Manganese has an oxidation state of +7 in potassium permanganate. When reacted with sulphuric acid, how does it tend to have +2 oxidation state by forming $\ce{MnSO4}$ along other products ?

In acid, there is already tendency to donate proton and not electrons. Yet it forms +2. While in basic medium Manganese forms +7 to +6.

As I think, it shouldve formed +2 in basic by gaining 5 electrons. Where have I gone in my opinion?

  • 2
    $\begingroup$ KMnO4 is not the strongest oxidazing agent. Reaction with H2SO4 does not produce MnSO4, but oily and highly unstable Mn2O7. $\endgroup$
    – Poutnik
    Nov 1, 2020 at 14:36
  • $\begingroup$ i doubt your answer. im sure it does produce mnso4 $\endgroup$
    – Abdullah
    Nov 1, 2020 at 14:54
  • $\begingroup$ I admire your confidence, but I have doubts if it is backup up by knowledge. Yes, if Mn2O7 is reduced, or decomposes, it may finally end as MnSO4. $\endgroup$
    – Poutnik
    Nov 1, 2020 at 15:00
  • $\begingroup$ OTOH, Mn(II) in alkaline environment forms Mn(OH)2. It is reducing agent, quickly reacting with dissolved oxygen, what is used in analytic chemistry. $\endgroup$
    – Poutnik
    Nov 1, 2020 at 15:31

1 Answer 1


KMnO4 is reduced in strongly acidic environment by reducing agents ( like oxalic acid) as this:

$$\ce{2 MnO4- + 5 (COO)2^2- + 16 H3O+ ->[catalyzed Mn^2+] 2 Mn^2+ + 10 CO2 + 24 H2O}$$

It means, acids themselves do not convert it to Mn(II), unless they have reducing properties as well, like hydrochloric acid.

In alkaline solutions, manganese tends to reach Mn(III,IV) oxidation state, if oxidized from Mn(II), or Mn(VI,IV) states, if reduced from Mn(VII).


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