[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 Pb_I
 - [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...

[Borek]

does that look ok?

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

[Borek]

how do uyou know that the [PbI2] = 1 ?

PbI₂ is solid, not dissolved. Activity of solids is always considered to be 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.

Coefficient in the Nernst equation is 0.059/n - where n is number of electrons transferred during reaction. In this case n=2, so the coefficent is 0.059/2 = 0.0285. You used in your calculations some strange value of  0.0127.

Ecell = Eo - 0.0591/n * log([cathode]/[anode])

These are not [cathode]/[anode] but Pb²⁺ concentrations in both cells.

0.066 = -0.13 -0.0591/2 * log(0.05/[anode])
0.22555 = log(0.05)-log(anode)

Where did you get 0.22555 from? Check your math.

Besides, are you sure it was 0.066 and not -0.066?

[sundrops]

ok i see,


\

These are not [cathode]/[anode] but Pb2+ concentrations in both cells

yes but i thought that might be a good way to distinguish the Pb2+ from solution from the Pb2+ from the PbI2 mixture.

now using Ecell = Ecathode - Eanode
0.066 = -.013-X
X=-0.196 = Eanode

so would Eo in the nernst equation be -0.196 instead of the -.13?
in that case:

0.066 = -.196 - 0.0591/2 * log(0.05/Pb2+)
-.2926 = log(0.05) - log[Pb2+]
-0.2926 = -1.3010 -  log[Pb2+]
-1.0084 =  log[Pb2+]
[Pb2+] = 10^(-1.0084)
[Pb2+] = 0.09808

does that look better? *crosses fingers*

[Borek]

does that look better? *crosses fingers*

Please stop guessing, try to understand what you are doing first.

Potential in concentration cell is given by

E = E₀ + 0.059/n log(c₁/c₂)

so only concentration ratio is important. Now put the numbers into equation:

0.066 = -0.13 + 0.059/2 log(c₁/c₂)

after rearrangement:

log(c₁/c₂) = 2(0.066 + 0.13)/0.0591 = 6.6

so

c₁/c₂ = 2.5*10^-7

Now, I never remember which concentration is which - but in this case it is not a problem, as the concentration in the half cell with PbI₂ must be much lower than the concentration in the second cell (the one with 0.05M Pb(NO₃)₂). If so:

c₁ = 2.5*10^-7 * 0.05M

is the only way to go.

Use this c₁ to calculate Ksp in the PbI₂ cell - [I^- ] can be asily calculated from the initial data, assuming that the precipitation was stoichiometric (it almost was).

The problem is - if you calculate Ksp this way you will be off by several orders of magnitude.

However, if you did a classic mistake measuring potential, and what you wrote as 0.066 was in reality -0.066, error in your Ksp determination will be much lower (although still too large for me).

[bubblyporcupine]

thanks Borek, you've been incredibly patient  

the 0.066volts was obtained experimentally and our machine did not go below zero, I don;t think it really matters as long as the mean class average had around the same voltage.  

thank you again.  

[Borek]

the 0.066volts was obtained experimentally and our machine did not go below zero

It doesn't matter. If you switch connectors potential will be reversed.