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March 29, 2017

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what is the pK (=-logK)at 25C for the reaction below?

Cl2(g)+6H2O(l)+5Cu2+(aq)<-->2ClO3-(aq)+12H+(aq)+5Cu(s)

E0=-1.133V

  • gen chem - ,

    I think I must have Ecell to calculate pK (or concns of all of the reagents).

  • gen chem - ,

    E cell is not given, just the E0 cell which is -1.133V

  • gen chem - ,

    That will be
    ln K = nFEo>sub>cell/RT
    Use 8.314 for R, 298 for T, 96,485 for F. Calculate K, then convert to pK.

  • gen chem - ,

    ln K = nFEo>sub>cell/RT should read
    ln K = nFEocell/RT

  • gen chem - ,

    what is "n"

    and thank you so much for all the help!

  • gen chem - ,

    Isn't n = 10?
    Cl2 ==> 2ClO3^-
    zero on the left to +10 (for both Cl atoms) on the right.
    5Cu^+2 ==>5Cu(s)
    10 on the left (for 5 Cu^+2) and zero on the right.
    Total 10 either way.

  • gen chem - ,

    with n=10, im getting an answer of 0 when I do e^(k)

    i got
    lnK= (10)(96485)(-1.133)/(8.314)(298)
    lnK=-441.228
    e^(-441.228)=0

  • gen chem - ,

    My calculator reads 2.38 x 10^-192.

  • gen chem - ,

    i tried that, apparently its not the right answer either =/.

  • gen chem - ,

    The question asks for pK. 2.38 x 10^-192 is K. pK = -log K = -log 2.38 x 10^-192.

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