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Setting up a galvanic cell (1.25V)

  1. Nov 28, 2004 #1
    This is what I have chosen

    I2 (s) + 2e- = 2I- (aq) reduction potential is +0.53
    Zn2+ (aq) + 2e- = Zn (s) reduction potential is -0.76

    using nerntz I get

    E = 1.29 - (0.0257/2)(ln (2.24/0.1)
    E= 1.2500

    What would could I use to get the I and Zn ions??? Also, would the Zn be the solution that should be 2.24 molar or the I???

    Would KCl be a good salt bridge?

    Thanks
    Nautica
     
  2. jcsd
  3. Nov 28, 2004 #2
    I am also finding E not, Delta G not, and K

    I calulated
    Enot=1.29V
    DeltaGnot = -nFEcell = -248,944J
    and for K I used 1.29V=0.0257V/2e- lnK and came up with 3.97 x 10^43 but this sounds like an extremely large number.

    thanks
    nautica
     
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