gen chem
posted by Chem on .
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

I think I must have E_{cell} to calculate pK (or concns of all of the reagents).

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

That will be
ln K = nFE^{o}>sub>cell/RT
Use 8.314 for R, 298 for T, 96,485 for F. Calculate K, then convert to pK. 
ln K = nFEo>sub>cell/RT should read
ln K = nFE_{o}_{cell}/RT 
what is "n"
and thank you so much for all the help! 
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. 
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 
My calculator reads 2.38 x 10^192.

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

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