# What are the oxidation numbers for the elements in the compound C2H5COOH?

Question ID
537944

Created
April 27, 2011 8:57pm UTC

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2

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1. I do these this way.
First, I keep H and O "normal" if I can and let C change to whatever I need.
The formula becomes C3H6O2
H = +1 each x 6 = +6
O = -2 each x 2 = -4
Which means C must be -2 for three of them or -2/3 each to make the compound zero (and all compounds are zero).
Here is a very good link.
http://www.chemteam.info/Redox/Redox-Rules.html

537968

Created
April 27, 2011 9:27pm UTC

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0

2. Oh, let's have some fun with oxidation numbers, shall we? In the compound C2H5COOH, we have our beloved carbon (C), hydrogen (H), and oxygen (O). Now, carbon, being the attention seeker it is, likes to show off its versatility. In this case, carbon has an oxidation number of +3. Hydrogen, on the other hand, is a bit of a pushover and always has an oxidation number of +1. Finally, oxygen, the diva of elements, usually has an oxidation number of -2. However, since we have two oxygen atoms here, let's do the math (don't worry, you'll need it): 2 x -2 = -4. Now, prepare yourself for a grand finale! Since the compound overall is neutral, we balance it out by assigning the remaining hydrogen atoms an oxidation number of +1. And there you have it, the oxidation numbers for C2H5COOH: carbon (+3), hydrogen (+1), and oxygen (-2). Bravo!