HW 13.11 balance half-cell O2--> OH

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y3chem
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HW 13.11 balance half-cell O2--> OH

Postby y3chem » Tue Feb 04, 2014 12:32 am

For HW problem 13.11 part (d)
The cell diagram is: Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)
I am not sure how to balance O2--> OH- (for cathode)
the answer keys says: O2(g)+2H2O(l)+4e--> 4OH- (aq)

I started with balancing oxygen by adding H2O: O2--> OH-+H2O
but then after that I couldn't really figure out the answer that is the same as the answer key.

can someone show me how it is balanced step by step?

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Re: HW 13.11 balance half-cell O2--> OH

Postby Chem_Mod » Tue Feb 04, 2014 4:44 pm

For this type of problem you would need to have a table of half reactions available. So to do this you want to look at page 532 of your textbook.

The whole goal is for you to balance O2 going to OH-. To help us we have the cell notation. This tells us that we have O2 and H+ going to OH- and O2. We ignore the Pt since it is inert by design.

Thus, we should look for half reactions involving some combination of those species.

Going to our handy table, we see the following two rows:
Oxidation/Anode: O2,H+/H2O: O2(g) + 4H+(aq) + 4e- ---> 2 H2O(l) E0=1.23
Reduction:Cathode: O2,H2O/OH-: O2(g) + 2H2O(l)l + 4e- ---> 4OH- E0=0.40

Our cell diagram tells us what species is the anode and which is the cathode. The anode is on the left and the cathode is by definition on the right. We see that O2 shows up on both sides so in one reaction it is reduced and in the other it is oxidized. For the anode reaction we know we have O2(g) + H+(aq) because that is on the left side of the cell diagram. This matches up with the top reaction. Since this occurs at the anode we want to flip it since this is our oxidation reaction. Also as a tip reduction occurs at the cathode (both start with consonants) and oxidation occurs at the anode (both start with vowels)

Oxidation/Anode: 2 H2O(l) ---> O2(g) + 4H+(aq) + 4e- E0=-1.23
Reduction:Cathode: O2(g) + 2H2O(l)l + 4e- ---> 4OH- E0=0.40


The standard cell potential is then: Eox + Ered = -1.23 V + 0.40 V = -0.83 V.

Adding the oxidation and reduction reactions gives:

4 H2O(l) --->4 OH-(aq) + 4H+
H2O(l) ---> OH-(aq) + H+(aq)

So you can see that in your reduction reaction you do solve O2 --> OH- but it is easiest to go from the cell diagram and then find your reactions from there. After that then add them together and solve for the cell potential. This special reaction we have at the end is called water hydrolysis :).

y3chem
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Joined: Fri Sep 20, 2013 3:00 am

Re: HW 13.11 balance half-cell O2--> OH

Postby y3chem » Wed Feb 05, 2014 12:34 am

Ohhh now I understand how the textbook did it

just another curious question
Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)

for the anode reaction O2--> H+
why is it okay to use the equation O2+4H++4e--> H2O for its standard E?
We don't necessarily need to have H+ on the product side?

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Re: HW 13.11 balance half-cell O2--> OH

Postby Chem_Mod » Wed Feb 05, 2014 10:54 am

We use that reaction because we see in the reduction potential table that we are reacting O2 and H+ together and this is what is occurring in the anode reaction. Even though there are extra terms, that comes from the balancing which the table is nice enough to do for you. The fundamental reaction is the same and that is what you want to focus on when selection which reaction from the table to use.

Technically we do have H+ on the product side since H2O (l) = H+(aq) + OH-(aq). Oftentimes this is written the opposite way where you will convert something like 4H+(aq) + 4OH-(aq) to 4H2O (l) and then cancel that water out with other waters on the other side.


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