Balacncing using redox methodology

Homework Statement

Balance the following redox equation.
KMnO4(aq) + H2S (aq) + H2SO4(aq) --> K2SO4 (aq) + MnSO4(aq) + S(s)

Homework Equations

Knowledge of oxidation numbers etc.

The Attempt at a Solution

I simply just cannot figure out where to start and would like basic guidance.
Other questions I've done only included 1 and/or two products or reactants so I could simply create 2 half-reaction equations and easily solve from there. However I can't find a way to derive half-reaction equations for a reaction with threes terms.

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symbolipoint
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First guess is that the sulfuric acid is present to maintain acidic condition. The half reactions are the reduction of permanganate and the oxidation of H2S.

You will need to calculate the "oxidation" numbers for the parts undergoing change in their state of reduction. You need to account for the charge of the appropriate ions and the expected charges of the appropriate atoms. This helps you determine how many electrons are going in which direction, product or reactant, for each half-reaction. You use the number of electrons of the two half reactions to BALANCE the two half reactions. You still may need to account for the number of oxygen atoms and hydrogen atoms and possibly water molecules if those need balancing.

Hi, thanks for a start on it.

However, I do have a question. What do I do if the sulfur is oxidized twice? As to my knowledge S -> -2 then to +6 and 0 as a lone element.

symbolipoint
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I will show here the "oxidation" states of what's needed, but you figure out yourself how many electrons in each halfreaction:

KMnO4 + (?electrons) ------> K+ + MnSO4
(+7) -------> (+2)
For the change in the Mn

H2S ----------> S + (?electrons)
(-2) ---------> (0)
For the change in the S

(I just hope those figures aligned properly)

Yeah, I understand those well. First is 5e on the reactants side and second is 2e on the product side.

My question is how come I don't have to worry about S changing to +6 as well in H2SO4 and K2SO4?

symbolipoint
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Yeah, I understand those well. First is 5e on the reactants side and second is 2e on the product side.
You are making progress in understanding. Now you need to balance these two half reactions based on the number of electrons.

My question is how come I don't have to worry about S changing to +6 as well in H2SO4 and K2SO4?
As I said, the H2SO4 seems to be present just to give the appropriate acidic environment. I might be wrong, but no change seems to be expected in the reduction-or-oxidation state of anything involving the sulfate. You still need to balance the H+'s.

Borek
Mentor
You should start assigning oxidation numbers to all elements to find out what is being oxidized and what is being reduced, that's the most general approach in such cases.

SO42- is a just a spectator. How do I know? Well, a little bit of applied psychology. Question already lists something that is being oxidized (S2-) and oxidation product (S). That means whoever asked the question was not interested in other possible reactions.

How would I know in reality? I would not know. Permanganate is a strong oxidizer and I am not sure what can be the final product of sulfides oxidation; could be it is a sulfuric acid in some cases. It is probably a function of pH, concentration, temperature and perhaps some other factors.

dextercioby
Homework Helper

Homework Statement

Balance the following redox equation.
KMnO4(aq) + H2S (aq) + H2SO4(aq) --> K2SO4 (aq) + MnSO4(aq) + S(s)
Since you've written an equation with electrically neutral molecules, I can ask you where did the hydrogen go ?? It should be in the RHS as well.

symbolipoint
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Gold Member
Since you've written an equation with electrically neutral molecules, I can ask you where did the hydrogen go ?? It should be in the RHS as well.
The reaction first needs to be balanced. THEN accounting for the sulfate and the H+1 may be performed.

Borek
Mentor
The reaction first needs to be balanced. THEN accounting for the sulfate and the H+1 may be performed.
No idea what you mean, but what you wrote doesn't make sense. As long as H+ are not balanced, reaction is not balanced. bigubau is perfectly right, there is a product missing. Obvious one, but it is not there.

symbolipoint
Homework Helper
Gold Member
No idea what you mean, but what you wrote doesn't make sense. As long as H+ are not balanced, reaction is not balanced. bigubau is perfectly right, there is a product missing. Obvious one, but it is not there.
Then I did not catch what it was. I solved the original problem and could find the paper on which I did the work. I first balanced the two half reactions according to number of electrons, and then I checked numbers of atoms. I found I needed to put in maybe 14 H+ ions on the products side. As best I remember, everything appeared balanced, including the sulfates.

Borek
Mentor
If you added H+ on the RHS, you added a product. Not the one I would add (I would go for a neutral molecule), but you added it.

symbolipoint
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I rechecked my work on paper about this problem, and there I find one big mistake, about sulfate. But there is still another mistake in my solving method, and I still have been unable to properly balance for the hydrogen ions.

For forum discussion, you can rely on my work just until balancing the two half-reactions for the electrons. I am still trying to sort through the flaws after that part.

symbolipoint
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Another check into the half reaction for MnO4-1 part:

2KMnO4 + 10e- + 3H2SO4 --------> K2SO4 + 2MnSO4

Borek
Mentor
No idea what you are trying to do - this is not a half reaction. Half reactions are balanced, this is not.

symbolipoint
Homework Helper
Gold Member
Another check into the half reaction for MnO4-1 part:

2KMnO4 + 10e- + 3H2SO4 --------> K2SO4 + 2MnSO4
No idea what you are trying to do - this is not a half reaction. Half reactions are balanced, this is not.
That is based on what half reactions I showed in post #4. For the permanganate half-reaction, I multiplied by 2 (because 5 electrons), and I would multply the sulfide sulfur half-reaction by 5 (because 2 electrons). I am then using sulfates and sulfuric acid, since we are given that the problem is in acidic solution using sulfuric acid. This whole reaction (both half-reactions) is a 10-electron change. The K+ and the Mn+2 are given the SO4-2 as their counter-ions. Maybe the problem could be handled better by eliminating any spectator ions; not sure, but I have looked at a similar example in an old General Chemistry book which eliminated spectator ions, although no mention of acidic conditions was shown in the example.

Borek
Mentor
As I wrote above - half reaction has to be balanced. Electrons exchanged by the half reaction are calculated not using oxidation numbers. You should first balance all elements, then use electrons to balance charge.

symbolipoint
Homework Helper
Gold Member
I checked in a standard table of reduction potentials and tried this again referring just to the half-reactions shown there, and as found there, without any inclusion of H2SO4 or K+ or SO4-2.

Acidic Permanganate:
MnO4- + 8H+ + 5e -----> Mn+2 + 4H20

Hydrogen Sulfide:
H2S -----> S0 + 2H+ + 2e

I then balanced based on the electrons (this is still not the way you say to do it, but I comment on this last/later). Balancing electrons and then combining the two half reactions gives:

2MnO4- + 16H+ + 5H2S -------> 2Mn+2 + 8H2O + 5S0 + 10H+

Notice, still no inclusion of sulfate, no potassium, no sulfuric acid shown; just shows acidic conditions using H+. No showing of MnSO4.

The last reaction seems to have too many hydrogen ions on both sides. Ten too many on each side, so,

2MnO4- + 6H+ + 5H2S -------> 2Mn+2 + 8H2O + 5S0

...But based on what I find as the same example in an old gen. chem textbook, that last one is still off, based on the hydrogen ions. The example I find in this book does not include both hydrogens on the H2S, only using the S-2. I will need to study that example myself more carefully.

Borek
Mentor
2MnO4- + 6H+ + 5H2S -------> 2Mn+2 + 8H2O + 5S0
And that's correct. K+ and SO42- are just spectators, what you have here is a perfectly valid net ionic reaction. It can be rewritten in terms of neutral species, but it won't change stoichiometry of permanganate/hydrogen sulfide reaction.

...But based on what I find as the same example in an old gen. chem textbook, that last one is still off, based on the hydrogen ions. The example I find in this book does not include both hydrogens on the H2S, only using the S-2. I will need to study that example myself more carefully.
We are talking about low pH solution and hydrogen sulfide is a weak acid, fully protonated. Some books ignore these things. IMHO they are wrong, although I am personally not 100% consistent.