Determining Anode and Cathode

[KaleenaJezycki_1I]

If both the species are given as "La3+/La and U3+/U" (problem 6M.11) then do you just choose which one you want to be the reducing and oxidizing agent? I don't understand this because they both would be written as reductions.

[Vincent Leong 2B]

Their reduction potentials give it away. Larger E (naught) = higher reduction potential = increased likelihood of reducing ; therefore, higher E (naught) species gets reduced while the lower E naught gets flipped to get oxidized. If you think about it logically, you can't have 2 reducing species in a galvanic cell or else there would be no galvanic cell to begin win. You need both compartments and a cell potential difference for their to be a cell.

[205291012]

You need to look for the reduction half reaction in the appendix at the back of the textbook. Look for the 2 equations that have both your species. The one with a more positive reduction potential value will be more likely to be reduced. So the other equation will be oxidized and you can flip it.

[KaleenaJezycki_1I]

Their reduction potentials give it away. Larger E (naught) = higher reduction potential = increased likelihood of reducing ; therefore, higher E (naught) species gets reduced while the lower E naught gets flipped to get oxidized. If you think about it logically, you can't have 2 reducing species in a galvanic cell or else there would be no galvanic cell to begin win. You need both compartments and a cell potential difference for their to be a cell.

Thank you so much!!

[005162902]

The higher Enaught value is the the cathode