[VinnyCee]

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.

[Alberto_Kravina]

2MnO4- + 3S2- + 4H2O  --->  2MnO2 + 3S8 + 8OH-

The sulfur atoms are not balanced

Hint: write "S" instead of S₈

[Albert]

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?

[VinnyCee]

Ok, I am going to spend another hour retyping this whole problem into the crappy interface, however, I will COPY it before trying to post this time!!!

I start with the first "half-reaction" equation:

MnO₄^-  -->  MnO₂

MnO₄^-  +  3e^-  -->  MnO₂

MnO₄^-  +  3e^-  +  4H⁺  -->  MnO₂

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


And the second:

S^2-  -->  S₈

S^2-  -->  S₈  +  2e^-

(S^2-  -->  S₈  +  2e^-) * 3



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!

[Albert]

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.

[VinnyCee]

Can you show me where I am missing water molecules? Even better, could you please show me how to do this correctly?

The S₈ is part of the original problem, therefore I cannot just "forget" about it! I know that the S's do not balance, but how would I make them do so? Maybe I have to balance the S's first, right?

[VinnyCee]

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^-

[VinnyCee]

Problem solved?:


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-  -->  S₈  +  2e^-) * 3

24S^2-  -->  3S₈  +  6e^-



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

2MnO₄^-  +  24S^2-  +  8H⁺  -->  2MnO₂  +  3S₈  +  4H₂O



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

2MnO₄^-  +  24S^2-  +  8H⁺  +  8OH^-  -->  2MnO₂  +  3S₈  +  4H₂O  +  8OH^-

2MnO₄^-  +  24S^2-  +  8H₂0  -->  2MnO₂  +  3S₈  +  4H₂O  +  8OH^-

2MnO₄^-  +  24S^2-  +  4H₂0  -->  2MnO₂  +  3S₈  +  8OH^-

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

2MnO₄^-  +  24S^2-  +  4H₂0  -->  2MnO₂  +  3S₈  +  8OH^-

[Albert]

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

[Borek]

24S^2-  -->  3S₈  +  6e^-

Charge is not balanced in this half reaction.

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

2MnO₄^-  +  24S^2-  +  4H₂0  -->  2MnO₂  +  3S₈  +  8OH^-

These two can't be correct at the same time - in one case 2MnO₄^- are used to oxidize 3S^2- in the second - to oxidize 24S^2-.

Vinny: your reaction is not balanced. Charge must be balanced just like atoms are.

There is a small problem with hydrogen/oxygen balancing. MnO₂ is one of the products, which suggests neutral solution (pH close to 7). Thus both approaches (use of H⁺ or OH^-) seems justified. I use H⁺ in such cases, but that's only personal preference.

[VinnyCee]

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".

[Albert]

Negative charges aren't balanced.

If you want to use S₈:

MnO₄^- + 3e- + 4H⁺ -> MnO₂ + 2H₂O

S₈ + 16e- -> 8S^2-


16Mno₄^- + 24S^2- + 64H⁺ -> 16MnO₂ + 3S₈ + 32H₂O

[Borek]

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.

[VinnyCee]

Thank you very much albert! Using your suggestion, I get the following:

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) * 16

16MnO₄^-  +  48e^-  +  64H⁺  -->  16MnO₂  +  32H₂O



Second "half-reaction"

8S^2-  -->  S₈

8S^2-  -->  S₈  +  16e^-

(8S^2-  +  -->  S₈  +  16e^-) * 3

24S^2-  -->  3S₈  +  48e^-



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

16MnO₄^-  +  24S^2-  +  64H⁺  -->  16MnO₂  +  3S₈  +  32H₂O



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

16MnO₄^-  +  24S^2-  +  64H⁺  +  64OH^-  -->  16MnO₂  +  3S₈  +  32H₂O  +  64OH^-

16MnO₄^-  +  24S^2-  +  64H₂0  -->  16MnO₂  +  3S₈  +  32H₂O  +  64OH^-

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

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

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


PS - The strikedthrough electrons are supposed to be 48e^-, yet it renders as 40e^-. Software limitation I guess!

       Not true -->  48e^-  =  40e^-

[Albert]

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

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

Yes, it is.