Equation for the oxidation reaction for KMnO4 solution

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The discussion focuses on the oxidation reaction of potassium permanganate (KMnO4) in dilute hydrochloric acid (HCl). The oxidation half-equation for chloride ions (Cl-) is established as 2Cl- → Cl2 + 2e-, while the reduction half-equation for permanganate is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. The complete balanced equation is 2MnO4- + 16H+ + 10Cl- → 2Mn2+ + 8H2O + 10Cl2. Participants clarify the roles of spectator ions and oxidation states, emphasizing the importance of balancing half-reactions correctly.

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agrocadabra
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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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methods
 


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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methods
 
  • #23


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


Yes.

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methods
 
  • #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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methods
 
  • #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?
 

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