Question

Balance the following using the ½ reaction method Permanganate ion plus bromide ion yields (MnO4) and BrO3 - (basic solution)

Answer 1

+7          -1          +4        +5

MnO₄^- + Br^- ----> MnO₂ + BrO₃^-

Oxidation half reaction is : Br^- ----> BrO₃^-   

Balance O atoms ( by adding H₂O on the deficient side ) : Br^- +3H₂O ----> BrO₃^-

Balance H atoms ( by adding H₂O on deficient side & OH^- on the other side) :

                            Br^- +3H₂O +6OH^- ----> BrO₃^- +6H₂O

Balance charge ( by adding electrons) : Br^- +3H₂O +6OH^- ----> BrO₃^- +6H₂O + 5e^- ------(1)

Reduction half reaction is : MnO₄^- ----> MnO₂

Balance O atoms ( by adding H₂O on the deficient side ) : MnO₄^- ----> MnO₂ +2H₂O

Balance H atoms ( by adding H₂O on deficient side & OH^- on the other side) :

                           MnO₄^- + 4H₂O ----> MnO₂ +2H₂O + 4OH^-     

Balance charge ( by adding electrons) : MnO₄^- + 4H₂O +3e^- ----> MnO₂ +2H₂O + 4OH^-    ----(2)

The combined equation is obtained by adding both half cells so that electrons cancel each other.

This can be done as follows :

[3xEqn(1) ] + [5xEqn(2)] gives

3Br^- +9H₂O +18OH^- +5MnO₄^- + 20H₂O +15e^- ----> 5MnO₂ +8H₂O + 20OH^- + 3BrO₃^- + 18H₂O + 15e^-

3Br^- +3H₂O +5MnO₄^- ----> 5MnO₂ + 2OH^- + 3BrO₃^-