# Cell potential

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!

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

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methods

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

Borek
Mentor
Strange.

solution of 0.270 M NaOH

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

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methods

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
Mentor
OK, 2x+0.27.

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methods

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

Borek
Mentor
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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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.