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Electrochemistry - Nernst Equation

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Electrochemistry - Nernst Equation

    Pb2+ + 2 e- → Pb (s) ξo = -0.13 V
    Ag+ + 1 e- → Ag (s) ξo = 0.80 V

    What is the voltage, at 298 K, of this voltaic cell starting with the
    following non-standard concentrations:

    [Pb2+] (aq) = 0.08 M
    [Ag+] (aq) = 0.5 M

    2. Relevant equations
    Use the Nernst equation:

    ξ = ξo - (RT/nF) ln Q

    3. The attempt at a solution
    there are 2 parts to this question one is to find Q and then use that to
    find E.
    ξo=0.80-(-0.13)=0.93V

    first I balanced the equation:

    2(Ag+ + 1 e- → Ag (s))
    Pb(s)→ Pb2+ + 2 e-
    --------------------
    2Ag+ + Pb(s) --> Pb2+ + 2Ag(s)

    Q = [products]^p/[reactants]^r
    so Q = 0.08/0.5^2 = 0.32

    i used ξ = ξo - (RT/nF) ln Q
    ξ = 0.93V - ((8.314)(298K)/(2)(96500)) ln0.32 = 0.9446V

    I also used E=Eo-(0.05916/n)logQ
    E = 0.93V - (0.05916/2)log0.32= 0.9446

    I am down to my last try and hoping if anyone can double check this for me. thanks!
     
    Last edited: Mar 25, 2008
  2. jcsd
  3. Mar 25, 2008 #2
    How can your answer be negative when the ln of a fraction is a negative, and subtracting a negative number from a positive number should make your answer positive?
     
  4. Mar 25, 2008 #3
    o thanks i edited it, i just typed in the - sign for some reason XD, but does the process seem correct?
     
  5. Mar 25, 2008 #4
    Yes.
     
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