oxide, a purplish smoke manifests. This suggests to the species $\ce{MnO4-}$. Yet, what is the balancing particle? Is it $\ce{Mn^2+}$, or rather $\ce{H}$? I have no clue. $$\ce{2 Mn_{(aq)}^2+ + 5 MnO2_{(s)} + 6 H3O+_{(aq)} -> 2 HMnO4_{(aq)} + 5 Mn^2+_{(aq)} + 8 H2O_{(l,g)}}$$

• A green to brown mixture

The initial mixture is green. This points to either $\ce{MnO4^2-}$ or $\ce{Mn2O7}$, even $\ce{MnO4^3-}$ could be it. Again, which one is it? Maybe it is all of the above? Upon standing, we observe a brown precipitate. The manganese(IV) oxide appears to reform … Speculation (more in the next chapter):

$$\ce{MnO4_{(aq)}^2- + H2SO4_{(aq)} -> H2MnO4_{(aq)} + SO4_{(aq)}^2-}\\\ce{3 H2MnO4_{(aq)} -> 2 HMnO4_{(aq)} + MnO2_{(s)} + 2 H2O_{(l,g)}}$$

$$\ce{3 MnO4^2- + 2 H2O_{(l)} -> 2 MnO4- + MnO2_{(s)} + OH_{(aq)}-}$$

• Oxygen, where's that from?

I did the classic match experiment; it did indeed burn more vigorously near the reaction mixture.

$$\ce{HMnO4_{(aq)} -> MnO2_{(s)} + 3 O2_{(g)} + 2 H2O_{(l,g)}}$$ $$\ce{2 MnO4^2- -> MnO4^3- + MnO2_{(s)} + O2_{(g)}}$$ $$\ce{2 Mn2O7_{(l)} -> 4 MnO2_{(s)} + 3 O2_{(g)}}$$

• I think you get the idea

Sorry I could not be of more assistance. The mixture is indeed complex with multiple processes happening in parallel. (This is actually the case with every reaction.) It should at least point you in some direction.

• Extra

• I was a bit loosey–goosey with the ionic equations; you should not do that ;). In my opinion, it made the process more clear.
• One might also wish to consider $\ce{Mn2MnO4}$ or $\ce{Mn3O4}$ which, I believe, are also brown.
• Also, as @orthocresol mentioned, there are too many combinations, generally, which is why there aren't many arrow-pushing mechanisms in inorganic chemistry. Glance at the link Orthocresol provided. Jun 2, 2016 at 20:37