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Redox equation


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#1 Ice-cream

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Posted 19 April 2005 - 12:28 PM

I've been given this redox equation to balance:

CH3CH2OH + Cr2O7(2-) --> CH3COOH + Cr(3+)

The 2 half-reactions are:
oxidation of ethanol: CH3CH2OH --> CH3COOH
reduction of acidic dichromate ion: Cr2O7(3-) --> Cr(3+)

What I got was:
Ox: CH3CH2OH + H2O --> CH3COOH + 4H+ + 4e
Re: 14H+ + 9e + Cr2O7(2-) --> 2Cr(3+) + 7H2O

I then added multipled the top equation by 9, the bottom by 4 and then added the 2 equations together to get:

9CH3CH2OH + 4Cr2O7(2-) + 20H+ --> 9CH3COOH + 8Cr(3+) + 19H2O

Does any1 agree with my answer? (I just think the numbers seem a bit big...)
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#2 neo007

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Posted 19 April 2005 - 03:08 PM

hmm....your answer does look a bit big. I think you've miscalculated the reduction half equation. I work out the oxidation number of Cr in Cr2O7(-2) to be +12 per Cr2 therefore +6 per Cr. This means that Cr(6+) is reduced to Cr(3+), with a 3 electrons being added, per Cr.

The reduction half equation should be(i think) :

14H+ + 12e + Cr2O7(2-) --> 2Cr(3+) + 7H2O

I think you should manage the rest.

Hope i've calculated the above correctly, seemed to make sense to me.
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#3 qwerty

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Posted 10 November 2005 - 01:45 AM

also the oxidation reaction, i think you've got 2 more H+ than you need? just a quick look but never mind me if im wrong
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#4 rthmjohn

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Posted 10 December 2005 - 11:28 AM

First off, here are the steps to balancing redox reactions in ACIDIC solution (apply each of the following steps to both the reduction and oxidation half reactions):

1. Balance the REACTING species (basically balance the ELEMENTS that undergo oxidation state change).
2. Balance the OXYGEN with H2O.
3. Balance HYDROGEN with PROTONS (H+)
4. Balance CHARGE with ELECTRONS (by charge I mean the net charge of each side of the reaction)
5. Find the Least Common Multiple of the number of ELECTRONS for both halves of the reaction. Combine the two half reactions, and the ELECTRONS MUST CANCEL OUT.

Here's what I get for your reaction (and I'm pretty sure this is correct):

Ox: CH3CH2OH + H2O --> CH3COOH + 4(H+) + 4e-
Re: Cr2O7(2-) + 14(H+) + 6e- --> 2Cr(3+) + 7H2O

The LCM of the number of electrons for the two half-reactions is 12, right?

So we multiple the Ox: by 3 and the Re: by 2 to get:

3CH3CH2OH + 3H2O + 2Cr2O7(2-) + 28(H+) + 12e- --> 3CH3OOH + 12(H+) + 12e- + 4Cr(3+) + 14H2O

Which can be simplified to:

3CH3CH2OH + 2Cr2O7(2-) + 16(H+) --> 3CH3OOH + 4Cr(3+) + 11H2O

Hope this helps. Don't mean to be rude, but I think qwerty and neo are incorrect.
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#5 AL

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Posted 10 December 2005 - 07:12 PM

Don't mean to be rude, but I think qwerty and neo are incorrect.

Neo007's reasoning was correct, he just made a typo. He even stated 3 electrons per Cr in the dichromate (which obviously implies 6 total), but then he ended up typing 12e.
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#6 rthmjohn

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Posted 12 December 2005 - 07:06 PM

my bad, I didn't bother to look at his reasoning, just went straight to his answer
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