6L.3 part d

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inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

6L.3 part d

Postby inlovewithchemistry » Tue Feb 26, 2019 6:38 pm

Write the half reactions and the balanced equation for the cell reaction for the following galvanic cells:

(d) Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)

I don't understand how to get the answer that is in the solutions manual or how to set up a half reaction with the cathode. Can someone explain what they did?

ariana_apopei1K
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Joined: Fri Sep 28, 2018 12:27 am
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Re: 6L.3 part d

Postby ariana_apopei1K » Tue Feb 26, 2019 7:02 pm

so for the cathode i did O2 --> OH- and treated it as a half reaction in a basic solution because of the OH-. once you balance both sides with H+'s, OH-'s, etc you should get 4e- + 2H2O + O2 -> 4OH-

For the anode side you would solve normally, like in an acidic solution. so O2 -> H+ would balance out to 4H+ + O2 + 4e- -> 2H2O. youd have to flip the anode rxn so the electrons cancel out, then add them together and get the overall rxn from the solutions manual.

Was there a specific part of the problem you had trouble with? or just the cathode thing

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

Re: 6L.3 part d

Postby inlovewithchemistry » Tue Feb 26, 2019 7:49 pm

I just dont understand why the solution manual writes the anode half rxn as O2 +4H+ + 4e- --> 2H2O when the two things listed in the left of the cell diagram are O2(g) and H+(aq). I tried to find the half rxn beginning with

O2 --> H+

but that apparently isnt right.


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