Answer
$H_2O + Br^- + 2MnO_4^- \rightarrow BrO_3^- + 2MnO_2 +2OH^-$
Work Step by Step
First, solve it as if it takes place in an acidic solution
Half reaction 1: MnO4- —> MnO2
Add 2 H2O on right side, add 4 H+ on left side, add 3e- on left side
3e- +4H+ +MnO4- —>MnO2 +2H2O
Half reaction 2: Br- —> BrO3 -
Add 3 H2O on left side, add 6H+ on right side, add 6e- on right side
3H2O + Br- —> BrO3- +6H+ + 6e-
Reaction 1 times 2 so that e- can be cancelled
Get: 2H+ +Br- +2MnO4 —> BrO3- +2MnO2 +H2O
Add 2OH- to both the reactants and the products side to neutralize H+, and cancel out water as needed
You get: H2O + Br- + 2MnO4- —> BrO3- + 2MnO2 +2OH-
