6K.5

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Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

6K.5

Postby Maggie Doan 1I » Wed Feb 27, 2019 8:29 pm

Balance each of the following skeletal equations by using oxidation and reduction half reactions.

a) O3 + Br- O2 + BrO3-

How do you balance the half reaction of O3 O2?

Manya Bali 4E
Posts: 66
Joined: Fri Sep 28, 2018 12:23 am

Re: 6K.5

Postby Manya Bali 4E » Wed Feb 27, 2019 10:41 pm

Hello. I have the 6th edition, but I believe that this is the same problem (in basic solution).

O3 -> O2
Add H2O on left side and 2 OH- on right side so that there are equal numbers of Os and Hs
Add 2 electrons to left side so charges are -2 on both sides
Half reaction: O3 + H2O + 2e- -> O2 + 2OH-

If you have trouble figuring out where to put the H2Os and OH-s for basic solutions, you can also first do it in an acidic solution and then "convert" to a basic solution by adding OH-s.
O3 -> O2
Add H2O on right side
Add 2H+ on left side
Add 2OH- to both sides so that there are no H+s. On the left, 2H+ and 2OH- will "combine" to form 2H2O. On the right, there will be 1 H2O and 2 OH-.
Eliminate the extra H2O and add electrons to have the same charge on both sides.
Half reaction: O3 + H2O + 2e- -> O2 + 2OH-

Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

Re: 6K.5

Postby Maggie Doan 1I » Wed Feb 27, 2019 11:33 pm

Thank you I forgot about the negative charge on OH so I put the electrons on the wrong side.


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