14.5 (a)

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Andy Liao 1B
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14.5 (a)

Postby Andy Liao 1B » Sat Mar 17, 2018 10:51 am

14.5 Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in basic solution. Identify the oxidizing agent and reducing agent in each reaction.
(a) Action of ozone on bromide ions:
O3(aq) + Br-(aq) -> O2(g) + BrO3-(aq)

I was wondering why the solutions manual wrote the reduction half-reaction as O3(aq) -> O2(g). Why is this true? Both O3(aq) and O2(g) have a charge of 0. Can someone please explain?

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Re: 14.5 (a)

Postby GabrielGarciaDiscussion1i » Sat Mar 17, 2018 11:44 am

This question was asked and answered on Chemistry community years ago and so I'll post the original link, but I'm gonna write the answer again here and fix it because it had mistakes in the terminology.
viewtopic.php?t=1577 (This answer says Br is being reduced and O3 is being oxidized, which is incorrect).

The best way to approach this problem is knowing that Br- is oxidized to BrO3- (Note, to see this you need to know the oxidation of Br in BrO3-, I think about it as follows: TOTAL CHARGE (on the molecule) = SUM OF CHARGES OF ATOMS. In our case, total charge is -1, and the sum of charges of the OXYGEN atom is (-2 x 3 = -6). Therefore, from the eqtn I wrote: -1 = -6 + BROMINE's CHARGE. Hence Br has a charge of +5, it is clearly being oxidized from Br- to Br(+5).) After you identify the 1/2 reactions, by process of elimination if Br- is oxidized then OZONE is reduced, follow the steps on pg 447. Balance the atoms, then balance the charges etc. EVERYTHING will fall into place! (NOTE: ozone has an overall neutral charge, however, if you draw the Lewis structure of ozone, you will see that one O atom has a formal charge of zero, another has a FC of +1 and the third O has a FC of -1).

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