[Burningkrome]

So, I’m very frustrated. I've been reading  this tutorial…

http://www.chem1.com/acad/pdf/c1xElchem.pdf

…thinking I’m understanding it completely; until I get to the following question. The answer in the book is either completely wrong, or I've completely misunderstood everything.


The question is, which way is the electron flow traveling in the following electrochemical cell equation?

Cu(s) | Cu²⁺ || Cl^– | AgCl(s) | Ag(s)

- (The left to right format of the equation is supposed to be arbitrary and not an indicator)
- The equation is at Std ATM of 1 and 25oC.
- The electrolytes solutions contain existing concentrations of Cl^- and Cu²⁺.

In the book, the author states the electrons are being withdrawn from Cu (anode) and traveling to Ag (cathode).

I think it should be the other way around.

Here’s why. The E^o of Cu(s)/Cu⁺² is +0.337 and "AgCl/Ag(s) Cl^-" = +0.222. Based on the standard reduction potentials table in the book, this means Cu²⁺ is a significantly greater electron receptor than AgCl. The author also states that any reducing agent that lays in the table “above” (lower voltage potential couple) an oxidizing agent can (will?) provide electrons to that oxidizing agent.

Well, the "AgVl/Ag(s) Cl^-" couple is a lower voltage potential ("higher" in the table at +0.222) than the "Cu(s)/Cu²⁺" couple. 

So, shouldn't Cu²⁺ end up working as the electron sink (being reduced)?

Thanks! 

[Borek]

In general you are right, although exact details will depend on the concentration of Cl^-.

[Hunter2]

In the book, the author states the electrons are being withdrawn from Cu (anode) and traveling to Ag (cathode).

Means through the wire not through the solution. Silver is more noble as copper, so therefor copper Cu²⁺ get dissolved, electrons will remain on that electrode (Anode) and travel through the wire to silver electrode and catch silver ions.