6L.3-d Water and oxygen cell reaction?

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6L.3-d Water and oxygen cell reaction?

Postby nicolely2F » Mon Feb 24, 2020 12:59 am

I'm not sure how to write the half-reactions for Pt(s) | O2(g) | H+(aq) || OH-(aq) | O2(g) | Pt(s)
I can see that we're going to use water, but trying to write the reactions is giving me inverted answers. Could someone please walk me through this?

Ashley Fang 2G
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Joined: Fri Aug 30, 2019 12:17 am

Re: 6L.3-d Water and oxygen cell reaction?

Postby Ashley Fang 2G » Mon Feb 24, 2020 10:05 am

O2 + 4H+ + 4e- -> 2H2O E(anode)=1.23V
O2 + 2H2O + 4e- -> 4OH E(cathode)=0.40V
Reverse the first equation since it's the anode and add the two equations together
You should get 4H2O -> 4H+ + 4OH- which simplifies to
H2O -> H+ + OH-
E(cell) will be 0.40V - 1.23V = -0.83V

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