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Mercury battery reaction

  1. Dec 7, 2005 #1

    Math Is Hard

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    Howdy. I'm looking at a mercury battery reaction which goes like so:

    HgO(l) + Zn(s) --> ZnO(s) + Hg(l)

    I need to figure out the oxidation half reaction. What I can tell so far is that Zn is oxidized, since on the reactant side it has a charge of zero, and on the product side, Zn has an individual charge of 2+ in the ZnO compound. It must be losing electrons.

    I believe the oxidation half reaction should have Zn on the reactant side and ZnO and some electrons (two?) on the product side, but I can't figure out how to work in the oxygen for the reactant side. Sorry, I'm totally new at this. And I kinda suck at it besides.:redface:

    Any help is appreciated. Thanks in advance!
     
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  3. Dec 8, 2005 #2

    Astronuc

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    HgO(l) + Zn(s) --> ZnO(s) + Hg(l)

    So one has

    Hg2+ + Zn(s) --> Zn2+ + Hg(l)

    and the oxygen does not change valence, it is simply transferred form Hg to Zn.

    Half cell reactions

    Hg2+ + 2 e- --> + Hg

    Zn --> Zn2+ + 2 e-
     
  4. Dec 8, 2005 #3

    Math Is Hard

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    Thanks, Astronuc. So when I write these half reactions, I shouldn't show the oxygen anywhere? I should only show the element that is getting oxidized or reduced? Sorry, I am still struggling with the concept of what I'm trying to represent.
     
  5. Dec 8, 2005 #4

    Astronuc

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    Yeah, that's pretty much what one does. One just writes the oxidation and reduction reactions.
    No problem, it took me awhile to get used to the conventions with respect to cathode/anode reactions.

    Here is a nice little tutorial - http://www.science.uwaterloo.ca/~cchieh/cact/c123/halfcell.html

    This might help too.

    Spontaneity of REDOX Systems - http://members.aol.com/logan20/elecspon.html

    http://www.life.uiuc.edu/crofts/bioph354/redox.html
     
  6. Dec 9, 2005 #5

    Math Is Hard

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    Thanks - I really appreciate the help. And the links are great!! Hopefully, I'll be able to finish up my homework now.
     
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