6K.5 part b

[Chris Tai 1B]

Could someone walk me through how to set up the two redox reactions for part b of 6K.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.
b. Reaction of bromine with itself (disproportionation) in aqueous solution: Br2(l) -> BrO3-(aq) + Br-(aq)

[ShastaB4C]

Hi!

The Br2 is acting as both an oxidizing and reducing agent. So the first half rxn is an oxidation from Br2–> BrO3 ^-2. The second half rxn is a reduction from Br2–> 2Br-

[ShastaB4C]

You solve these how you normally would balance in a basic solution. Also, If you need help figuring out why each species in the product is which, look at how the oxidation number of the Br in them change.
I hope this helps!!

[J_CHEN 4I]

Notice the Br2(l) is both the oxidizing and reducing agent. Therefore, you have two half-reactions: Br2(l) -> BrCO3-(aq) and Br2(l) -> Br-(aq). To balance Br2(l) -> Br-(aq), all you need to is balance Br-. For Br2(l) -> BrCO3-(aq), notice the oxygen. You need to add H2O and H+ to balance this half-reaction. Once you are done, add the electrons.

[Katherine Wu 1H]

rxn of bromine w/ itself (disproportionation) in aqueous solution:
Br2(l)→BrO3^-(aq)+Br^-(aq)

Br2(l)+2e-→2Br^-(aq) (balanced half rxn)

Br2(l)+6H2O(l)→2BrO3^-(aq) (O balanced)
Br2(l)+6H2O(l)+12OH^-(aq)→2BrO3^-(aq)+12H2O(l) (H balanced)

Br2(l)+12OH^-(aq)→2BrO3^-(aq)+6H2O(l)+10e- (e-s balanced, the H2O cancel)

Combining half rxns yields
5[Br2(l)+2e-→2Br^-(aq)]= 5Br2(l)+10e-→10Br^-(aq)
Br2(l)+12OH^-(aq)→2BrO3^-(aq)+6H2O(l)+10e-

6Br2(l)+12OH^-(aq)+10e-→2BrO3^-(aq)+10Br^-(aq)+6H20(l)+10e-

cancel the e-s

6Br2(l)+12OH^-(aq)→2BrO3^-(aq)+10Br^-(aq)+6H2O(l)