## 14.5

Michele 2C
Posts: 23
Joined: Fri Sep 29, 2017 7:06 am

### 14.5

The directions are to balance the redox reaction in basic solution. Part a writes: Action of ozone on bromide ions, O3 (aq) + Br - (aq) --> O2 (g) + BrO3 - (aq).

Why does the solutions manual write the first half-reaction as: O3 (g) --> O2 (g)?

Don't both O3 and O2 have oxidation numbers of 0?

Sarah_Stay_1D
Posts: 57
Joined: Sat Jul 22, 2017 3:00 am
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### Re: 14.5

Michele 2C wrote:The directions are to balance the redox reaction in basic solution. Part a writes: Action of ozone on bromide ions, O3 (aq) + Br - (aq) --> O2 (g) + BrO3 - (aq).

Why does the solutions manual write the first half-reaction as: O3 (g) --> O2 (g)?

Don't both O3 and O2 have oxidation numbers of 0?

The first reaction is O3 (g) --> O2 (g) because the oxygen is being oxidized. The left side of the half equation has a total negative charge of -6 (3 O atoms x charge of -2). The right side of the equation has a total charge of -4 (2 O atoms x charge of -2). Therefore, the O3 is losing electrons so it is being oxidized.

Marina Georgies 1C
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

### Re: 14.5

The reason O2's oxidation number is 0 is that oxidation states of atoms in their elemental states is zero. Oxygen gas is diatomic, often found in nature as O2, so that is why its oxidation number is zero. The same cannot be said for O3 however, and its oxidation number is -6 (3 x 2 O's)

Lindsay H 2B
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

### Re: 14.5

I'm confused on this question too. Are both Br and O getting oxidized? Br seems to go from -1 in Br- to +5 in BrO3-, but according to this post, O3 is also being oxidized from -2 to 0.