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:
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.
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.
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.
To determine whether the given half-cell would act as an anode or a cathode in a galvanic cell, you need to compare the standard reduction potentials (E°) of the half-reactions involved.
In this case, the half-cell consists of Pt (platinum), O2 (oxygen gas at 800 torr), and HCl (hydrochloric acid at 2.5E-5 M concentration).
Step 1: Identify the relevant half-reaction(s).
First, let's identify the possible half-reactions that involve the species in the given half-cell:
1. The reduction half-reaction for O2: O2 + 4e- -> 2O2-
2. The reduction half-reaction for H+ (from HCl): 2H+ + 2e- -> H2
Step 2: Determine the standard reduction potentials (E°).
Next, we need to determine the standard reduction potentials (E°) for each half-reaction. You can find a table of standard reduction potentials for various half-reactions.
For example:
- E°(O2/O2-) = +1.23 V
- E°(H+/H2) = 0.00 V
Step 3: Compare the standard potentials to determine the anode and cathode.
Since the anode is the electrode where oxidation occurs, it will have a more positive reduction potential (E°). On the other hand, the cathode is the electrode where reduction occurs and will have a less positive reduction potential (E°).
In this case, since E°(O2/O2-) = +1.23 V and E°(H+/H2) = 0.00 V, O2/O2- would act as the cathode and H+/H2 would act as the anode.
Step 4: Calculate the cell potential (Ecell).
The cell potential can be calculated using the Nernst equation:
Ecell = E°cathode - E°anode
In this case, since E°cathode = +1.23 V and E°anode = 0.00 V:
Ecell = (+1.23) - (0.00) = +1.23 V
So, the cell potential for the given galvanic cell is +1.23 V.
Remember to check the signs of the standard reduction potentials and make sure they are correctly assigned to the anode and cathode based on their relative values.