6K. 5b

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Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

6K. 5b

Postby Lindsey Chheng 1E » Thu Feb 27, 2020 9:52 pm

Balance Br2 → BrO3 + Br- in basic solution.
Answer: 3Br2 + 6OH- → 5Br- + BrO3- + 3H2O
I know that Br2 is both the reducing and oxidizing agent, but I can't get the two half reactions right. These are the ones I got:
2 x (12OH- + Br2 → 2BrO3- + 12H + 6H2O + 5e-)
5 x (Br2 + 2e- → 2Br-)
Can someone please tell me what I'm doing wrong and which equations I'm supposed to get? Thanks!

Jessica Li 4F
Posts: 115
Joined: Fri Aug 09, 2019 12:16 am

Re: 6K. 5b

Postby Jessica Li 4F » Thu Feb 27, 2020 10:25 pm

I believe in your first half-cell equation, there should be 10 electrons transferred because you have 2 atoms of Br instead of 5. Hope that helps!

Daria Azizad 1K
Posts: 116
Joined: Thu Jul 25, 2019 12:15 am

Re: 6K. 5b

Postby Daria Azizad 1K » Thu Feb 27, 2020 10:56 pm

Balance the electrons in the oxidation reaction because the stoichiometric coefficient applies to the # of e- as well

Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

Re: 6K. 5b

Postby Anthony Hatashita 4H » Fri Feb 28, 2020 12:59 am

The equations should be as follows:
12OH- + Br2 → 2BrO3- + 6H2O + 10e-
5 x (Br2 + 2e- → 2Br-)
I'm not sure where you got the H from when you did your calculations, maybe you forgot that this was in a basic solution? You use H2O and OH- to balance the H on each side.

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