Posted 19 April 2005 - 12:28 PM
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...)
Posted 19 April 2005 - 03:08 PM
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.
Posted 10 November 2005 - 01:45 AM
Posted 10 December 2005 - 11:28 AM
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.
Posted 10 December 2005 - 07:12 PM
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.
Don't mean to be rude, but I think qwerty and neo are incorrect.
Posted 12 December 2005 - 07:06 PM
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