Textbook 6K.5a

[Jeffrey Hablewitz 2I]

How do you divide the reaction given in 6K.5a (O₃ + Br^- -> O₂ + BrO₃^-) into reduction and oxidation half-reactions? I understand that the Br is oxidized and the O is reduced, but I don't get how these two processes can be divided as they occur in the same molecule.

[ALee_1J]

I believe O₃ --> O₂ and then Br- --> BrO₃^- + 6e-? I didn't check so my numbers may be off but that's who you should be dividing it. The oxygens have an oxidation state of 0. The Br- gets oxidized into BrO₃.

[Sara_Lim_2C]

It think should be O₃->O₂ and Br^-->BrO₃^-, where the Br^- is the reducing agent and O₃ is the oxidizing agent.

edit: I had to edit this like 2-3 times because I keep making typos rip

[Stuti Pradhan 2J]

Since it is a redox reaction, you know you can add water and H+ to balance out any existing oxygens and hydrogens, so you can split the reaction up. The half-reactions would be O3-->O2 and the Br- -->BrO3- because the molecules with Br must be in a single half-reaction, and you can add waters to balance the oxygen in Br- -->BrO3-.

Hope this helps!

[Malakai Espinosa 3E]

I also believe if you look through the appendixes there are half reactions with values for both of them.

[Kaiya_PT_1H]

how is O3-->O2 the reduction reaction if both sides have the same oxidation number of 0?

[RitaThomas_3G]

An easy way for me to look at it was to deal with Br- -> BrO3- first because it is more clear that is the oxidation reaction because Br is losing electrons. Then, we can assume that the other reaction with O3 -> O2 must be the reduction reaction.