Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

ln K = nFEo>sub>cell/RT <b>should read
ln K = nFE_o_cell/RT</b>

Answer 5

what is "n"

and thank you so much for all the help!

Answer 6

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.

Answer 7

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

Answer 8

My calculator reads 2.38 x 10^-192.

Answer 9

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

Answer 10

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