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sundrops:
Determination of Ksp of PbI2.

make PbI2 by adding:
 1mL of 0.050M Pb(NO3)2
 3mL of 0.050 M KI

fill one well with the PbI2 and an adjacent well with 0.050 M Pb(NO3)2

Use polished Pb electrodes in both cells.
Use a string soaked in KNO3 solution as a salt bridge.


So experimentally I found that the cell potential was 0.066 volts.
that the:
Pb(NO3)2 was the cathode, so the Pb2+
and the: PbI2 was the anode.


I need to determine the:
 - [Pb2+] in the equilibrium with PbI
 - [I-] in solution
 - the calculated Ksp value
 - the theoretical Ksp value

well the theoretical Ksp is easy - you just look it up and it is 8.3E-9

would Ksp be = [Pb2+][I-]2  ?

now having that, i just need the concentrations!  :-\

now I know that I can determine the Pb2+ concentration using the nerst equation:

0.66V = Eo - (0.0591/2) * log( [Pb2+] / [anode] )

problem is - i dont know the concentration of PbI2 nor do I know what Eo could be.

is there anyone out there who could give me a hand?

Borek:
PbI2 activity is 1, but it doesn't matter.

You have two identical halfcells, with Pb electrode and Pb2+ in solution. The only difference is in the [Pb2+ ]. IIRC it is called gradient cell.

sundrops:
ok I tried to determine the concentration of the PbI cell. Lets hope you can follow my reasoning...

my concentration of Pb2+ is 0.050M (from the Pb(NO3)2 soln)

Ecell = Eo(cathode) - Eo(anode)
0.066V = -0.13V - X
0.196V = X

so the voltage of the PbI2 half-cell is 0.196V

then I use the nernst equation:

0.066V = 0.196V - ((8.3145)(294.65) / (2)(96485)) * ln([anode]/[cathode])
0.066V = 0.196V - 0.0127 * ln([PbI2]/0.050)
-0.13 / -0.0127 = ln [PbI2] - ln (o.o5o)
10.24 = ln [PbI2] - ln (0.050)
ln(10.24) = e^ (PbI) - e^0.050
2.33 = e^(PbI2) - 1.0512
3.38 = e^(PbI2)
ln (3.38) = [PbI2]
1.217 = [PbI2]

does that look ok?


I sure hope so... ;D

Borek:

--- Quote from: sundrops on November 27, 2005, 02:36:51 AM ---does that look ok?
--- End quote ---

No. At first sight, due to several reasons.

First, you are not interested in the PbI2 concentration, but in Pb2+ concentration.

Second, what is 0.0127? It should be close to 0.059/n=0.0295.

And third - as I already mentioned in my previous post activity of PbI2 is 1 - so it can't be 1.217 at the same time.

Try to write down Nernst equation for concentration cell (not a gradient cell, sorry for that). Check it here.

sundrops:
how do uyou know that the [PbI2] = 1 ?
also how did you determine the number of electrons ais 0.0295? wouldn't it be 2e-? and therefore n=2? I'm still very confused.

I'll try the question again, and hopefully this time around it'll be a bit better...

Ecell = Eo - 0.0591/n * log([cathode]/[anode])
0.066 = -0.13 -0.0591/2 * log(0.05/[anode])
0.22555 = log(0.05)-log(anode)
0.22555 = -1.30 - log[anode]
1.527 = -log[anode]
[anode] = 10^(1.527)
[anode] = 33.65 = [Pb2+] but that is not possible....  :-\

i'm getting beyond frusterated

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