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Homework Help: Cell potential

  1. Apr 11, 2010 #1
    What is the potential of this cell ( in V) at 25 oC if the copper electrode is placed in a solution in which [Cu2+] = 5.1×10-10 M?

    I found this answer to be 0.065 which is correct. The second part:


    If the copper electrode of the above cell is placed in a solution of 0.270 M NaOH that is saturated with Cu(OH)2, what is the potential of the cell (in V, measured in the same direction as in question 5) at 25 oC? Ksp = 1.6×10-19.

    I'm using the Nernst equatuion.
    Here's how I think it goes:
    The NaOH is soluble so it dissolves but the Cu(OH)2 isn't. It's Ksp is very small but it will dissociate a little. So the dissociation equation I got is:

    Cu(OH)2 ---> Cu2+ + 2OH-
    Ksp = [Cu2+][OH-]^2
    = [x][2x]^2
    x = 3.42e-7

    To find Ecell I said Cu gets oxidized and OH reduced and my Ecell = -0.74
    Plugging into the Nernst equations:

    E=-0.74-(0.0592/4)*log[Cu2+][OH-]^2
    E=-0.64

    This is wrong. Can someone explain to me please. Thanks!
     
  2. jcsd
  3. Apr 11, 2010 #2
    Can anyone please help me, or just give me a hint...
     
  4. Apr 12, 2010 #3

    Borek

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    Staff: Mentor

    In the first part you have calculaed potential of the copper half cell, why do you try to incorporate OH- concentration into Nernst equation now? Were you told anything about change of the other half cell?

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  5. Apr 14, 2010 #4
    Ok, I have it set up as

    Cu(OH)2 ---> Cu2+ + 2OH-

    Ksp = 1.6e-19 = [x][2x+0.1]^2

    Now this might sound stupid but how do I solve this equation. I'm stuck since we get an x^3 and x^2 term...
     
  6. Apr 14, 2010 #5

    Borek

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    Staff: Mentor

    Sorry, I posted something else at first, but then realized you did something strange. Why 2x+0.1?

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    methods
     
  7. Apr 15, 2010 #6
    2x+0.1 because of the .1 from the NaOH
     
  8. Apr 15, 2010 #7

    Borek

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    Staff: Mentor

    Strange.

    Is concentration of dissolved copper hydroxide comparable with concentration of NaOH?

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    methods
     
  9. Apr 15, 2010 #8
    I mean 2x+0.27 this is from the answer key that the professor gave me. I don't know how to solve the equation though.
     
  10. Apr 15, 2010 #9

    Borek

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    Staff: Mentor

    OK, 2x+0.27.

    Can you answer the question I have asked?

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    methods
     
  11. Apr 16, 2010 #10
    From what I understood the 0.27 is from the NaOH that dissociates.
     
  12. Apr 17, 2010 #11

    Borek

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    Staff: Mentor

    That's not answer to the question I asked, to speed things up asnwer two questions now:

    Is concentration of dissolved copper hydroxide comparable with concentration of NaOH?

    Why 2x+0.27 for the OH- concentration?

    And we won't move froward till you answer them both, so dodging won't help.

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  13. Apr 19, 2010 #12
    I'm not trying to evade any question. Nevermind my chem exam is over, I don't feel like caring about it anymore... thanks for your help.
     
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