Equation for the oxidation reaction for KMnO4 solution

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Ok so I get: 2MnO4- + 16H+ + 5S+4 ---> 2Mn2+ + 4H2O + 5S+6
 
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No, you can't remove single atom from the complex ion and use it to balance balance reaction. You start with SO32- and you end with SO42-.

agrocadabra said:
Why is it +6, why not something else, something higher?

Because these are properties of sulfuur, properties that you can check looking at the periodic table.

I'm using ON numbers because I don't know how else to go about it.

I gave you a link to a page where half reactions are explained.

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Ok so the third is KMnSO4 solution mixed with dilute H2O2 in the presence of dilute H2SO4:

now the H2O2 is the reductant right.. so we have H+ ---> Hn+ + n electrons.. how do we know what it is oxidised to?
 


You start with SO32- and you end with SO42-. How are there electrons then to cancel out the 5 from the MnSO4?
 


KMnSO4? No such animal.

Strange as it sounds, hydogen peroxide is oxidised to oxygen.

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2MnO(4-) + 16H(+) + 5SO3(2-) + 50(2-) ---> 5SO4(2-) + Mn(2+) + 4H2O ?
 


ok so mix diluted hadfield solution (steel dissolved in
nitric acid followed by treatment with ammonium
peroxydisulfate to oxidize carbon and treated with Sn(2+)
to reduce the Fe(3+) to Fe(2+) ) in a conical flask with
some H2SO4 and some KIO4.. the whole idea is to find out
the amount of manganese in the steel.. What is the
oxidation half equation that converts Mn(2+) to MnO4(-)?
 


I thought Mn(2+) + IO4(-) ---> MnO4- + I- ?
 


Strange, as far as I know Mn2+ can be oxidized to permanganate by ammonium peroxydisulfate in the presence of catalytic amounts of Ag+; no need for further Fe and oxidation by periodate.

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well my notes seem to indicate IO4 has something to do with the oxidation..
 


It is a strong oxidizer, but I don't see a point in using it here. Perhaps you have not shown complete procedure.

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