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Chem 102 Lab Question:

  1. Mar 14, 2007 #1
    1. The problem statement, all variables and given/known data
    The question asks: Using the measured cell potential and the Nernst equation, calculate the [Ag+] in the test solution. However, you guys will need a little bit of background information to answer this question since it is a lab. First of all this reaction was:

    Ag2CO3(s) <-> 2Ag+(aq) +CO3^2-(aq)

    Also the Nernst equation we are using is the following:
    Ecell = E^0cell -0.05916V/n logQcell

    And, the Ecell potential that I calculated was 0.280V

    2. The attempt at a solution

    So I've made an attempt at this question, but I'm not too positive if I'm solving it correctly. From the reaction we know that n=1 (1 electron being transfered between the reactants and products). Therefore:

    0.280V = 0.800V-0.468V(Standard potentials off CDS)- 0.05916V/1 log[Ag+]^2
    0.280V = 0.332V-0.5916V log[Ag+]^2
    1.026 = log[Ag+]^2
    1.013 = log[Ag+]
    [Ag+] 0.00562 = 5.63X10^-3 mol/L

    However, I'm not too positive if this is correct or if I should be looking in the Chemistry Data Sheet for the standard potentials of the cathode and anode. Thanks for any help guys. :D
  2. jcsd
  3. Mar 14, 2007 #2
    Actually, here n will be 2 since the total number of electrons traded is 2. Your calculation looks wrong. If 1.013=log[Ag+], then [Ag+] should be [tex]10^{1.013}[/tex] which should be greater than 10, not smaller.
    Last edited: Mar 14, 2007
  4. Mar 14, 2007 #3
    Ahhh okay, yeah I'm not sure why I said 1, it clearly is n=2 for the number of electrons being traded. That makes a bit more sense. Thank you for your help.

    My only other question is whether or not I got the standard cell potentials right from the chemistry data sheet. Because I know that the cathode is Ag in this case, so looking on the CDS, we find it is 0.800V and then the anode I said was from the equation Ag2CO3(s) + 2e^- -> 2Ag(s) + CO3^2- which is 0.468. I'm assuming those are correct for this case, and that it's just an error in my calculation. Thanks again though.
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