Hi,
I'm not sure where to start when writing the reduction half reaction, in which O3 is oxidized from 0 to 2- (BrO3^2-). I just don't see a way to balance out Br.
My oxidation half reaction: 6OH- + Br- -> BrO3 2- + 3H20 + 6e-
14.5a [ENDORSED]
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Annie Chang 3G
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Re: 14.5a [ENDORSED]
Begin with O3 --> O2
Work to balance out the O's by adding H20 on the right side:
O3 --> O2 + H2O
Then, balance out the H's by adding H20 on the left side and OH- on the right side:
2H2O + O3 --> O2 + H2O + 2OH-
Cancel out the H2O:
H2O + O3 --> O2 + 2OH-
Finally, balance the charges:
H2O + O3 + 2e- --> O2 + 2OH- (Reduction half-reaction)
When combining the two half-reactions, multiply the reactions by the factor that will result in the same number of electrons in both half-reactions. So, multiply the reduction half reaction by a factor of 3 and then add the half-reactions. The electrons will cancel out to give the balanced equation for the whole reaction.
Work to balance out the O's by adding H20 on the right side:
O3 --> O2 + H2O
Then, balance out the H's by adding H20 on the left side and OH- on the right side:
2H2O + O3 --> O2 + H2O + 2OH-
Cancel out the H2O:
H2O + O3 --> O2 + 2OH-
Finally, balance the charges:
H2O + O3 + 2e- --> O2 + 2OH- (Reduction half-reaction)
When combining the two half-reactions, multiply the reactions by the factor that will result in the same number of electrons in both half-reactions. So, multiply the reduction half reaction by a factor of 3 and then add the half-reactions. The electrons will cancel out to give the balanced equation for the whole reaction.
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