[kbakersr]

I am having trouble understanding how to find the product side of the two half reactions for: MnO₄^- + SO₃^2- if they are reacted with eachother in HCL(aq).

The two half reactions are:
SO₃^2- --> SO₄^2-

MnO₄^- --> Mn²⁺

I don't understand why the manganese ends up with no oxygen and a 2+ charge. Also, did the sulphur tetroxide product get its four oxygen from the manganese reactant? I'm understanding how to balance everything after I get the two half reactions, but if I'm not given the product side of the reaction I usually have a lot of trouble figuring it out.

[Hunter2]

You have to add water and H⁺ in your equations. SO₄^2- is called Sulfate and not sulfurtetroxide.
So try to balance the first one by adding water on left side, what is the result on right side. You also have to balance the charges, this is done by electrones for negative charge and H⁺ for positiv charges

[kbakersr]

I know how to do the balancing part. i was wondering how to get the SO₄^2-  and Mn²⁺  on the product side of the equation if they are not given. Lets say i was only given SO₃^2-  +  MnO₄^-  and was told they react in together in HCL(aq).

[Borek]

SO₃^2- can't be oxidized to many things, SO₄^2- is (almost) the only possible product.

Permanganate is a strong oxidizer. Depending on pH it gets reduced to manganate, manganese dioxide or Mn²⁺. That you will have to remember.

[kbakersr]

thank you very much