6L.5 Part D [ENDORSED]

[VSU_3F]

Hello!

For part d, I wrote the oxidation half-reaction as Au+ (aq) --> Au3+ (aq)+ 2e-. The solution manuel had their oxidation reaction as Au(s) --> Au3+(aq) + 3e- (aq). I got the same overall reaction as the textbook solution, but it wasn't very intuitive for me to have Au(s) on the reactant side of the oxidation reaction. I suspect that they got this from the table of standard reduction potentials. As such, I did not include Au(s) on the anode side of my cell diagram. Did anyone else approach this problem like the textbook solution did? Thanks!

[Chem_Mod]

The given redox reaction is not balanced.

Using the standard reduction potentials one obtains a positive voltage with Au(s) oxidized to Au³⁺(aq) and Au⁺ reduced to Au(s).

After balancing (multiplying reduction reaction by 3) and adding to two half reactions the Au(s) on the left (reactant) side cancels out and 2 Au(s) remain on the right (product) in the overall redox reaction.

The two half reactions have Au(s) involved in both.
The half reactions are 'shown' in the cell diagram and therefore Au(s) is the cathode and the anode.

But this is not a concentration cell as the species in solution are different and therefore the reactions at the anode and cathode are different.

[VSU_3F]

Thank you so much for your help, Dr. Lavelle!

[Chem_Mod]

Welcome