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Predict the products of electrolysis

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    What products are expected in the electrolysis of CuBr2(aq)?


    2. Relevant equations



    3. The attempt at a solution
    This is actually a solved example from my textbook that I have a question about. It says to first consider the possible reactions at the cathode.

    Since reduction happens at the cathode, it says the possible reductions are:
    Cu2+(aq) + 2e- --> Cu(s) E = +0.34
    2H2O(l) + 2e- --> H2(g) + 2OH- E= -0.83

    The first one happens because its E is the most positive.

    What I'm wondering is why we didn't include the reduction of Br2. Is it because Br- is an anion that we don't consider it?

    And, how did they know to use the reaction for water that I gave and not 2H20(l) --> O2(g) + 4H+ + 4e-? (Because it's oxidation, which can't happen at the cathode, right?)
     
  2. jcsd
  3. Apr 21, 2009 #2

    Borek

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    Because there is no Br2?
     
  4. Apr 21, 2009 #3
    Wouldn't you get the dissociation of CuBr2 --> Cu2+ + 2Br-

    And so need to take into account that 2Br2- + 2e- --> Br2? Why can't that be happening at the cathode?
     
  5. Apr 21, 2009 #4

    symbolipoint

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    What does a Standard Reduction Table tell you about
    Br2 + 2e- -------> 2Br- ?
    Note that you are interested in the oxidation, not the reduction.
     
  6. Apr 21, 2009 #5
    It has a value of +1.07.

    But I don't understand what that means for this problem =(

    There's something I'm missing here...
     
  7. Apr 23, 2009 #6

    symbolipoint

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    You would determine or calculate the E for any pair of halfcell reduction-oxidation pairs. You know what the possible reduction half cells could be to choose from; you know the possible oxidation half cells could be to choose from. If needed, choose to examine all possible combinations of redox pairs and determine the whole cell E's of each combination. The most positive value might correspond to the most favorable whole cell redox reaction. Just remind yourself that you need an oxidation to occur at the anode and you need a reduction to occur at the cathode. Fit the halfcell values accordingly.
     
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