6L.3D

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Savannah Mance 4G
Posts: 107
Joined: Fri Aug 30, 2019 12:17 am

6L.3D

Postby Savannah Mance 4G » Mon Mar 02, 2020 7:53 pm

Can someone please explain the oxidation rxn with the anode for this one? I'm confused because the anode is O2 and H+ , how would this make up the anode rxn if it's not H20 to H+?

Charlyn Ghoubrial 2I
Posts: 50
Joined: Wed Nov 13, 2019 12:26 am

Re: 6L.3D

Postby Charlyn Ghoubrial 2I » Mon Mar 02, 2020 9:04 pm

Your half reactions should be
Reduction: O2 + 2H2O + 4e- —> 4OH-
Oxidation: 2H2O —> O2 + 4H+ + 4e-
Cancel out the O2, 4e-
Balanced equation: 4H2O —> 4H+ + 4OH- which reduces to H2O —> H+ + OH-

Savannah Mance 4G
Posts: 107
Joined: Fri Aug 30, 2019 12:17 am

Re: 6L.3D

Postby Savannah Mance 4G » Mon Mar 02, 2020 9:34 pm

So, from the cell diagram, the two elements don't have to be on opposite sides of the rxn? Like since O2 and H+ on the anode side of the cell diagram doesn't have to have a half rxn of O2--->H+?

Angela Wu-2H
Posts: 50
Joined: Fri Aug 30, 2019 12:16 am

Re: 6L.3D

Postby Angela Wu-2H » Wed Mar 04, 2020 1:51 am

Savannah Mance 4G wrote:So, from the cell diagram, the two elements don't have to be on opposite sides of the rxn? Like since O2 and H+ on the anode side of the cell diagram doesn't have to have a half rxn of O2--->H+?


I have the same question. Shouldn't the half reaction go from O2 to H+ because the cell diagram says O2(g)|H+(aq)? Where did they randomly get the H2O from and how was I supposed to know to use H2O?


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