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Topic: Please help with this REDUCTION REACTION: MnO4- + S2- --> MnO2 + S8  (Read 33154 times)

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VinnyCee

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Re:Please help with this REDUCTION REACTION: MnO4- + S2- --> MnO2 + S8
« Reply #15 on: March 05, 2006, 09:40:37 AM »
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!
« Last Edit: March 05, 2006, 09:43:26 AM by VinnyCee »

Offline Albert

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Re:Please help with this REDUCTION REACTION: MnO4- + S2- --> MnO2 + S8
« Reply #16 on: March 05, 2006, 10:02:22 AM »
The number of electrons has to be, in the end, the same in both of the half reactions.

VinnyCee

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Re:Please help with this REDUCTION REACTION: MnO4- + S2- --> MnO2 + S8
« Reply #17 on: March 05, 2006, 10:15:28 AM »
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- + S2-  --->  MnO2 + S8

MnO4- + S2-  --->  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"

S2-  -->  S

S2-  -->  S  +  2e-

(S2-  +  -->  S  +  2e-) * 3

3S2-  -->  3S  +  6e-



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

2MnO4-  +  3S2-  +  8H+  -->  2MnO2  +  3S  +  4H2O



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

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

2MnO4-  +  3S2-  +  8H20  -->  2MnO2  +  3S  +  4H2O  +  8OH-

2MnO4-  +  3S2-  +  4H20  -->  2MnO2  +  3S  +  8OH-



Now change back into an equation using the S8:

(2MnO4-  +  3S2-  +  4H20  -->  2MnO2  +  8OH-) * 8

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

16MnO4-  +  24S2-  +  32H20  -->  16MnO2  +  3S8  +  64OH-
« Last Edit: March 05, 2006, 10:33:14 AM by VinnyCee »

Offline Alberto_Kravina

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Re:Please help with this REDUCTION REACTION: MnO4- + S2- --> MnO2 + S8
« Reply #18 on: March 05, 2006, 11:07:58 AM »
Quote
16MnO4-  +  24S2-  +  32H20  -->  16MnO2  +  3S8  +  64OH
seems OK

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