# Which element is being reduced in this reaction?

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

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