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Topic: Galvanic cell concept clarification?  (Read 1416 times)

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Offline Bublik

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Galvanic cell concept clarification?
« on: December 22, 2012, 11:49:41 AM »
Hello,
I'm having trouble, and getting confused, about the concept by which a galvanic cell generates electricity. I read my textbook about all of it, and I have a few questions which I'm hoping to get clarified.

1) If you put zinc metal in zinc sulfate solution, the metal will be releasing some zinc ions into the solution, as well as electrons (2 electrons per zinc ion released, Zn(s)  :rarrow: Zn2+ + 2e-
    If you put copper metal in copper sulfate solution, the exact same thing happens. Copper ions are released into the copper sulfate solution, and electrons are released as well.
My question is, why is there a transfer of electrons when you connect the two solutions by a wire? Each solution can release ions from its corresponding solid, as well as combine the extra electrons with the ions to rebuild the solid (kind of like an equilibrium).

I'm not sure if what I just said is correct, and I don't understand why electrons would move through a wire. What motivates them to do so?

2) http://www.sagepub.com/liustudy/chapters/02/Abstracts/Electrochemistry.pdf

I ran by this review sheet when doing research to answer my own questions. The very first statement on that list confused me: "In standard reduction potential tables the species with the highest Eo value is the anode."

However, when I solved the sample problems in my textbook about galvanic cells, they specifically said that cell potential must always be positive. And since Ecell=Ecathode-Eanode, it was always easy to determine which is the cathode and which is the anode (to make their difference positive).

for example, the redox reaction below
Ag+ + e-  :rarrow: Ag      Eo = 0.80V
Fe3+ + e-  :rarrow: Fe2+      Eo = 0.77V

Since a positive Ecell value is required, reaction (2) must run in reverse, so that the cell potential is a positive 0.03V. this is 0.80V + (-0.77V), so clearly 0.80V is the cathode and not the anode. Doesn't that contradict the first statement of that link?

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