Calculating Cell Potential at 25oC with Cu2+ and NaOH

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    Cell Potential
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Discussion Overview

The discussion revolves around calculating the cell potential of a copper electrode in different solutions, specifically focusing on the effects of copper ion concentration and the presence of NaOH and Cu(OH)2. Participants explore the application of the Nernst equation in this context, addressing both theoretical and practical aspects of electrochemistry.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant calculates the potential of a copper half-cell at a specific concentration of Cu2+ and finds a value of 0.065 V, but questions the potential when Cu(OH)2 is involved.
  • Another participant questions the inclusion of OH- concentration in the Nernst equation, suggesting that the other half-cell conditions were not considered.
  • There is confusion regarding the correct formulation of the Ksp equation for Cu(OH)2, with one participant expressing difficulty in solving the resulting cubic equation.
  • Participants discuss the concentration of dissolved copper hydroxide in relation to the concentration of NaOH, raising questions about the assumptions made in the calculations.
  • One participant refers to an answer key provided by a professor, indicating a specific formulation for the OH- concentration that includes the NaOH concentration.
  • Another participant expresses frustration over the ongoing questions and the complexity of the problem, indicating a lack of clarity in the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the cell potential, with multiple competing views and unresolved questions regarding the assumptions and formulations used in the calculations.

Contextual Notes

There are limitations regarding the assumptions made about the concentrations of ions in solution and the dependence on the definitions of solubility products. The discussion reflects uncertainty in the mathematical steps involved in solving the Ksp equation.

kooombaya
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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!
 
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Can anyone please help me, or just give me a hint...
 
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...
 
Sorry, I posted something else at first, but then realized you did something strange. Why 2x+0.1?

--
methods
 
2x+0.1 because of the .1 from the NaOH
 
Strange.

kooombaya said:
solution of 0.270 M NaOH

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

--
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.
 
OK, 2x+0.27.

Can you answer the question I have asked?

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methods
 
  • #10
From what I understood the 0.27 is from the NaOH that dissociates.
 
  • #11
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.

--
 
  • #12
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.
 

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