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A question about oxidization

  1. Oct 28, 2004 #1
    All right.

    Well, I just would like a clarification about Oxidization. I already know that it is the gain of charge in a redox reaction, however, I have heard it used elsewhere too.

    Oxidized metal, for instance. I know that this is metal that has been exposed to oxygen and has thusly changed color. Does this occur because of a redox reaction taking place or is it based on something else?

    I've learned that by oxidizing an alcohol you can get a carboxylic acid. So, is this a redox reaction or just pumping oxygen through the alcohol?

    What is happening on the chemical level there?

    The reason of this question springs from the sentence in a lab that I am doing "Iron(II) in solution is easily oxidized to Fe(III)". By oxidized, does it mean, in this case and generally, blowing oxygen through the liquid?

    Last edited: Oct 28, 2004
  2. jcsd
  3. Oct 28, 2004 #2


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    Oxidization is in a molecular level; if you are impatient, then you can try blowing oxygen gas from iron(II) solution to oxidize it quickly to iron(III).

    Oxidation of inorganic substances generally involves electron loss from the ion, but some may accept oxygen atoms to form oxides. Oxidation of organic substances, however, generally involves proton loss from the molecule. For example, a primary alcohol, written as R-CH2OH, loses two protons to give an aldehyde, R-CHO. The oxygen may have come from the oxidizing agent, very frequently chromium(VI) ions such as Cr2O72- are used for this purpose.

    So, "pumping" of oxygen may be considered in some cases, but rather reactive oxygen species (oxidants) are responsible for these reactions.
  4. Oct 28, 2004 #3
    Briefly, oxidation is as you describe... When a metal loses electron it will be oxidized. Color change is something that is not directly related to the reaction. Color change could occur say, when e.g. reduction happens...

    In organic reactions, as the H moiety is decreased as the O moiety increases then you can point an oxidation. For example when ethanol is oxidized, it will turn to acetic acid. Here, the carbon's hybridization is changed from sp3 to sp2 and one H is removed while an O is introduced. This is an oxidation reaction. However, this cannot be regarded as just oxygen blowing.
  5. Oct 28, 2004 #4


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    The term "oxidation" is misleading as it implies that oxygen itself must be used in the process, which is not true. The best way to think of oxidation on a molecular level is through transfer of electrons. The mechanism for a chromium oxidation involves the coordination of the alcohol to the chromium and then removal of one of the hydrogens (as H+) on the carbon of the alcohol group. That would leave you with a carbanion adjacent to oxygen (not good), but those electrons in the anion (which were previously in the C-H bond) can move up to form a double bond between oxygen and carbon. This, however, would make oxygen violate the octet rule. To address this, the oxygen-chromium bond must be broken with both electrons from that bond going onto the chromium. At this stage, the substrate molecule has lost two electrons and has therefore be oxidized. Note that there is no need for O2 in this mechanism!

    My point is that oxidation is an electronic process. You could write out redox half reactions for this process, but that analysis wouldn't tell you anything about the actual mechanism of the reaction; it would just tell you that the net effect is that the substrate molecule has lost two electrons (was oxidized) and chromium has gained two electrons (was reduced).

    For completeness, the oxidation to the carboxylic acid occurs by water attacking the aldehyde that was formed by the first oxidation and then the resulting geminal-diol is oxidized by a mechanism similar to that described above. Again, none of the oxygens that are in the product molecule come from molecular oxygen.

    This is not to say that molecular oxygen can't oxidize organic molecules (it can, and it does) but typically these reactions are not useful synthetically because they are typically not well defined.
  6. Oct 28, 2004 #5


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    A brilliant explanation indeed. Thank you.
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