Indicate whether the following half-cells would act as an anode or a cathode when coupled with a S.H.E. in a galvanic cell. Calculate cell potential:

Question

Pt/O2(800 torr)/HCl(2.5E-5 M)

I have not figured out a way to approach this problem. I know the nernst comes in, but I am at a lost on how to set it up.

Answer 1

I assume the half cell for the Pt/O2 couple is 4H^+ + 4e + O2 ==> 2H2O E = 1.229 v as a reduction.
Add that to the H2 couple of
H2 ==> 2H^+ + 2e E = 0 as an oxidation.

The cell reaction then is
2H2 + 4H^+ + 4e + O2 ==> 2H2O + 4H^+ 4e
The 4e cancel.
I've arranged this so the O2 couple is reduction and the H2 is oxidation. That gives a + voltage of 1.229 v which makes the SHE the anode (negatively charge).
Then Ecell = Eocell - (O.06/4)log(Q)
Be careful with Q. At first glance it appears that the H^+ will cancel but the way I read the problem the H^+ is not the same concn in SHE as it is in the O2 so I think they must stay separate.

Answer 2

What goes into Q exactly? I know activities are involved, but what species are in the Q expression? Would it be Cl-/O2? This is where I'm getting confused.