Cell Notation and Oxidation

Question

Consider a cell based on the following half- reactions:

Au+3 + 3e− →Au ε° = 1.50 V

Fe+3 + e− →Fe+2 ε° =0.77 V

a.)Write the cell notation (being sure to put the anode on the left and cathode on the right) to explain this cell.

b.) Where in the cell is oxidation taking place? Where in the cell is reduction taking place?

Answer 1

We'll assume this is a galvanic cell. In which case we want E^o_cell > 0 so that ΔG < 0 for the reaction (recall that ΔG^o = -nFE^o).

Since E^o_cell = E^o_cathode - E^o_anode we should manipulate our half reactions to get a positive result.

Note that in the equation above, the subtraction sign is necessary since one of the reactions is an oxidation. If we designate the Au half reaction as the cathode we have:

E^o_cell = 1.50V - 0.77V = +0.73V

You can see that if we chose the Fe half reaction as the cathode, we would get a negative answer, so we know we have the correct cathode and anode at this point. The cell notation is then

Pt | Fe²⁺, Fe³⁺ || Au³⁺ | Au

Where the anode half reaction is on the left and the cathode half reaction is on the right. Since both Fe²⁺ and Fe³⁺ are ions, they are in solution and we use a Pt electrode.

Oxidation takes place at the anode and in this case is happening around the Pt electrode. Reduction takes place at the cathode and is happening around the Au electrode.

Answer 2

Pt | Fe³⁺, Fe²⁺ || Au³⁺ | Au

oxidation is taking place at the anode and reduction is taking place at the cathode.