Electrochemistry Ecell vs emf question

Started by mcheung
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mcheung

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Hey guys, I'm just a bit stumped right now after doing the practice passage in TBR for the last chapter of general chemistry. I would be so appreciative if someone could lend me a helping hand!

The question that stumped me was passage 13 on page 300, question 83.

This is what is given:

Cu2+ + 2e- --> Cu(s) E= 0.34V
Cu2+ + 1e- --> Cu+ E= 0.15V

Q: What the Gibb's free energy change for the reaction: Cu (s) ---> Cu+ + 1e- ???

Instead of doing their long way of finding the change in Gibb's free energy of the two separate reactions and adding them up, I decided to sum up the E values to find Ecell, and then attempt to find G by using the equation G=-nfE.

Can someone please explain to me why my approach is wrong? :S

My Ecell was calculated to be = -0.34V + 0.15V = -0.19V

Usually Ecell = Ered + Eoxi , without needing to take into account the number of e- . For example, in the 2nd reaction above, we would still use E= 0.15V, instead of 0.30V even though the first reaction has 2e- and the 2nd reaction has 1e-

I'm just really frustrated that I can't figure out what went wrong with my logic..
Someone, please help haha!
Thanks in advance!!!
 
I'm sorry haha I swear I'm going full derp here... can you just elaborate on that a little? Like when can you add them and when can't you add them?
 
Nvm! Found a good explanation online.
For anyone else that is wondering the same thing as me, the page says:

"...if we are combining two half reactions to obtain a third half reaction, the E° values are not additive, since this third half-reaction is not accompanied by another half reaction that causes the charges to cancel. Free energies are always additive, so we combine them, and use ΔG° = –nFE° to find the cell potential"

Link: http://chemwiki.ucdavis.edu/Analyti...emistry_3:_Cell_potentials_and_thermodynamics

Thanks guys! Happy studying!