1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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?

    --
     
  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

    User Avatar

    Staff: Mentor

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

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

    Borek

    User Avatar

    Staff: Mentor

    Strange.

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

    --
    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

    User Avatar

    Staff: Mentor

    OK, 2x+0.27.

    Can you answer the question I have asked?

    --
    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

    User Avatar

    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.

    --
     
  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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook