6L.3 Part (d) Question [ENDORSED]

[Stella Nguyen 1J]

Hi everyone!

For part (d) of this problem, the only way that we would be able to figure out the oxidation reaction that occurs at the anode is if it was provided for us correct? I'm not sure how you would be able to write it out with just the information provided by the table.

6L.3(d): Pt(s)|O2(g)|H+(aq)∥OH−(aq)|O2(g)|Pt(s)

Thank you!

[Chem_Mod]

All half reactions are given in the list of standard reduction potentials.
Students do not need to 'know/memorize' the reactions.
This makes it much easier and straight forward to figure out the anode (oxidation half-reaction), the cathode (reduction half-reaction), and the overall redox reaction.

When (aq) is stated then that means water is present and can be part of the redox reaction.
The oxygen atom in water is 2- oxidation state.
The oxygen atom in O2(g) is zero oxidation state.
See the list of standard reduction potentials showing the balanced half reactions involving water.

Therefore in the reaction you ask about:

Anode: O2(g)|H+(aq) Only way to show oxidation is to have the O (2-) in H₂O oxidized to O2(g) under acidic conditions.

Cathode: OH−(aq)|O2(g) Only way to show reduction is to have O2(g) reduced to H₂O under alkaline conditions.