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Ethanoic acid dissociates according to the following equation: $$\ce{CH3COOH + H2 O -> CH3COO- + H3O+}$$ I want to know if it is a redox reaction or not. In the ethanoic acid I obtained that the oxidation number of carbon in it is 0 by calculation, but it is (-2). What is the error in my answer?

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    $\begingroup$ Redox involves movement of electrons. No electrons are moving in the dissociation reaction, only a proton is transferred. Draw the Lewis structures to see the bonds to the carbons. That should help you see the oxidation state better. $\endgroup$ – Shafter Aug 18 '15 at 18:01
  • $\begingroup$ Isn't carbon oxidized? $\endgroup$ – user225430 Aug 18 '15 at 18:03
  • $\begingroup$ No and I have no idea why you'd think so. $\endgroup$ – Mithoron Aug 18 '15 at 18:09
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    $\begingroup$ Perhaps you could explain how you calculated the oxidation states. This would help us to answer your question better. $\endgroup$ – bon Aug 18 '15 at 18:17
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    $\begingroup$ Self-answers are welcome on StackExchange! Just so you'd know. $\endgroup$ – M.A.R. Aug 18 '15 at 19:08
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You're getting 0 ox. state for carbon because of treating both carbons equally - 0 is mean average value from oxidation numbers of both carbons. The one in carboxylic group has three bonds with more electronegative oxygen therefore has +3. The one in methyl group has -3 because it's connected with three hydrogens which have lower electronegativity.

See https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure - there's example with acetic acid!

In organic chemistry trying to get average value of ox. state usually doesn't make much sense as it's rare to have all carbons equivalent. @Shafter gave you good advice - without knowing the structure you were making mistake.

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Redox reactions involve increase in oxidation state of one chemical species and decrease in the oxidation state of the other through loss and gain of electrons respectively but since this does not occur here then it's not a redox reaction but rather water acts as a weak base to accept the proton donated by ethanoic acid

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  • $\begingroup$ My point is not knowing if it is a redox or not, but I am lost in the oxidation numbers since I don't know why I am having the oxidation number of C as 0. $\endgroup$ – user225430 Aug 18 '15 at 19:35

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