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Balancing the reduction of o3->o2 in basic soln

Posted: Wed Feb 19, 2020 12:03 am
by Skyllar Kuppinger 1F
So the oxidation half rxn is the oxidation of bromide and the reduction half that needs to be balanced is this:
O3->O2
It's in BASIC solution.
So here's what I was doing, based on the steps in the book:
Step 1= balance oxygen by adding one water to the right side: O3->O2 +H2O
Step 2= balance H by adding H2O to the left and OH- to the right: O3 + H2O ->O2+H2O+OH-
But now it's not balanced. There are 4 oxygen on each side so that's fine , but on the left side there are 2 hydrogen on the left and 3 on the right. How do I fix this??

Re: Balancing the reduction of o3->o2 in basic soln

Posted: Wed Feb 19, 2020 1:12 am
by Benjamin Feng 1B
Before the last step, there are 2 Hydrogen atoms present in the H2O. This means you must add 2 OH- and H2O on opposite sides to balance. Also, you can cancel out the waters on both sides until there is only water on 1 side remaining.