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Topic: Redox equation: reduction of acidic dichromate ion (Read 8361 times)

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Ice-cream

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Redox equation: reduction of acidic dichromate ion
« 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 --> CH3COOHreduction of acidic dichromate ion: Cr2O7(3-) --> Cr(3+)What I got was: Ox: CH3CH2OH + H2O --> CH3COOH + 4H+ + 4eRe: 14H+ + 9e + Cr2O7(2-) --> 2Cr(3+) + 7H2OI 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+) + 19H2ODoes any1 agree with my answer? (I just think the numbers seem a bit big...) « Last Edit: April 24, 2005, 04:09:11 PM by Mitch » Logged

Garneck

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Re:Redox equation
« Reply #1 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 be14H+ + 6e + Cr2O7(2-) --> 2Cr(3+) + 7H2OBecause:14 - 6 - 2 = 66 = 6 « Last Edit: April 19, 2005, 02:35:11 PM by Garneck » Logged

Ice-cream

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Re:Redox equation
« Reply #2 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+) + 11H2Oany1 agree with this answer? Logged
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