Asked by Ana
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
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
Answers
Answered by
DrBob2222
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.
Answered by
Ana
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...
Answered by
DrBob222
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
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
Answered by
Ana
Thank you for such a great response! You are a legend!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.