14.5 a, balancing redox O3 to O2?

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Adriana Rangel 1A
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Joined: Fri Sep 29, 2017 7:04 am

14.5 a, balancing redox O3 to O2?

Postby Adriana Rangel 1A » Wed Feb 21, 2018 3:46 pm

I balanced the elements and have:
H20 + O3 --> O2 + 2OH
But the solution manual has + 2e- on the left side. If both O3 and O2 have a neutral charge how do you know which is oxidized/reduced?? Shouldn't the two electrons be on the right side because O3 has a charge of -6 and O2 has a charge of -4 (-4 + -2 is -6?)

Dylan Davisson 2B
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Joined: Thu Jul 27, 2017 3:00 am
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Re: 14.5 a, balancing redox O3 to O2?

Postby Dylan Davisson 2B » Wed Feb 21, 2018 4:09 pm

The way I worked out the problem, saw that the Bromine ion was oxidized, so it only followed that the oxygen had to be reduced. And any half reaction for reduction has the electrons added to the left side. You can come to this same answer of the electrons being on the left, because when you take the oxygen half reaction and balance it out with water and hydroxide and then add electrons to balance the charge, the electrons will be added to the left side.

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