*Update* I tried following a similar example in my textbook using the Nernst equation but still got the wrong answer
log(xM/0.135M) = 2(0.180V)/-0.0592 = -6.081081081
xM/0.135M = 10^-6.081 = 8.29E-7(0.135M) = 1.12E-7M = x = [Ag+]anode
1.12E-7(2) = 2.24E-7M
Ksp= 1.12E-7(2.24E-7)^2 = 5.62E-21
My book gave this explanation for the similar example: "Because the electrodes in this cell are identical, the standard electrode potentials are numerically equal and subtracting one from the other leads to the value Eo = 0.000V. However, because the ion concentrations differ, there is a potential difference between the two half cells (non-zero nonstandard voltage for the cell). [Pb2+] = 0.100 M in the cathode compartment, while the anode compartment contains a saturated solution of PbI2 .
We use the Nernst equation (with n = 2 ) to determine Pb2+ in the saturated solution." Not entirely sure how this Nernst equation works because my prof has not covered this yet; nonetheless my assignment is due tonight.
A voltaic cell, with Ecell = 0.180 V at 298K, is constructed as follows:
Ag(s)|Ag+(satdAg3PO4)||Ag+(0.135 M )|Ag(s)
What is the Ksp of Ag3PO4?
2 answers
Could you type in the example problem of PbI2 and the answer they gave for that example. If so perhaps I can help. Don't make a new post out of it; use this post for that.