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You would have to determine which of the reactions is more easily reduced and that will be your cathode. You will have to compare the absolute values of the cell potentials. A helpful tip to remember is that you don't want a negative E naught.
Given the reduction half reactions, you will use the appendix to find E naught values. The lower E naught value is the anode because it is the stronger reducing agent. Then, to get E naught of the cell you do cathode - anode = E naught cell
You arrange it so that Enot is the most positive. So you should assign the half reaction with the larger Enot as cathode, and assign the half reaction with the smaller Enot as the anode. This way, Enot(cathode)-Enot(anode) will be the largest possible.
when you find the two half reactions in the from of reductions from an original equation, whichever one has the reactants and product on the same side as the original is the cathode and the one where the reactants and products are flipped is the anode.
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