Use the following information to answer this question:

Cu+(aq) + e- → Cu(s) E° = 0.521V
Cu2+ (aq) + e- → Cu+ (aq) E° = 0.153 V.
Given these half cell reactions, an aqueous solution of Cu+ ion in the absence of O2(g) :
A. will be thermodynamically stable
B. will almost entirely oxidize to form Cu2+
C. will almost entirely reduce to form Cu(s)
D. will disproportionate to form 50% Cu(s) and 50% Cu2+
E. will reduce water to H2 gas

Chemistry - Science (Dr. Bob222) - DrBob2222, Saturday, May 3, 2014 at 12:12pm

I am torn between answers C and D. I rule out D on the basis of the 50% so I would choose C. I calculated K for C and it is about 10^8 while K for D is about 10^6 if I didn't goof on the math. Check out my thinking since I have this doubt in my mind.

Chemistry - Science (Dr. Bob222) - Ana, Saturday, May 3, 2014 at 12:17pm
the correct answer is D, but i am confused on the rational here. Not sure what the professor wants me to understand here, the concept behind...

Chemistry - Science (Dr. Bob222) - DrBob222, Saturday, May 3, 2014 at 11:08pm
Thanks. I just didn't think the 50% could be right.
The concept behind disproportionation is that an element in an "intermediate" oxidation state (in this case Cu^+ is halfway between Cu below it and Cu^2+ above it) can "react with itself" if the potentials are right. In the case of Cu we would have

Cu^+ ==> Cu^2+ + e E = -0.153
Cu^+ + e ==> Cu E = +0.521
So the cell reaction will be
Cu^+ + Cu^+ ==> Cu + Cu^2+ with a positive voltage of 0368v.
In other words one Cu^+ is oxidized and one Cu^+ is reduced.
I started to leave the question but thought better about it IF I noted I was unsure about the answer. The reason I was unsure was because of the 50% statement so I calculated K and found K for this reaction was about 100 times less than for K for answer C. What I forgot was that C couldn't be right with the possibility of disproportionation since SOME of it would disproportionate so C couldn't be right. And the 50% statement is true since for every Cu^+ that gets reduced another one gets oxidized. 50% on the nose. Cu^+ isn't the only ion that does this. I think Hg2Cl2 does it. Hydrogen peroxide disproportionates. You can read more about it here.


Chemistry - Science (Dr. Bob222) - Ana, Sunday, May 4, 2014 at 9:26am

Thank you for such a great response! You are a legend!

Good morning,

Could you please let me know how did you calculated K for answer D and C. Thank you.

RTlnK = nFEo

To answer this question, we need to analyze the given half-cell reactions and determine the possible outcomes for an aqueous solution of Cu+ ions.

The first half-cell reaction is Cu+(aq) + e- → Cu(s) with a standard reduction potential (E°) of 0.521V. This means that Cu+ ions can be reduced to form solid Cu.

The second half-cell reaction is Cu2+(aq) + e- → Cu+(aq) with an E° of 0.153V. This means that Cu2+ ions can be reduced to form Cu+ ions.

Based on these two reactions, we can conclude that Cu+ ions can exist in the presence of Cu(s) and Cu2+ ions. This indicates that Cu+ ions can both oxidize to form Cu2+ and reduce to form Cu.

The correct answer is option D - an aqueous solution of Cu+ ions will disproportionate to form 50% Cu(s) and 50% Cu2+. This is because one Cu+ ion gets reduced to Cu(s) while another Cu+ ion gets oxidized to Cu2+ ion, resulting in a 1:1 ratio of Cu(s) and Cu2+ ions.

It is important to note that disproportionation reactions occur when an element in an intermediate oxidation state reacts with itself. In this case, Cu+ is in an intermediate oxidation state between Cu and Cu2+, allowing it to undergo disproportionation. This concept is useful in understanding the behavior of certain ions and compounds.

If you have further questions or need more clarification, feel free to ask!