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Topic: Balancing this Redox Reaction.  (Read 5512 times)

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bugsmenot

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Balancing this Redox Reaction.
« on: May 06, 2006, 07:26:25 PM »
I need help balancing this redox reaction. The question is asking me to balance a bunch of reactions and the last one is stumping me. It is also asking to balance them using the half-reaction method and oxidation number method, btw.

S2O42- + O2 -> SO42- (BASIC)

I've managed to balance it with the oxidation number method.

6e /S2O42- X 2
4e/O2 X 3

4 OH + 2 H2O + 2S2O42- + 3O2 -> 4SO42- + 4H + 4OH
4 OH + 2S2O42- + 3O2 -> 4SO42- + 2H2O

I hope that that is correct. Now the problem is the half-reaction method. No matter what I do I just can't seem to balance the reaction. I've done it quite a few times, and I still can't get it. I think the problem is that the elements being oxidized and reduced are in the same compound. I'm really stumped. Help, please.


Offline Borek

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Re: Balancing this Redox Reaction.
« Reply #1 on: May 07, 2006, 04:52:43 AM »
Show your half reactions.
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bugsmenot

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Re: Balancing this Redox Reaction.
« Reply #2 on: May 07, 2006, 01:27:15 PM »
4 [8OH + S2O4 + 4H2O-> 2SO4 +6e + 8H2O]
3 [2O2 + 8e-> SO4]

Offline Alberto_Kravina

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Re: Balancing this Redox Reaction.
« Reply #3 on: May 07, 2006, 01:36:26 PM »
3 [2O2 + 8e-> SO4]
???

Quote
4 [8OH + S2O4 + 4H2O-> 2SO4 +6e + 8H2O]
This one seems to be OK.

Offline Borek

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Re: Balancing this Redox Reaction.
« Reply #4 on: May 07, 2006, 01:47:46 PM »
Quote
4 [8OH + S2O4 + 4H2O-> 2SO4 +6e + 8H2O]
This one seems to be OK.

Water?

Second one is a disaster. Try O2 -> OH-.
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