2. Balance the redox equations by the ion-electron method:
Br2 --> Br2O3- + Br-(in basic solution)?


It is easier to balance equations in basic solution by first balancing them in acidic solution.

1. Break up the reaction into two half-reactions.Br2 -----> Br2O31-Br2 -----> Br1-

2. Balance the main atoms; in this case, BrBr2 -----> Br2O31-
(Br already balanced)Br2 -----> 2Br1-

3. Balance the O by adding H2OBr2 + 3H2O -----> Br2O31-Br2 -----> 2Br1-

4. Balance the H by adding H+Br2 + 3H2O -----> Br2O31- + 6H+Br2 -----> 2Br1-

5. Balance the charges by adding e- to the more positive sides of each half-reaction.
Br2 + 3H2O -----> Br2O31- + 6H+ + 5e- (oxidation)Br2 + 2e- -----> 2Br1- (reduction)

6. Make the e- gained and lost equal. Multiply the top equation by 2 and the bottom one
by 5.2Br2 + 6H2O -----> 2Br2O31- + 12H+ + 10e-5Br2 + 10e- -----> 10Br1-

7. Add the two half-reactions.7Br2 + 6H2O -----> 2Br2O31- + 10Br1- + 12H+
The equation is balanced for an acidic solution.

8. To convert the equation to a basic solution, add the same number of OH- to each side of the
equation as there are H+.12OH- + 7Br2 + 6H2O -----> 2Br2O31- + 10Br1- + 12H+ + 12OH-

9. Combine the H+ and OH- to make water.
12OH- + 7Br2 + 6H2O -----> 2Br2O31- + 10Br1- + 12H20

10. Simplify the H2O12OH- + 7Br2 -----> 2Br2O31- + 10Br1- + 6H20
The equation is balanced for a basic solution. Note that both sides are -12 in charge and that
all atoms are balanced.