problem 14.5 6th edition

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505166714
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Joined: Fri Sep 28, 2018 12:17 am

problem 14.5 6th edition

Postby 505166714 » Sat Feb 23, 2019 7:10 pm

can someone help me out with this question? how can I find the skeletal reaction when the product is both produced by the oxidizing agent and the reducing agent?
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megansardina2G
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Re: problem 14.5 6th edition

Postby megansardina2G » Sat Feb 23, 2019 11:42 pm

You can balance the two skeletal equations using H2O, since they are aqueous solutions.
The first equation is O3 --> O2
Adding H2O and OH- will balance the oxygens and hydrogens on either side, and then you balance the charge to result in: H2O + O3 + 2e- --> O2 + 2OH-

The second equation is Br- --> BrO3-
Again, adding H2O and OH- will balance the oxygens and hydrogens on either side, then balance charge to result in: 6 OH- + 3 H2O + Br- --> BrO3- + 6 H2O + 6 e-


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