Balancing Redox Equations by the Half Reaction Method in acid or base

This method balances all redox equations by the half reaction method, as if the reactions take place in acidic solution. Reactions in basic solution are handled last (if needed).

1) Start with an overall skeleton reaction. We will use the following as an example throughout:

MnO₄^- (aq)  + Br^- (aq) à MnO₂ (s) + BrO₃^- (aq) (the example takes place in basic solution)

Write skeleton half reactions from the overall skeleton. They are:

MnO₄^- (aq) àMnO₂ (s)     and

Br^- (aq) à BrO₃^- (aq)

2) Balance all atoms which are not hydrogen or oxygen (already done in our example)

3) Balance oxygen by adding water, get

MnO₄^- (aq)  à MnO₂ (s) + 2 H₂O (l)                    

3 H₂O (l) + Br^- (aq) à BrO₃^- (aq) 

4) Balance Hydrogen by adding H⁺

4 H⁺ (aq) + MnO₄^- (aq)  à MnO₂ (s) + 2 H₂O (l)             

3 H₂O (l) + Br^- (aq) à BrO₃^- (aq) + 6 H⁺ 

5) Balance charge by adding electrons (can also look at oxidation numbers and how they change and add electrons that way) to get the balanced half reactions in acid:

3 e^- + 4 H⁺ (aq) + MnO₄^- (aq)  à MnO₂ (s) + 2 H₂O (l)                      (Reduction)

3 H₂O (l) + Br^- (aq) à BrO₃^- (aq) + 6 H⁺ (aq) + 6 e^-                (Oxidation)

6) Multiply the Reduction (Mn) half reaction by 2 to get 6 electrons on each side (so they'll cancel) and add the two half reactions:

6 e^- + 8 H⁺ (aq) + 2 MnO₄^- (aq)  à 2 MnO₂ (s) + 4 H₂O (l)    (Reduction)

3 H₂O (l) + Br^- (aq) à BrO₃^- (aq) + 6 H⁺ (aq) + 6 e^-                (Oxidation)

2 H⁺ (aq) + 2 MnO₄^- (aq) + Br^- (aq) à 2 MnO₂ (s) + H₂O (l) +BrO₃^- (aq)       (Sum)

If the reaction takes place in acidic solution, you are done at this step. It is always a good idea to check that all atoms and charge are balanced on both sides.

For basic solutions (which we have in our example) there is one more step involved.
7) Now we recall this is really in BASIC SOLUTION, so we add enough hydroxide ion (OH^-) to both sides of the equation to neutralize all of the acidic protons (H⁺), making H₂O from the protons and adding hydroxide to the other side of the equation.  

Recall our balanced equation from step 6 is:

2 H⁺ (aq) + 2 MnO₄^- (aq) + Br^- (aq) à 2 MnO₂ (s) + H₂O (l) +BrO₃^- (aq)       (Sum)

In this case we have 2 H⁺ , so we add 2 OH^- to each side.  Recall that H⁺  + OH^- -----> H₂O:

2 H⁺ (aq) +2 OH^- (aq) + 2 MnO₄^- (aq) + Br^- (aq) à 2 MnO₂ (s) + H₂O (l) +BrO₃^- (aq) + 2 OH^- (aq)

Rewrite (reactants) 2 H⁺  + 2 OH^- as 2 H₂O:

2 H₂O (aq) + 2 MnO₄^- (aq) + Br^- (aq) à2 MnO₂ (s) + H₂O (l) +BrO₃^- (aq) + 2 OH^- (aq)

Cancel excess waters on both sides, and we are done. It is always a good idea to check that all atoms and charge are balanced on both sides:

H₂O (aq) + 2 MnO₄^- (aq) + Br^- (aq) à 2 MnO₂ (s) +BrO₃^- (aq) + 2 OH^- (aq)

Final notes: There are other methods of balancing redox equations, especially in basic solution. You may use any method you like as long as you show all of your work (and get the proper balanced equation in the proper solution at the end).

Chapter 7 of the text is on this topic, although it focuses more on electrochemistry. Practice Problems taken from Darrell D. Ebbing, General Chemistry, 4^th ed. Houghton-Mifflin: Boston, 1993.

Balance the following reactions, which all take place in water and are acidic or basic as noted.

1) HI + HNO₃ à I₂ + NO

2) Ag + H₂SO₄ à Ag₂SO₄ + SO₂

3) MnCl₂ + KMnO₄ + KOH à MnO₂ + KCl

4) H₃AsO₄ + Zn + HNO₃ à AsH₃ + Zn(NO₃)₂

5) SnCl₂ + O₂ + HCl à H₂SnCl₆

6) K₂MnO₄ à MnO₂ + KMnO₄ + KOH

7) Cr₂O₇^2- + C₂O₄^2- à Cr³⁺ + CO₂  (acid)

8) Cu + NO₃^1- à Cu²⁺ + NO  (acid)

9) MnO₂ + HNO₂ à Mn²⁺ + NO₃^1-  (acid)

10) Mn²⁺ + H₂O₂ à MnO₂ + H₂O  (base)

11) MnO₄^1- + NO₂^1- à MnO₂ + NO₃^1-   (base)

12) Cl₂ à Cl^1- + ClO₃^1-  (base)

ANSWERS

1) 6 HI + 2 HNO₃ à 3 I₂ + 2 NO + 4 H₂O 

2) 2 Ag + 2 H₂SO₄ à Ag₂SO₄ + SO₂ + 2 H₂O 

3) 3 MnCl₂ + 2 KMnO₄ + 4 KOH à 5 MnO₂ + 6 KCl + 2 H₂O 

4) H₃AsO₄ + 4 Zn + 8 HNO₃ à AsH₃ + 4 Zn(NO₃)₂ + 4 H₂O 

5) 2 SnCl₂ + O₂ + 8 HCl à 2 H₂SnCl₆ + 2 H₂O 

6) 3 K₂MnO₄ + 2 H₂O  à MnO₂ + 2 KMnO₄ + 4 KOH

7) Cr₂O₇^2- + 3 C₂O₄^2- + 14 H⁺ à 2 Cr³⁺ + 6 CO₂ + 7 H₂O

8) 3 Cu + 2 NO₃^1- + 8 H₂O  à 3 Cu²⁺ + 2 NO + 4 H₂O

9) MnO₂ + HNO₂ + H⁺ à Mn²⁺ + NO₃^1-  + H₂O

10) Mn²⁺ + H₂O₂ + 2 OH^1- à MnO₂ + 2 H₂O 

11) 2 MnO₄^1- + 3 NO₂^1- + H₂O à 2 MnO₂ + 3 NO₃^1- + 2 OH^1-

12) 3 Cl₂ + 6 OH^1- à 5 Cl^1- + ClO₃^1-  + 3 H₂O