[VinnyCee]

Thanks alot! +(1 X 10⁹) Scooby Snacks!

Anyways, I know that there are two methods of verification for this answer:

1) The overall charges on both sides are equal (at -64).
2) Something else, maybe the oxidation states of each side are to be equal also? (5 for left, 3 for right: not equal)

How would I verify the second option? I tried but I don't think the oxidation states are equal!

[Albert]

The number of electrons has to be, in the end, the same in both of the half reactions.

[VinnyCee]

So the oxidation numbers do not have to match in the final equation? They match in both of the half-reactions already, I think.

Anyways, I tried your original suggestion of using S instead of S₈ and got the same thing!:


Initial equation

MnO₄^- + S^2-  --->  MnO₂ + S₈

MnO₄^- + S^2-  --->  MnO₂ + S



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"

S^2-  -->  S

S^2-  -->  S  +  2e^-

(S^2-  +  -->  S  +  2e^-) * 3

3S^2-  -->  3S  +  6e^-



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

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



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

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

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

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



Now change back into an equation using the S₈:

(2MnO₄^-  +  3S^2-  +  4H₂0  -->  2MnO₂  +  8OH^-) * 8

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

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

[Alberto_Kravina]

16MnO4-  +  24S2-  +  32H20  -->  16MnO2  +  3S8  +  64OH

seems OK