Post a New Question

(19) Chemistry - Science (Dr. Bob222)

posted by .

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!

  • (19) Chemistry - Science (Dr. Bob222) -

    Good morning,

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

  • (19) Chemistry - Science (Dr. Bob222) -

    RTlnK = nFEo

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question