Balancing Redox Reactions

Question

Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction.

MnO4 -(aq) +Br-(aq) --> MnO2(s) + BrO3 -(aq)

Answer 1

Write the half reactions

MnO4^1- → MnO2 reduction

Br^1- → BrO3^1- oxidation reacxtion

Balance elements undergoing redox reaction. They are already balanced.

Balance oxygen by adding water and hydrogens by adding H¹⁺

MnO4^1- + 4H¹⁺ → MnO2 + 2H2O

Br^1- + 3H2O → BrO3^1- + 6H¹⁺

Add OH^1- ions to both side equal to the number of H¹⁺ ions.

4OH^1- + MnO4^1- + 4H¹⁺ → 4OH^1- + MnO2 + 2H2O

6OH^1- + Br^1- + 3H2O → 6OH^1- + BrO3^1- + 6H¹⁺

Combine H¹⁺ and OH^1- where possible to produce H2O.

4H2O + MnO4^1- → 4OH^1- + MnO2 + 2H2O

6OH^1- + Br^1- + 3H2O → 6H2O + BrO3^1-

Add electrons to balance negative charges

4H2O + MnO4^1- + 3e^- → 4OH^1- + MnO2 + 2H2O

6OH^1- + Br^1- + 3H2O → 6H2O + BrO3^1- + 6e^-

Multiply the coefficients for the reaction showing 3 e^- by 2 to make the e^- equal to 6, so that they will cancel when the two half reactions are added.

8H2O + 2MnO4^1- + 6e^- → 8OH^1- + 2MnO2 + 4H2O

6OH^1- + Br^1- + 3H2O → 6H2O + BrO3^1- + 6e^-

Add the two half reactions.

8H2O + 2MnO4^1- + 6e^- + 6OH^1- + Br^1- + 3H2O → 8OH^1- + 2MnO2 + 4H2O + 6H2O + BrO3^1- + 6e^-

Simplify by cancelling e^- , H2O, and H2O to get the smallest number of each on both sides.

2MnO4^1- + Br^1- + H2O → 2MnO2 + BrO3^1- + 2OH^1-

Check that: all elements are balanced.

charges are equal on both sides.

Check all my steps for yourself.