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Balancing Redox Reaction by Half Reaction Method

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    CuCl2(aq)+Al(s)→AlCl3(aq)+Cu(s)


    2. Relevant equations



    3. The attempt at a solution

    I'm just not sure what to do with a situation involving compounds. I can do simple balancing without a problem. Here is my attempt. Please, please let me know where I went wrong...

    Oxidation: Al→Al+3+3e-
    Reduction: Cu+2+2e-→Cu
    Reduction: 2Cl-2→3Cl-3 (?????)

    I'm confuzzled. I don't know where to go from here, or even if I did the above correctly. The book is pretty awful in describing these half-reaction equations, so I've been searching online but without any applicable results. I didn't think I should split the reduction into two parts, but maybe I do. I don't know. I also don't know what to do with the equation concerning the Chlorine. All around, I'm confused.

    I can balance the equation just fine with the "trial and error" method, and it matches the back of the book, I just don't know how to get there by half-reactions.

    Thank you so much for any help.
     
  2. jcsd
  3. Feb 16, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    Cl- is just a spectator and doesn't change during the reaction. Yes, you should split.

    Check ChemBuddy page on the half reaction method. If it doesn't help, ask for clarification (here or there).
     
  4. Feb 16, 2012 #3
    Well, gee. That was simple! Thank you very much, and thank you for the link. I suppose what got me then was my assumption that I understood spectator ions pretty well. I guess not! In this case, is Cl- a spectator because we don't need to balance the charge since it stays the same throughout the reaction?
     
  5. Feb 16, 2012 #4

    Borek

    User Avatar

    Staff: Mentor

    It is a spectator because it doesn't take part in the reaction. It doesn't take part in the reaction - so it doesn't change its charge.
     
  6. Feb 16, 2012 #5
    Got it. Thanks.
     
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