14.5

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Allison Young 1K
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Joined: Fri Sep 29, 2017 7:06 am

14.5

Postby Allison Young 1K » Thu Feb 08, 2018 9:11 pm

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.

skalvakota2H
Posts: 52
Joined: Sat Jul 22, 2017 3:01 am

Re: 14.5

Postby skalvakota2H » Thu Feb 08, 2018 9:34 pm

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.

Allison Young 1K
Posts: 23
Joined: Fri Sep 29, 2017 7:06 am

Re: 14.5

Postby Allison Young 1K » Fri Feb 09, 2018 2:43 am

Ohh that makes sense. Thanks :-)

Yutian Zhao -1J
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

Re: 14.5

Postby Yutian Zhao -1J » Sat Feb 10, 2018 5:02 pm

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.

Tiffany Dao 1A
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

Re: 14.5

Postby Tiffany Dao 1A » Sat Feb 10, 2018 6:19 pm

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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