Electrochemistry homework help problem

In summary, the Tl+/Tl couple can be prepared by saturating 0.100 M KBr with TlBr and allowing Tl+ ions to equilibrate. The couple has a potential of -0.443V with respect to the Pb2+/Pb couple. Given EoPb2+/Pb = -0.126V and EoTl+/Tl= -0.336V, the Ksp of TlBr can be calculated using the cell potential and assuming Br- concentration is 0.1M. The calculated value is 1.124 x 10-5, which may differ from the value in tables due to ionic strength not being taken into account.
  • #1
zorro
1,384
0

Homework Statement



Tl+/Tl couple can be prepared by saturating 0.100 M KBr with TlBr and alowing Tl+ ions from the insoluble bromide to equilibrate. This couple was observed to have a potential of -0.443V w.r.t Pb2+/Pb couple in which Pb2+ was 1 M. Ksp of TlBr is ?

Given: EoPb2+/Pb = -0.126V and EoTl+/Tl= -0.336V


The Attempt at a Solution



I have a problem understanding the question. Can somebody explain me?
 
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  • #2


You have a cell made of two half cells - one is Pb/Pb2+, the other Tl/Tl+. You are given everything but [Tl+], and that's the only thing you need to calculate to find Ksp of TlBr.
 
  • #3


The potential -0.443V given in the question confuses me. It simply means that the cell reaction is not spontaneous. Which one is cathode and which one is anode?

We have Ecell = -0.443V and Eocell = -0.21 (if we assume Tl+/Tl is the cathode)

Is this right?
 
  • #4


To be honest - I am not sure which way it goes off the top of my head, but if you take the sign wrong, you will get absurd results.

You will need E0 for Pb/Pb2+ as well.
 
  • #5


Cathode- Tl+ + e ----> Tl
Anode- Pb------> Pb2+ + 2e

Ecell = Eocell - 0.059/2*log[Pb2+/[Tl+]2]

We have Pb2+=1M
What is the use of 0.1M KBr?

E0 for Pb/Pb2+ is given in the question.
 
  • #6


Abdul Quadeer said:
What is the use of 0.1M KBr?

Have you read the question?
 
  • #7


I read it but did not understand fully.
Br- concentration is 0.1M

Let us assume that Tl+ produced from TlBr is x moles
Ksp = x(0.1+x)

How to find 'x'?
 
  • #8


Abdul Quadeer said:
Br- concentration is 0.1M

Right.

Let us assume that Tl+ produced from TlBr is x moles
Ksp = x(0.1+x)

How to find 'x'?

From the cell potential. It will be so low you can easily ignore x in 0.1+x part.
 
  • #9


Borek said:
From the cell potential. It will be so low you can easily ignore x in 0.1+x part.

Plugging the values in the equation in my post #5, I got x as 1.124 x 10-4
So Ksp = 1.124 x 10-5. This is not the correct answer.
 
  • #10


Thallium chemistry.

Prof. is showing off? :biggrin:
 
Last edited:
  • #11


epenguin said:
Thallium chemistry.

Prof. is showing off?

:biggrin:

@Borek- where is the mistake?
 
  • #12


Your Ksp looks OK to me, unless you are expected to take ionic strength of the solution into account.
 
  • #13


I just have the answer for it.
3.6 x 10-6
There is a big difference.
I think the answer given is incorrect.
 
Last edited:
  • #14


No my fault. I know answer calculated is different from the Ksp value from tables, but as far as I can tell data given yields 1.12e-5.
 

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