I need someone to be more specific because I still don't understand. Here is the problem:
Cu+2 + 2e- ---> Cu 0.34
Zn+2 + 2e- ---> Zn -0.76
A galvanic cell is made using 2M Zn+2 and 1M Cu+2 solutions. The emf of the cell is:
(a) less than 1.1 V
(b) 1.1 V
(c) greater than 1.1 V
(d) the answer can't be determined
The answer is (a) because Zn is the product and we since we increase the concentration of the product, we lower the emf. My question is, how do we know that Zn is the product and that we lower the emf?
The one with the lower concentration has to be at the anode. So the 1M Cu solution has to be at the anode. Since oxidation occurs at the anode ("An Ox"), we know the reaction at the anode is:
Cu --> Cu+2 + 2e-
This implies the reaction at the cathode is the opposite reaction, i.e.:
Zn+2 + 2e- ---> Zn
So that's how we know Zn is a product formed at the cathode. If we are building up more and more product, then the reaction is not going to want to go forward in that direction as much (think LeChatelier's principle). This translates into a lower potential. Just think of it very generally as-- whatever happens to stimulate the reaction = higher potential for the reaction.
Going back to an earlier key pt of the problem- how do we know we want the lower concentration at the anode? Again, it's sorta like LeChatelier's problem. The two half cells have different concentrations, and the one that has less wants to have more.
The only way to get the higher [metal cations] is to run the reaction in the direction where a solid metal is getting oxidized to produce metal cations and electrons. So we know the one with the lower concentration solution should be the one getting oxidized, so as to produce more metal cations. Oxidation occurs at the anode, so this must be happening at the anode. This works out, because the electrons, as products of that reaction, get sort of "drawn out" from the anode to the cathode, so this drives the reaction forward (as products "disappear").
So remember, the [cations at the cathode] = reactants (since they're reacting with the incoming electrons to form a solid metal on the cathode), whereas [cations at the anode] = products (since the solid metal of the anode is getting oxidized and producing the cations as well as electrons). You could use this with the Nernst equation by plugging in Q = x/y, where x = [products] = cations at anode] and y = [reactants] = [cations at cathode]. So in this case, y > x, and Q < 1. This means a lower potential than what you'd have if x = y, s.t. Q = 1.
Basically there are just a few facts to keep in mind in these problems:
1. "An Ox" (oxidation occurs at anode)/"Red Cat" (reduction occurs at cathode)
2. electrons always go from anode to cathode (in both galvanic & electrolytic cells)
3. if concentrations of 2 half cells are not equal, lower concentration solution should be at anode, if you want the reaction to run on its own.
