Calculating Cell Potential at 25oC with Cu2+ and NaOH

[kooombaya]

What is the potential of this cell ( in V) at 25 oC if the copper electrode is placed in a solution in which [Cu2+] = 5.1×10-10 M?

I found this answer to be 0.065 which is correct. The second part:


If the copper electrode of the above cell is placed in a solution of 0.270 M NaOH that is saturated with Cu(OH)2, what is the potential of the cell (in V, measured in the same direction as in question 5) at 25 oC? Ksp = 1.6×10-19.

I'm using the Nernst equatuion.
Here's how I think it goes:
The NaOH is soluble so it dissolves but the Cu(OH)2 isn't. It's Ksp is very small but it will dissociate a little. So the dissociation equation I got is:

Cu(OH)2 ---> Cu2+ + 2OH-
Ksp = [Cu2+][OH-]^2
= [x][2x]^2
x = 3.42e-7

To find Ecell I said Cu gets oxidized and OH reduced and my Ecell = -0.74
Plugging into the Nernst equations:

E=-0.74-(0.0592/4)*log[Cu2+][OH-]^2
E=-0.64

This is wrong. Can someone explain to me please. Thanks!

 

[kooombaya]

Can anyone please help me, or just give me a hint...

 

[Borek]

In the first part you have calculaed potential of the copper half cell, why do you try to incorporate OH^- concentration into Nernst equation now? Were you told anything about change of the other half cell?

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[kooombaya]

Ok, I have it set up as

Cu(OH)2 ---> Cu2+ + 2OH-

Ksp = 1.6e-19 = [x][2x+0.1]^2

Now this might sound stupid but how do I solve this equation. I'm stuck since we get an x^3 and x^2 term...

 

[Borek]

Sorry, I posted something else at first, but then realized you did something strange. Why 2x+0.1?

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methods

 

[kooombaya]

2x+0.1 because of the .1 from the NaOH

 

[Borek]

Strange.

solution of 0.270 M NaOH

Is concentration of dissolved copper hydroxide comparable with concentration of NaOH?

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methods

 

[kooombaya]

I mean 2x+0.27 this is from the answer key that the professor gave me. I don't know how to solve the equation though.

 

[Borek]

OK, 2x+0.27.

Can you answer the question I have asked?

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methods

 

[kooombaya]

From what I understood the 0.27 is from the NaOH that dissociates.

 

[Borek]

That's not answer to the question I asked, to speed things up asnwer two questions now:

Is concentration of dissolved copper hydroxide comparable with concentration of NaOH?

Why 2x+0.27 for the OH^- concentration?

And we won't move froward till you answer them both, so dodging won't help.

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[kooombaya]

I'm not trying to evade any question. Nevermind my chem exam is over, I don't feel like caring about it anymore... thanks for your help.