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Oxidation + Reduction Half-Reactions

  1. Jan 30, 2004 #1

    JDK

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    Hello,

    I was curious if anyone could explain to me the simple procedure/guidleines to carrying out oxidation (half-reactions), reduction (half-reactions) and overall net reactions. I've read my Chem book over and over, but I don't quite understand it. Even a pointer to a good site with a decent explanation would be great. Any help is awesome. Thanks so much.

    Here's a simple example I need some help with...

    Li + O2
     
  2. jcsd
  3. Jan 30, 2004 #2

    GCT

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    Oxidation reduction reactions involve an certain element obtaining an electron(s) from a second element. In this case one element becomes the reducing agent, gives its electrons away and the element recieving this electron becomes the oxidizing agent (since it oxidizes the other element).

    You need not to be concerned about the exact mechanism of the reaction. First, try writing out the balanced equation.

    4Li + O2 --->2Li2O

    So which element was oxidized? Li does not have a charge while LiO is an ionic compound in which electrons are "shared" according to electronegativity. In this case oxygen is much more electronegative and it essentially obtains two electrons from Li through bonding.

    Now write out the half reactions. Oxygen is reduced.

    O2 + 4e- ---> 2O (with a 2- charge)

    Li is oxidized.

    4Li--->4e- + 4Li (with a 1+ charge)

    *So overall

    O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e-

    A third reaction is also important. The products of the latter two steps combine to form 2Li2O.

    2O (2-) + 4Li (1+) ---> 2Li2O

    Combine this with the above equation and you will return to the original balanced equation.

    Hope this answers your question.
     
  4. Jan 30, 2004 #3

    JDK

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    Hmmm... *reads over information*

    It somewhat does. I have a question... can you explain this line and how you got to it...

    4Li + O2 --->2Li2O

    Normally I would see Li and 02 occuring together like so... Li2O (lithium oxide). In this case I'm stumped. I have a hunch it has to do with the fact that oxygen is diatomic. If you can explain that line for me I have a feeling I might get the rest of it. Excuse my inability to catch onto this stuff quickly. Chem 20 was relatively easy up until this. Thanks again.
     
  5. Jan 30, 2004 #4

    GCT

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    You don't need to worry about the specifics in how the reactions occur. A chemical equation simply pertains to the initial and final states (at least for now; you will learn about intermediates later in reaction kinetics). In this case we know we have a reaction between Li and Oxygen gas.

    If we wrote out the reaction as

    Li + O2 --->Li2O, the equation would not balance indicating that such reactions simply do not.

    It is logical that since the reactants makeup the products the number of moles would have to be equivalent on both sides.

    Thus we know the initial and final states of the reaction as

    4Li(g) + O2 (g)--->2Li2O (g)

    Just tell me if you are still confused.
     
  6. Jan 30, 2004 #5

    JDK

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    Ahhh, I get it. Alright, I understand everything up until this line...

    O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e-

    Why does the 4e- still remain in the products and then disappear afterwards? Thank-you.
     
  7. Jan 30, 2004 #6

    GCT

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    O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- is essentially

    O2 + 4Li ---> 2O (2-) + 4Li (1+)

    The products of this reaction combine to form 2 molar equivalents of Li2O.

    2O (2-) + 4Li (1+)--->2Li2O

    We are just summarizing the important steps (states) of the reaction.
     
  8. Feb 2, 2004 #7

    JDK

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    Thanks so much for your wonderful responses. I think I get it all completely now. Your help has been appreciated. :smile:
     
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