Copper Solubility in Water with pH 5

AI Thread Summary
The discussion revolves around calculating copper solubility in water with a pH of 5 and a reduction potential of +0.30 V. The user attempts to apply the Nernst equation and outlines their calculations, but expresses uncertainty about the relevance of pH in the context of the problem. They derive a concentration of copper ions but doubt the accuracy of their result, suggesting confusion over the problem's requirements. Ultimately, they question whether pH affects the calculation or if it can be disregarded in favor of the reduction potential. The conversation highlights the complexities of electrochemistry and the need for clarity in problem statements.
Kristina_elt
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Hello,

I have a problem with this task:

"Water with pH 5 and reduction potential +0.30 V flows through copper pipe at 25 °C. Calculate how much copper dissolves (in ppm) in water due to corrosion of this pipe. Standard reduction potential of copper Cu0/Cu2+ is 0.34 V."

I spent lot of hours with lot of calculations without any glory :( Please, is there anybody who could try to help me?

This figure could be helpful:

http://www.advancedaquarist.com/2003/5/chemistry_album/figure3.gif

I tried to use Nernst equation:

Eh = E0 - (RT / nF)*(lnQ).

The main equation I used was:

Cu + 2H+ -> Cu2+ + H2.
 
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Write Nernst equations for both half cells. At the equilibrium both reduction potentials will be identical. Potential of the hydrogen cell is easy to calculate, that leaves you with just one unknown - concentration of the copper.
 
I do not think so :( ...
 
Have you tried to follow what I wrote?

If so, can you show the steps?
 
Borek said:
Have you tried to follow what I wrote?

If so, can you show the steps?

Ok. Firstly I wrote this:

Cu -> Cu2+ + 2e- (E0 = -0.34 V)
H2 -> 2H+ + 2e- (E0 = 0.00 V)

The following reaction was:

Cu + 2H+ -> Cu2+ + H2 (E0 cell = 0.34 V)

Next step:

0.3 V = 0.34 V - (RT / 2F) * ln ([Cu2+] / [10^-10])

Then I got:

3.11 = ln ([Cu2+] / [10^-10])

And for [Cu2+]:

22.42 * 10^-10 = approx. 2.24E-09

In ppm:

2.24E-09 * 1000 * 63.5 = 1.42E-04 ppm (I think this result is incorrect)
 
Sorry. On the second read question no longer makes any sense to me. Apparently I was way too fast in my answer now (and I realized why I didn't answer the first time I have seen the question elsewhere).

I have no idea what they mean by "Water with pH 5 and reduction potential +0.30 V". Unless pH is completely irrelevant and the only thing that is important is the potential - regardless of its source (most likely some not mentioned, but present, redox system). If so, question can be solved by ignoring pH (just by applying Nernst equation to copper). If not - I have no idea what to do.
 
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