[Silent]

I've been working on this question for a while and can't figure out the answer, any help would be appreciated.

I need to find the strongest oxidizing and reducing agents and write the chemical equations to represent the cathode, anode and net cell reactions for

Pt_(s) l Na⁺_(aq), Cl^-_(aq),O_2(g), H₂O_(l) ll Al³⁺_(aq) l Al_(s)

I know that the platinum is an inert electrode, that the strongest reducing agent is Al_(s) and that the chemical equation for the anode is Al_(s) -> Al³⁺_(aq) + 3 electrons. The cation chemical reaction is what has me confused.

[Kryolith]

If you have to decide which species will be reduced you have to consider the redox potentials and the overpotential at the specified electrode.

[Silent]

I never got taught anything about overpotentials, but I did get an answer on another forum I posted this question on. Thanks for the help though!

In case anyone wants to know the answer was:

Since the Al is losing e-, the other electrode must have something gaining e-.  Of the three substances in contact with the Pt (which is inert as you said), the only one that will gain e- is the O2.  The 1/2 reaction that will occur there is

O2  + 4e-  +  2H20  -->   4 OH-

Just find multiply the oxidation reaction by 4 and the reduction reaction by 3 to insure the # of e-'s gained and lost are the same and then add the two 1/2 reactions to get the net cell reaction.