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Skyllar Kuppinger 1F
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Joined: Thu Jul 25, 2019 12:16 am


Postby Skyllar Kuppinger 1F » Tue Feb 25, 2020 5:06 pm

In part D I just have no idea what's going on in this cell. Both the cathode and anode are exactly the same except there's an H+ on the left and an OH- on the right. What is being reduced?? What is being oxidized??? Why are there 3 diff species on each side instead of 2???

Posts: 95
Joined: Wed Sep 11, 2019 12:17 am

Re: 6L.3

Postby BNgo_2L » Tue Feb 25, 2020 11:44 pm

H20 is involved in both redox half reactions. If you look in the appendix, you would use O2 + 2 H2O + 4 e2 --> 4 OH2 and O2 + 4 H1+ + 4 e2 --> 2 H2O. Also, for cell diagrams, the left is always the anode (oxidized) while the right is always the cathode (reduced).

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