Which element is reduced in this reaction: $$\ce{2KMnO_4 +3NaSO_3 +H_2O ->2MnO_2 +3NaSO_4 +2KOH}$$

I am trying to figure out which elements lose their electrons, but how am I supposed to know when elements are neutral and when they are ionized? I have tried the following, but I don't believe it to be correct

$$\ce{2K^+Mn^+O_4^{-2} +3Na^+SO_3^{-2} +H_2O ->2Mn^{+2}O_2^{-2} +3Na^{+}SO_4{-2} +2K^+OH^-}$$

My reasoning being that:

  1. Oxygen atoms are always doubly anionized
  2. Sulfate and sulfite are also doubly ionized
  3. Sodium and potassium are always singly cationized

Based on these rules I tried to make the bonds neutral, but my assumptions must be incorrect. Any help would be greatly appreciated


1 Answer 1


Your reasoning is not entirely right. In those compounds, O is always (-II), K and Na always (+I). However, it's not true for S, which can have different oxidation states. So does Mn.

So, in KMnO4, Mn must be (+VII). In MnO2, it is (+IV). Thus, Mn is reduced.

For sulfur: in NaSO3, it must be be (+V). In NaSO4, it must be (+VII). Thus, sulfur is oxidized.

However (big however, in fact): the sulfate anion is not SO4, but SO42–. Similarly, sulfite is SO32–, not SO3. So, your equation should be instead:

$$\ce{2KMnO_4 +3 Na_2SO_3 +H_2O -> 2MnO_2 +3Na_2SO_4 +2KOH}$$

which means that sulfur is oxidized from (+IV) to (+VI).


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