My book states that $\ce{2AgNO3(aq) +Cu ->Cu(NO3^-)2(aq) +2Ag(s)}$. In addition to the formula, it also shows the following image:

Image as taken from the textbook

As in the image it shows the salmon colored copper atoms are donating an electron to the $\ce{Ag+}$ ions. In the 2nd picture on the page it shows the $\ce{Ag}$ atoms on top of the salmon colored copper atoms. I would assume that the $\ce{Cu}$ and $\ce{Ag}$ form somewhat of an ionic bond, is this correct to say? If that is the case, why does the formula not indicate a bond between the copper and silver instead the overall product is just copper nitrate plus silver. I would think it would be something like $\ce{CuAg}$?


Actually the copper and silver aren't bonded. The silver that precipitates is pure. Also, a bond between two metals could not be ionic. Ionic bonds are always been metals and non-metals. Looking at the picture again, I think what may have thrown you off is that it looks like the silver at the end is Ag+. However, the silver is actually Ag+ in the beginning of the reaction when it is in the silver nitrate. That's how it is bonded with the nitrate. When the copper donates an electron to the silver, the silver goes from having a +1 charge to having no charge. It becomes neutral, and hence it doesn't bond with the copper.

  • $\begingroup$ Ionic bonds can also be between metal ions (e.g. $\ce{Cs+Au-}$), or between non-metal ions, such as pyridinium chloride. $\endgroup$
    – bon
    Jul 18 '16 at 12:47
  • $\begingroup$ There are exotic exceptions like( K-18-crown-6)${^+}$Na$^-$, but they can be ignored for most practical cases. Ag is just depositing on the surface of Cu, no ionic bond there. $\endgroup$
    – sixtytrees
    Jul 19 '16 at 14:16
  • $\begingroup$ @Noel how does Cu end up as cu2+? Also, in the third picture, (where it is blue) it depicts that in the free floating solution there are now cu2+ ions and NO3-, how does the copper become detached from the other other copper atoms that are aggregated together? From what I see they were originally fitted together or bonded in the coil? $\endgroup$ Jul 23 '16 at 6:17
  • $\begingroup$ The Cu ends up as Cu2+ because it donates electrons to the silver. If it 'gives away' two electrons, its charge will increase by 2. As for your second question, I'm afraid I don't know the answer. My guess would be that since it is suspended in the solution, the copper ions come in contact with the nitrate ions, and they only 'leave' the rest of the copper when they react with the nitrate. They don't freely float around as shown in the diagram.@Atticus283blink $\endgroup$
    – Noel
    Jul 24 '16 at 1:57
  • $\begingroup$ @Noel Thanks so much! Yes, I knew that the loss of electrons makes it more positive but now i see that it donates 2 electrons to two Ag's. Yes, i assumed that was the case but wasn't sure, the pictures are very deceiving. $\endgroup$ Jul 25 '16 at 2:08

The hint to answering your question yourself is that the book (correctly) does not speak of any ‘$\ce{CuAg}$ phase’ but of $\ce{Cu}$ and $\ce{Ag}$. There is no bond between the two.

The silver atoms merely aggregate on the copper metal surface to form metallic silver — it is the fluff you can see in the image. It depends on the exact conditions but it can probably easily be shaken off, would then fall to the ground of the tube and a fresh copper surface would appear.

Intermetallic phases can be made, but the phase diagram of copper/silver phases (see below) shows that no intermetallic copper/silver phases $\ce{Ag_xCu_y}$ exist.

Cu/Ag phase diagram

The eutectic point E ($71.9~\%\ \ce{Ag}$) is merely the lowest possible co-melting point of copper/silver mixtures.


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