14.5a
The reduction half-reaction for O3 -> O2 is: H2O + O3 + 2e- -> O2 + OH-
Why do you add 2 electrons if the oxidation numbers of O3 and O2 are 0 and the charge doesn't change for O2? I'm confused as to where those 2 electrons are going.
14.5
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Re: 14.5
The half reaction would need to be balanced first, so the equation would look like H2O (l) + O3 (g) + 2e- --> O2 (g) + 2OH- (aq). As mentioned, both O3 and O2 have no charge, but since the products contains two hydroxide ions, the overall charge needs to be balanced by adding two electrons to the left side.
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Re: 14.5
But why can't we assume the reduction of O3 to BrO3-? I thought the oxidation number would then be decreased from 0 to -2. Can the BrO3- be a product of both oxidation and reduction? I am a little bit confused.
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Re: 14.5
I think you can reduce O3 to BrO3- becuase you need to oxidize Br-, so if the Bromine is losing electrons, it can't also be accepting it.
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