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Posted: Thu Feb 22, 2018 2:04 am
How come the anode rxn's Eº = +1.23 V and not -1.23 V? In the book the reaction is written O2 + 4H+ + 4e- → 2H2O with Eº = +1.23 V. Since they've switched the reaction here, why isn't it -1.23?
Posted: Thu Feb 22, 2018 10:03 am
You don't have to switch the sign of Eº if when calculating Eºcell you always use Eºcathode-Eºanode.
If you prefer to use the method where you add the Eºcathode and Eºanode you must flip the Eºanode.
Posted: Thu Feb 22, 2018 10:13 am
Dr. Lavelle discussed 2 methods to calculate Ecell. The first method is to use Eocell = Eocathode - Eoanode, and you wouldn't switch the signs for the Eo here. The second method is to add Eo of the two half-reactions together, and you would switch the sign for the oxidation reaction Eo because the standard reduction potential tables lists the reduction half-reactions.