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Author Topic: Why does the copper (II) ion form more readily than the copper (I) ion?  (Read 16471 times)

Offline Chemistry4me

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Seeing as this board has been quite quiet lately, and also to help out our dear friend, I have this udder  [:-'(] question:

Cu2+ electron configuration: 1s22s22p63s23p63d9
Cu+ electron configuration: 1s22s22p63s23p63d10

Looking at that, the Cu+ ion looks much more stable.
Obviously they have different electrode potentials, Cu+/Cu = +0.52 V, Cu2+/Cu+ = +0.15 V, Cu2+/Cu = +0.34 V.
But WHY is Cu+ it a better reducing agent? Why does it not want its 3d orbital to be filled?


 

Offline Bored chemist

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The electrode potential is only half the story.
The other side of the balance is how much do you gain by making that dobly charged ion. The hydration energy of crystal field energy of a doubly charged ion is a lot bigger than that of a singly charged ion. In the case of copper it is sufficient to overcome the second ionisation potential.
On a good day with a following wind you can take the third and fourth electron off it too.
Of course, to do this you need a counter ion with no great enthusiasm for giving the electron back.
For copper (II) you can't use iodide because it gets oxidised. With copper (III) even chloride can't hold the electron tightly enough; but you can get fluoride and oxide complexes. Once you get to copper (IV) only fluoride is a poor enough reducing agent to survive.
 

Offline Chemistry4me

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The hydration energy of crystal field energy of a doubly charged ion is a lot bigger than that of a singly charged ion. In the case of copper it is sufficient to overcome the second ionisation potential.
Ah okay, I see now, thank you Bored Chemist.
 

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