Equation for the oxidation reaction for KMnO4 solution

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The discussion revolves around the oxidation reaction of KMnO4 in a dilute HCl solution, focusing on the oxidation of chloride ions (Cl-) to chlorine gas (Cl2). The oxidation half-equation is established as 2Cl- → Cl2 + 2e-, while the reduction half-equation for permanganate is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. Participants clarify the balancing of these half-reactions to achieve a net ionic equation, ultimately arriving at 2MnO4- + 16H+ + 10Cl- → 2Mn2+ + 8H2O + 10Cl2. Further discussions touch on other reactions involving KMnO4 and different reductants, emphasizing the importance of understanding oxidation states and balancing equations correctly. The conversation highlights the complexities of redox reactions and the need for careful attention to detail in chemical equations.
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Ok so the question goes as such: Write the half equation for the oxidation reaction and then the full net ionic equation for KMnO4 solution mixed with dilute HCL. My main problem is what is getting oxidised?
 
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Chlorides.

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chlorides are getting oxidised? how so?
 


It is just a matter of potentials, permanganate in solutions of pH low enough is a very strong oxidiser.

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ok so in this case.. the chlorine will have an oxidation state of minus one to begin with and an oxidation state of -2 to end with?
 


What is oxidation?

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sorry its a loss of electrons.. so the Cl- must lose and electron to become Cl?
 


so the oxidation half equation would be Cl- ---> Cl + e ?
 
  • #10


why Cl2 as opposed to Cl? So it oxidises chlorides to chlorine.. why?
 
  • #11


So the oxidation half equation would be:
2Cl- --> Cl2 + 2e ?
 
  • #12


Which gases are diatomic?

Edit: yes to your oxidation half reaction.

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  • #13


But the reduction equation will be: MnO4- + 8H+ + 5e ---> Mn2+ + 4H2O ... when the oxidation and reduction equations are added together the electrons won't cancel..
 
  • #15


uhh so multiply all the left side by 2 and the right by 5? then 10 electrons on each and they will cancel?
 
  • #16


oh no not that easy.. sorry.. i shall read over that link.. thankyou sooo much foir your help so far.. absolute champion!
 
  • #17


agrocadabra said:
uhh so multiply all the left side by 2 and the right by 5? then 10 electrons on each and they will cancel?

That will probably do - it may happen that coefficients in the reaction after will be not the smallest ones, but at least reaction would be correctly balanced.

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  • #18


Ok so I get..

2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) ---> 2Mn2+(aq) + 8H2O(l) + 10Cl2(g)

I've got three more of these to do.. ill be up all night.. argh!
 
  • #19


agrocadabra said:
2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) ---> 2Mn2+(aq) + 8H2O(l) + 10Cl2(g)

Divide everything by 2.

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  • #20


Ok so the next one is KMnO4 and Na2SO3 in the presence of dilute H2SO4.. however what the heck is the oxidation state of Na2SO3?
 
  • #21


The oxidation state of the SO3 should be 0.. but what about the Na2? Would the Na2 be assigned a negative oxidation state because of its greater affinity for electrions? so Na is usually + would the oxidation state of Na2 in this case be -2?
 
  • #22


Don't think in terms of ON, if you are going to balance thorugh half reactions.

Besides, there is no such thing as oxidation state of Na2SO3 - oxidation state is assigned to individual atoms, not molecules.

Na2SO3 is just a salt - it dissociates giving 2Na+ and SO32-. Na+ are just spectators.

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  • #23


so the Na and K ions are spectators in this one?
 
  • #24


Yes.

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  • #25


So sulphur is going to be oxidised.. what is the sulfur trioxide's formal charge? is it -2 or 0? if it is -2 then the sulfur has a ON of +4 how the heck do I know what its oxidation number goes to?
 
  • #26


is the sulfur oxidised to s+6 ions?
 
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  • #27


There is no trioxide here, I have alredy explained what is the ion that will be oxidised.

Yes, +6 will be the final ON of sulfur. No idea why you are using ON - they are not necessary when balancing through half reactions.

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  • #28


it seems weird that the sulfur would be oxidised... would it be that the SO3 ion is a spectator as well and that the hydrogen ions are oxidised to hydrogen gas?
 
  • #29


Why is it +6, why not something else, something higher? It just seems so unconcrete.. I'm using ON numbers because I don't know how else to go about it.
 
  • #30


So it would be S+4 ---> S+6 + 2e?
 
  • #31


Ok so I get: 2MnO4- + 16H+ + 5S+4 ---> 2Mn2+ + 4H2O + 5S+6
 
  • #32


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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  • #33


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?
 
  • #34


You start with SO32- and you end with SO42-. How are there electrons then to cancel out the 5 from the MnSO4?
 
  • #35


KMnSO4? No such animal.

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

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  • #36


2MnO(4-) + 16H(+) + 5SO3(2-) + 50(2-) ---> 5SO4(2-) + Mn(2+) + 4H2O ?
 
  • #38


agrocadabra said:
50(2-)

Where did you got it from?

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  • #39


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(-)?
 
  • #40


I thought Mn(2+) + IO4(-) ---> MnO4- + I- ?
 
  • #41


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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  • #42


well my notes seem to indicate IO4 has something to do with the oxidation..
 
  • #43


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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