Balancing Redox Reactions
Procedure for Balancing Redox Equations Using Oxidation Numbers
The general procedure follows:
1.	Write as much of the overall unbalanced equation as possible.
2.	Assign oxidation numbers and identify the atoms/ions whose oxidation numbers change.
3.	Using the change in oxidation numbers, write the number of electrons transferred per atom.
4.	Using the chemical formulas, determine the number of electrons transferred per reactant. (Use the formula subscripts to do this.)
5.	Calculate the simplest whole number coefficients for the reactants that will balance the total number of electrons transferred. Balance the reactants and products.
6.	Balance the O atoms using H2O (l), and then balance the H atoms using H+ (aq).

For basic solutions only,
7.	Add OH- (aq) to both sides equal in number to the number of H+ (aq) present.
8.	Combine H+ (aq) and OH- (aq) on the same side to form H2O (l), and cancel the same number of H2O (l) on both sides.

               WO3 (s) +      H2 (g) →  W (s) +      H2O (g)
              6 e g/W      1e/H
              6 e/WO3     2 e/H2   
                 X1                x3

               WO3 (s) +    3  H2 (g) →  W (s) +  3  H2O (g)

0

+6

-2

0

+1

-2

             AgNO3 (aq) +  Cu (s) → Cu(NO3)2 (aq) + Ag (s)
            1 e g/ Ag        2 e l/Cu
            1 e g/AgNO3  2 e l/Cu
               x 2                  x1

            2 AgNO3 (aq) +  Cu (s) → Cu(NO3)2 (aq) + 2 Ag (s)

0

0

-2

+5

+2

+1

-2

+5

Writing Half-Reaction Equations
The general procedure follows:
1.	Write the chemical formulas for the reactants and products, breaking it into unbalanced oxidation and reduction half reactions. 
2.	Balance by inspection all elements in each half reaction except O and H.
3.	Balance O by adding H2O (l).
4.	Balance H by adding H+ (aq).
5.	Balance the charge in each half reaction by adding e- as products or reactants.
6.	Balance the e- transfer by multiplying the balanced half reactions by appropriate intergers.
7.	Add the resulting half reactions and eliminate any common terms.

For basic solutions only,
8.	 Add OH- (aq) to both sides to equal the number of H+ (aq) present.
9.	Combine H+ (aq) and OH- (aq) on the same side to form H2O (l). Cancel equal amounts of H2O (l) from both sides.

Al (s) + MnO4- (aq) → MnO2 (s) + Al(OH)4- (aq) 
Step 1:	                         Al (s) → Al(OH)4- (aq) 
		MnO4- (aq) → MnO2 (s) 

Step 2: 		         Al (s) → Al(OH)4- (aq) 
		MnO4- (aq) → MnO2 (s) 

Step 3: 	         4 H2O (l) + Al (s) → Al(OH)4- (aq) 
		  MnO4- (aq) → MnO2 (s) + 2 H2O (l) 

Step 4: 	        4 H2O (l) + Al (s) → Al(OH)4- (aq) + 8 H+ (aq)
	4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l) 

Step 5: 	                   4 H2O (l) + Al (s) → Al(OH)4- (aq) + 8 H+ (aq) + 7 e-  (x3)
   	3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)                 (x7)

   	                     12 H2O (l) + 3 Al (s)  → 3 Al(OH)4- (aq) + 24 H+ (aq) + 21 e-
	21 e- + 28 H+ (aq)  + 7 MnO4- (aq) → 7 MnO2 (s) + 14 H2O (l)

3 Al (s) + 4 H+ (aq)  + 7 MnO4- (aq)  → 3 Al(OH)4- (aq) + 24 H+ (aq) + 7 MnO2 (s) + 2 H2O (l)

I- (aq) + ClO- (aq) → I3- (aq)  +  Cl- (aq)     (in basic conditions)

Step 1:              I- (aq) → I3- (aq)  
                      ClO- (aq) →  Cl- (aq)  

Step 2:           3 I- (aq) → I3- (aq)  
                      ClO- (aq) →  Cl- (aq)  

Step 3:           3 I- (aq) → I3- (aq)  
                      ClO- (aq) →  Cl- (aq)  + H2O (l)

Step 4:                         3 I- (aq) → I3- (aq)  
                   2 H+ (aq) + ClO- (aq) →  Cl- (aq)  + H2O (l)

Step 5:                               3 I- (aq) → I3- (aq) +  2 e- 	(x1)
                  e- + 2 H+ (aq) + ClO- (aq) →  Cl- (aq)  + H2O (l)	(x2)

                                                  3 I- (aq) → I3- (aq) +  2 e- 	
                  2 e- + 4 H+ (aq) + 2 ClO- (aq) →  2 Cl- (aq)  + 2 H2O (l)

2 ClO- (aq) + 3 I- (aq) + 4 H+ (aq) +  4 OH- (aq) →  2 Cl- (aq) + I3- (aq) + 2 H2O (l) +  4 OH- (aq) 

		           4 H2O

2 ClO- (aq) + 3 I- (aq) + 4 H2O (l) →  2 Cl- (aq) + I3- (aq) + 2 H2O (l) +  4 OH- (aq) 

2 ClO- (aq) + 3 I- (aq) + 2 H2O (l) →  2 Cl- (aq) + I3- (aq) + 4 OH- (aq) 

Alternatively the balancing of both oxygen and hydrogen is done by adding different compounds containing oxygen and hydrogen depending on the situation (acidic / basic) that the redox reaction is occurring in.

and 
then

To balance H:
Add H+

To balance O:
Add H2O

In Acidic Solutions

and 
then

To balance H:
For each H needed:
(1)	add one H2O to side needing H and
(2)	add one OH- to the other side

To balance O:
For each O needed:
(1)	add two OH- to side needing O and
(2)	add one H2O to the other side

In Basic Solutions

In basic conditions, hypochlorite ions (ClO-) oxidize chromite ions (CrO2-) to chromate ions and are reduced to chloride ions. Write the balanced net ionic equation for this reaction.
CrO2- (aq)  + ClO- (aq) → CrO42- (aq) + Cl- (aq)  
Step 1: 	 CrO2- (aq)  → CrO42- (aq)  
   	   ClO- (aq) → Cl- (aq)  

Step 2: 	 CrO2- (aq)  → CrO42- (aq)  
   	   ClO- (aq) → Cl- (aq)  
 
Step 3: 	 4 OH- (aq) + CrO2- (aq)  → CrO42- (aq) + 2 H2O (l) 
      	         H2O (l) + ClO- (aq) → Cl- (aq) + 2 OH- (aq) 

Step 4: 	 4 OH- (aq) + CrO2- (aq)  → CrO42- (aq) + 2 H2O (l)   
      	         H2O (l) + ClO- (aq) → Cl- (aq) + 2 OH- (aq) 

Step 5: 	 4 OH- (aq) + CrO2- (aq)  → CrO42- (aq) + 2 H2O (l) + 3 e-         (x 2)
      	         2 e- + H2O (l) + ClO- (aq) → Cl- (aq) + 2 OH- (aq)             (x 3)

Step 6: 	               8 OH- (aq) + 2 CrO2- (aq)  → 2 CrO42- (aq) + 4 H2O (l) + 6 e-   
      	         6 e- + 3 H2O (l) + 3 ClO- (aq) → 3 Cl- (aq) + 6 OH- (aq) 

8 OH- (aq) + 2 CrO2- (aq)  + 3 H2O (l) + 3 ClO- (aq) → 2 CrO42- (aq) + 4 H2O (l) + 3 Cl- (aq) + 6 OH- (aq)  

2 OH- (aq) + 2 CrO2- (aq)  +  3 ClO- (aq) → 2 CrO42- (aq) + H2O (l) + 3 Cl- (aq)