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Chemistry - Redox Disproportionation Balancing?

by escryan
Tags: disproportionation
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escryan
#1
Sep20-08, 12:52 PM
P: 13
1. The problem statement, all variables and given/known data

Balance the equation for disproportionation reactions.

2. Relevant equations

S8(s) ----> S2-+S2O32- (basic solution)

3. The attempt at a solution

I attempted in to do this with the ion electron method, obtaining an answer:

3S8 + 24OH- ----> 16S2- + 4S2O32- + 12H2O

However, I didn't follow the conventions that I was given, like using systematic steps to come to an answer... I was basically just guessing and checking, but I would like to know how to do it systematically.
Apparently though, I'm supposed to be doing my redox equations with the half reaction method instead.. which rather confuses me. When I attempted to do the half reaction...

reduction: S8 + 16e- ----> 8S2-
oxidation: 2S8 + 24OH- (???) ----> 4S2O32- + 8e- + 12H2O

The oxidation portion of the half reaction just doesn't seem to add up when I'm trying it.
I'd like to know if there is a systematic method to figuring out these sorts of equations so in the future I can have some sort of guideline in solving them. Help would be greatly appreciated :)!
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Borek
#2
Sep20-08, 03:15 PM
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Quote Quote by escryan View Post
oxidation: 2S8 + 24OH- (???) ----> 4S2O32- + 8e- + 12H2O
Why do you start with 2S8? Try just S8.
escryan
#3
Sep20-08, 04:40 PM
P: 13
I think that might've been a typo on my part, sorry.

I had written on my paper:
2S8 + 48OH- ----> 8S2O32- + 8e- + 24H2O

as well as:

S8 + 24OH- ----> 4S2O32- + 8e- + 12H2O

and neither of the charges worked out to them being equal on either side..

Another typo I noticed (eek.)
I attempted in to do this with the ion electron method, obtaining an answer:
I meant to say I used another method, not the ion electron method/ half reaction, to come to that answer. Not sure that that would matter any..?

Borek
#4
Sep20-08, 04:58 PM
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Chemistry - Redox Disproportionation Balancing?

Quote Quote by escryan View Post
S8 + 24OH- ----> 4S2O32- + 8e- + 12H2O

and neither of the charges worked out to them being equal on either side.
Which means that you have to change number of electrons. Everything else seems to be balanced.
escryan
#5
Sep20-08, 05:23 PM
P: 13
Ahh.. that's the part that confuses me.. how do you come to find the number of electrons you must add to one side?

I've always been under the impression that the electrons added were based sheerly on the oxidation numbers. In this equation, S8 has 0 charge, and an individual S has a 0 charge. Comparing this to the S in the S2O32- which has a charge of 2+.. multiplied by four by balancing only gives a value of 8e-....
Borek
#6
Sep20-08, 05:28 PM
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If you are balancing with half reactions method, you use electrons to balance charge. Don't even think about ON.
escryan
#7
Sep20-08, 05:43 PM
P: 13
Oh wow, that's a really helpful guide! Thanks so much for your help :)!

Just one last question, is there any reason as to why the half reaction method is recommended/preferred over the one involving oxidation numbers? I'm just curious because I'd rather use the oxidation numbers one (because that's the one I first learned), but my current teacher only teaches the half reaction method.
Borek
#8
Sep20-08, 06:06 PM
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In short: oxidation numbers don't exist. They are just an accounting device and they don't refer to any real chemical (measurable) property of atoms in compounds. Half reactions method is much closer to the real chemistry - they refer to existing processes, parameters of which can be measured and/or calcuated (like potential of half cell given by Nernst equation). It doesn't necesarilly mean that half reaction occurs exactly as its reaction equation describes - for example, it can have several steps.


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