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Cell Potential with Nernst Equation

  1. Feb 5, 2014 #1
    1. The problem statement, all variables and given/known data
    PjU9hR8.png


    2. Relevant equations
    Nernst Equation where Q = [product]/[reactants], R = 8.314, F = 96500
    nernst.gif


    3. The attempt at a solution
    I got E° to be 0.012 V because E° of cathode - E° of anode. -0.131 - (-0.143) = 0.012.

    The balanced equation I got was: Pb2+ + Sn → Pb + Sn2+

    For the values of Q, I added 0.349 M to the concentration of Sn2+ as it's the product, and I subtracted 0.349 M from the concentration of Pb2+ as it's the reactant.

    Plugging everything into the equation:
    E = 0.012 V - ((8.314 * 271)/(2 * 96500)) * ln(1.318/0.576) = 0.0023 V. This answer is wrong, but I cannot figure out what I'm doing wrong here. Please help!
     
    Last edited: Feb 5, 2014
  2. jcsd
  3. Feb 5, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    I can't find anything wrong.

    Not that I am infallible, but there are two of us stumped now.
     
  4. Feb 5, 2014 #3
    The answer I put in is what the system shows as the correct answer now that the homework has closed. I don't understand why it was marked incorrect, but I'll just have to email my professor.
     
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