Cell Potential Example in Notes

[Fran L 1J]

On page 52 of the notes, there is an example showing how to calculate the cell potential of a redox reaction with nitric acid and gold ions in solution. The cell potential value for the nitrate (reduction) equation is given as E= 0.96V and for the gold ion (oxidation) equation as E= -1.50 V. The calculations show that Eredox= 0.96- 1.50= -0.54V. If the formula given to find Eredox= Ecathode- E anode, why did the example add the two values instead of doing: 0.96- (-1.50)? Thanks.

[Chem_Mod]

The table of potentials is always given as positive values for the reduction reactions. In the course reader, the Au half reaction is an oxidation reaction so its potential has already been negated for you. Since it is already negated, you can just add the two half reaction potentials rather than subtract the anode potential from the cathode potential.