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Nernst Equation

  1. Nov 17, 2007 #1
    Can anyone check if I am doing this right?

    Given:
    Pb2+ + 2e- ---> Pb(s) E standard = -0.13V
    Ag+ + 1e- ---> Ag(s) E standard = 0.80V

    [Pb2+] = 0.05M
    [Ag+] = 0.5 M these are non standard concentrations

    Temp = 298K

    Using the Nernst equation, find E


    My answer:

    Pb oxidized and Ag is reduced ( is this right?.... because Ag has the higher reduction potential compared to Pb)

    E = E standard - (RT/nF) ln Q

    Q = [products]^p/[Reactants]^r

    the reaction is: 2Ag+ + Pb(s) + 2e- ---> 2Ag(s) + Pb2 + 2e- (the electrons cancels out)

    therefore, Q = (0.05M) / (0.5)^2

    Q= 0.20

    the E standard = 0.80 - (-0.13) = 0.80 + 0.13 = 0.93V
    E = E standard - (RT/nF) ln Q
    E = 0.93 - (8.314*298K/2*9.649E4) ln (0.20)
    E= 207.56 V

    Can anyone check the calculations for me and see if I am doing this right please?

    Thank you!!
     
  2. jcsd
  3. Nov 20, 2007 #2

    Gokul43201

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    Staff Emeritus
    Science Advisor
    Gold Member

    Looks good up to here.

    That's way off. Chug those numbers again.
     
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