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Question
Balance the following reaction by oxidation number method.
\[\ce{MnO^-_{4(aq)} + Br^-_{ (aq)}->MnO2_{ (s)} + BrO^-_{3(aq)}(basic)}\]
Solution
\[\ce{MnO^-_{4(aq)} + Br^-_{ (aq)}->MnO2_{ (s)} + BrO^-_{3(aq)}(basic)}\]
Step 1: Write a skeletal equation and balance the elements other than O and H.
\[\ce{MnO^-_{4(aq)} + Br^-_{ (aq)}->MnO2_{ (s)} + BrO^-_{3(aq)}}\]
Step 2: Assign the oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.
Increase in oxidation number:
(Increase per atom = 6)
Decrease in oxidation number:
(Decrease per atom = 3)
To make the net increase and decrease equal, we must take 2 atoms of Mn.
\[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)}->2MnO2_{ (s)} + BrO^-_{3(aq)}}\]
Step 3: Balance 'O' atoms by adding H2O to the right-hand side.
\[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)}->2MnO2_{ (s)} + BrO^-_{3(aq)} + H2O_{(l)}}\]
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H+ on the left-hand side.
\[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)} + 2H^+_{ (aq)}->2MnO2_{ (s)} + BrO^-_{3(aq)} + H2O_{(l)}}\]
Add OH− ions equal to the number of H+ ions on both sides of the equation.
\[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)} + 2H^+_{ (aq)} + 2OH^-_{( aq)}->2MnO2_{ (s)} + BrO^-_{3(aq)} + H2O_{(l)} + 2OH^-_{( aq)}}\]
The H+ and OH− ions appearing on the same side of the reaction are combined to give H2O molecules.
\[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)} + 2H2O_{(l)}->2MnO2_{ (s)} + BrO^-_{3(aq)} + H2O_{(l)} + 2OH^-_{( aq)}}\]
\[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)} + H2O_{(l)}->2MnO2_{ (s)} + BrO^-_{3(aq)} + 2OH^-_{( aq)}}\]
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: \[\ce{2MnO^-_{4(aq)} + Br^-_{ (aq)} + H2O_{(l)}->2MnO2_{ (s)} + BrO^-_{3(aq)} + 2OH^-_{( aq)}}\]
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