Find the emf for the cell
Cr(s) |Cr3+(aq, 0.37 M) ||
Pb2+(aq, 0.0095 M) | Pb(s).
Answer in units of V
I don't know what to do
I got
.6-(.0592/6)*log((.37)^2/(.0095)^3)
which ends up being .558V
Is that right?
Correct the reduction potential for Cr^+3/Cr for the 0.37 M.
E = Eo - (0.05916/n)*log (Cr)/(Cr^+3)
Correct the Pb reduction potential for 0.0095M
E = Eo-(0.05916/n)*log(Pb)/(Pb^+2)
Then plug in the numbers for
2Cr + 3Pb^+2 ==> 2Cr^+3 + 3Pb(s)
(You will need to change the sign for the Cr couple since that is being oxidized. You do NOT multiply potentials by the coefficients.).
0.558 is correct
That isn't the emf of the cell. You didn't follow my instructions. Those "corrections" I told you to make are separate corrections. You've tried to do it all together and you appear to be confusing this with the equilibrium constant.
.73 - (.05916/6)*log(2)/(2) for Cr?
-.13 - (.05916/6)*log(3)/(3) for Pb?
No.
n is 3 for Cr
n is 2 for Pb
(Cr) is 1 (That's the standard state).
(Pb) is 1. Standard state.
(Cr^+3) = 0.37M from the problem.
(Pb^+2) = 00095M from the problem.
The reduction potential for Cr is -0.73 so you should change that.
The Pb reduction potential is ok.
When you finish the above revisions, THEN
you reverse the sign of the Cr, keep the sign for Pb, and add them together WITHOUT REGARD TO THE NUMBER OF ELECTRONS. That will give you the EMF for the cell that you have set up.
So Dr. Bob the equilibrium constant has K and this does not
We are not confused. And it should be all in one.