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




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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