14.5

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April P 1C
Posts: 38
Joined: Thu Jan 12, 2017 3:01 am

14.5

Postby April P 1C » Wed Feb 21, 2018 7:17 pm

Br2(l)-> BrO3-(aq) + Br-(aq)

I'm having a hard time finding numbers for H2O and OH- when balancing the Hydrogens. Does anyone have any tips/tricks??
I know that since it's basic, to balance the hydrogens I have to add H20 to the side that needs H, and then add OH- to the other side.

So far I have ?OH + 6H2O + Br2 -> 2BrO3 + ?H2O

Nicole Anisgard Parra 2H
Posts: 39
Joined: Sat Jul 22, 2017 3:01 am

Re: 14.5

Postby Nicole Anisgard Parra 2H » Wed Feb 21, 2018 9:50 pm

The trick I use is to first balance everything else. So, add a two in front of Br03.
Br2--> 2BrO3
Now we have to balance the oxygens. Add H20 to the side that needs oxygen to balance it out.
6H2O+Br2--> 2BrO3 (6 oxygens total, so we add 6 H2O molecules to balance it out)
Now we have a surplus of hydrogen that we need to balance out. The trick I use is to add one H2O molecule for each H needed to the side of the equation that needs the hydrogens. Then, for each H2O you add, you need to add the same number of OH- molecules to the opposite side.
Here, we need 12 hydrogens total to balance out the 6H2Os we have on the reactants side, so we add 12H2O molecules to the products side:
6H2O+Br2--> 2BrO3+12H2O
and then we add 12OH- to the opposite side.
12OH- +6H2O+Br2--> 2BrO3+12H2O
Now, we can cancel out like species on opposite sides of the arrow, to obtain:
12OH- +Br2--> 2BrO3+6H2O (fully balanced, minus the electrons of course!)


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