can somebody please explain the rationale here.
At which electrode is oxygen gas liberated?
A . 4 only
B . 1 and 2 only
C . 2 and 4 only--->ans
D . 2, 3, and 4 only
i understood the part that 1 and 3 are cathodes and 2 and 4 are anodes. but how do you figure where O2 is liberated. I picked 4, but I really don't understand how to solve this problem. Thanks a lot!!
The last two equations on the chart do not seem to be balanced. For the fourth equation, I'm not sure where the SO42- is coming from in "O2 + 2H2O + 4e -> 4OH- + SO42-" It would make sense if it was "O2 + 2H2O + 4e > 4OH-."
If you look at the left cell, you see that you have your pick of Na+, OH-, and H2O (or H+ and OH- since they do dissociate slightly). Since electrons leave electrode 4, you know at one of these substances has to get oxidized here. The one which will get oxidized is the one with the lowest reduction potential. Now, to find which substance will get oxidized, the important thing here is to
only look at the products of all those reactions on the table. Those products are
reduction products and therefore they are
oxidation reactants. Considering that you only have Na+, OH-, and H+ in your cell, the only thing that's even on the products side of the table is OH-. So OH- will get oxidized. What happens then? According to the reduction reaction "O2 + 2H2O + 4e > 4OH-," you can see that OH- will give off O2 when oxidized.
On the right cell, you see that since Cu+ has the highest reduction potential at 0.34 V, it will get reduced by those electrons from electrode 4 at electrode 1. What is going to get oxidized at electrode 2? Again, consider what you have in this cell. You have Cu+, SO42-, H+ and OH- (the last two are from H2O). Again, the only thing that makes sense to be oxidized is OH-. You could figure this out by looking at the table and finding the compound with the lowest reduction potential, but it's quicker to realize that you simply don't have any other oxidize-able compound in your right cell that's listed on this table except OH-. This will give off O2 again. Hope that helps.
Also, I'm not sure how it works out that both oxidation and reduction are taking place in the same cell. This question seems to want you to understand that, for example, oxidation at electrode 4 will lead to reduction at electrode 3 or something. Does anyone know how this is possible?
