CHEMISTRY COMMUNITY

in this equation

MnO4- + 8H+ + 5Ce^3+ --> 5Ce^4+ + Mn^2+ + 4H2O

how would you identify the cathode and anode? i read in the book that the half rxn with the more positive reduction potential is where reduction occurs. however, permanganate potential is more positive, but it is the anode in the equation. why??

When the overall equation is given to you, it has already been determined which reaction occurs in the anode and which reaction occurs in the cathode. In the case of MnO4- + 8H+ + 5Ce^3+ --> 5Ce^4+ + Mn^2+ + 4H2O, Mn goes from an oxidation state of +7 to +2 so it is reduced, and Ce goes from an oxidation state of +3 to +4 so it is oxidized. This means that Ce(3+) is at the anode and MnO4- is at the cathode.

When you are not given the overall equation, just the two half reactions, you use the reduction potentials to determine which reaction is occurring at the cathode and which reaction is occurring at the anode. This is when the reaction with the higher reduction potential occurs at the cathode.

It is true that we want the cathode to have a greater positive reduction potential so that the galvanic cell will be able to do useful work (positive cell potential). However, the question never specified that the cell had to have a positive cell potential. It merely asks you to calculate the standard cell potential for the given equation. If you assign oxidation numbers to the reactants and products, you would be able to see that MnO4- is reduced while Ce3+ is oxidized. Therefore, the half reaction containing MnO4- takes place at the cathode (reduction) and the half reaction containing Ce3+ takes place at the anode (oxidation).

I'm not quite sure where you're getting your information from as Ce4+ + e- --> Ce3+ has a more positive standard reduction potential (+1.61V) than MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O (+1.51V). Also, the solutions manual has the permanganate half-reaction as the cathode, which is correct.

Matthew Tran 1H wrote:It is true that we want the cathode to have a greater positive reduction potential so that the galvanic cell will be able to do useful work (positive cell potential). However, the question never specified that the cell had to have a positive cell potential. It merely asks you to calculate the standard cell potential for the given equation. If you assign oxidation numbers to the reactants and products, you would be able to see that MnO4- is reduced while Ce3+ is oxidized. Therefore, the half reaction containing MnO4- takes place at the cathode (reduction) and the half reaction containing Ce3+ takes place at the anode (oxidation).

I'm not quite sure where you're getting your information from as Ce4+ + e- --> Ce3+ has a more positive standard reduction potential (+1.61V) than MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O (+1.51V). Also, the solutions manual has the permanganate half-reaction as the cathode, which is correct.

Does a greater positive reduction potential always signify that you are looking at the cathode?

If you want the cell to be able to do useful work (have a positive cell potential), then yes, you want the more positive reduction potential at the cathode. However, there are cases such as the problem mentioned in this thread where you are given a reaction and you have to go off of that in terms of what is being oxidized and what is being reduced. Sometimes you will not get a positive potential, but that's ok if you're going off the reaction specified by the problem.