I have an initial equation of:

MnO₄^- + S^2-  --->  MnO₂ + S₈

I then go through the "half-reaction" stuff to get this as the final equation:

2MnO₄^- + 3S^2- + 4H₂O  --->  2MnO₂ + 3S₈ + 8OH^-

Is this correct? If not, I can post more information on how I got to this answer if you need. Thanks.

No, it's not correct. Ok, first of all, you don't have to write S as S₈: when it comes to reductions S is sufficient.

In order to help you, could you write the two half-reactions?

Now just add the two together and add OH^- to make waters and move all waters to one side and I get the equation that I have for the answer in the first post, thanks again!

You miss a certain number of molecules of water in the former. Who told you to add OH^-?
If you find out the correct chemical equation for Mn, you'll solve the problem.
Oh, and forget about S₈, please.

I redid the problem over at the other forum (which is easier to edit cuz it has LaTeX). I just changed the first S^2- to 8S^2- to balance the right sides S₈ and recalculated.

Can anyone check it out to see if I did it right?

Thanks.

Edit: Here is the second "half-reaction" fixed (hopefully):

8S^2-  -->  S₈

8S^2-  -->  S₈  +  16e^-

(8S^2-  -->  S₈  +  16e^-) * 3


Or would it be this because only 2 electrons are needed for the oxidation state of the left side to balance?:

8S^2-  -->  S₈

8S^2-  -->  S₈  +  2e^-

(8S^2-  -->  S₈  +  2e^-) * 3

24S^2-  -->  3S₈  +  6e^-

It may be considered correct. However, I still prefer this way:

MnO₄^- + 3e- + 4H⁺ -> MnO₂ + 2H₂O

S + 2e- -> S^2-


2MnO₄^- + 3S^2- + 8H⁺ -> 2MnO₂ + 3S + 4H₂O

How about now?:

Initial equation, which still needed to be balanced (I hope)

MnO₄^- + S^2-  --->  MnO₂ + S₈

MnO₄^- + 8S^2-  --->  MnO₂ + S₈



First "half-reaction"

MnO₄^-  -->  MnO₂

MnO₄^-  +  3e^-  -->  MnO₂

MnO₄^-  +  3e^-  +  4H⁺  -->  MnO₂

(MnO₄^-  +  3e^-  +  4H⁺  -->  MnO₂  +  2H₂O) * 2

2MnO₄^-  +  6e^-  +  8H⁺  -->  2MnO₂  +  4H₂O



Second "half-reaction"

8S^2-  -->  S₈

8S^2-  -->  S₈  +  2e^-

(8S^2-  +  16H⁺  -->  S₈  +  2e^-) * 3

24S^2-  +  48H⁺  -->  3S₈  +  6e^-



Adding the two "half-reactions" while eliminating the electrons on each side

2MnO₄^-  +  24S^2-  +  56H⁺  -->  2MnO₂  +  3S₈  +  4H₂O



Adding hydroxide ions to each side to make water molecules and combine them onto the left side

2MnO₄^-  +  24S^2-  +  56H⁺  +  56OH^-  -->  2MnO₂  +  3S₈  +  4H₂O  +  56OH^-

2MnO₄^-  +  24S^2-  +  56H₂0  -->  2MnO₂  +  3S₈  +  4H₂O  +  56OH^-

2MnO₄^-  +  24S^2-  +  52H₂0  -->  2MnO₂  +  3S₈  +  56OH^-

Is this the final equation, and is it correct?:

2MnO₄^-  +  24S^2-  +  52H₂0  -->  2MnO₂  +  3S₈  +  56OH^-

I know that this equation is not correct, but why? I think it may have something to do with the steps taken to balance the second "half-reaction", right? I think the first "half-reaction" is correct (double check anyone?), but what about that pesky little second one? About changing the S₈ to a plain old S, what allows one to do this change? I think that the S₈ has to stay that way because the problem includes it in the initial equation that is given. Why do you think we should make the change?

The problem states that the reaction is "ran in a basic solution".

How about now?:

2MnO₄^-  +  6e^-  +  8H⁺  -->  2MnO₂  +  4H₂O

This one is OK

8S^2-  -->  S₈  +  2e^-

No. Once again charge is not balanced. -16 on the left, -2 on the right.

The problem states that the reaction is "ran in a basic solution".

OK, so use OH^-. AFAIR in basic solution MnO₄^- gets reduced rather to MnO₄^2-, not to MnO₂, but it may depend on other factors as well.

Is this the final equation, and is it correct?:

16MnO₄^-  +  24S^2-  +  32H₂0  -->  16MnO₂  +  3S₈  +  64OH^-

Yes, it is.