Textbook 6K.5a
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Textbook 6K.5a
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.
Re: Textbook 6K.5a
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.
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Re: Textbook 6K.5a
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
edit: I had to edit this like 2-3 times because I keep making typos rip
Last edited by Sara_Lim_2C on Fri Mar 12, 2021 12:36 pm, edited 2 times in total.
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Re: Textbook 6K.5a
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!
Hope this helps!
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Re: Textbook 6K.5a
I also believe if you look through the appendixes there are half reactions with values for both of them.
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Re: Textbook 6K.5a
how is O3-->O2 the reduction reaction if both sides have the same oxidation number of 0?
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Re: Textbook 6K.5a
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.
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