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

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.

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...

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.
http://en.wikipedia.org/wiki/Disproportionation

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

To answer this question, we need to compare the standard reduction potentials (E°) of Cu+(aq) and Cu2+(aq) to determine which species is more likely to be reduced.

First, let's compare the E° values:
Cu+(aq) + e- → Cu(s) E° = 0.521V (1)
Cu2+(aq) + e- → Cu+(aq) E° = 0.153 V (2)

In a redox reaction, the species with a higher reduction potential (E°) will be reduced, while the species with a lower reduction potential will be oxidized.

Comparing the two equations (1) and (2), we can see that Cu+(aq) has a higher E° value (0.521V) than Cu2+(aq) (0.153V). Therefore, Cu+(aq) is more likely to be reduced, and Cu2+(aq) is more likely to be oxidized.

Based on this information, we can deduce that in the absence of O2(g), an aqueous solution of Cu+ ion will almost entirely reduce to form Cu(s). Therefore, the correct answer is option C: will almost entirely reduce to form Cu(s).