Chemical Forums

Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Ice-cream on April 19, 2005, 08:26:19 AM

Title: Redox equation: reduction of acidic dichromate ion
Post by: Ice-cream on April 19, 2005, 08:26:19 AM

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...)

Title: Re:Redox equation
Post by: Garneck on April 19, 2005, 02:28:39 PM

Quote from: Ice-cream on April 19, 2005, 08:26:19 AM

Re: 14H+ + 9e + Cr2O7(2-) --> 2Cr(3+) + 7H2O

This is wrong.

It should be
14H+ + 6e + Cr2O7(2-) --> 2Cr(3+) + 7H2O

Because:
14 - 6 - 2 = 6
6 = 6

Title: Re:Redox equation
Post by: Ice-cream on April 24, 2005, 02:12:08 AM

Ahh...i c, so is this right then?

I get 12 electrons in both equations and then add them:

3CH3CH2OH + 2Cr2O7(2-) + 16H+ --> 3CH3COOH + 4Cr(3+) + 11H2O

any1 agree with this answer?