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Balancing Redox Reactions

 

How do we balance redox reactions? Here is a simple example:

__Al + __Cu2+ --> __Cu + __Al3+

 

1. Start by writing half reactions (Oxidation and reduction)

(Electrons go on the more positive side) 

Oxidation:    Al --> Al 3+ + 3e-

Reduction:    2e- + Cu2+ --> Cu

2. Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 

The common multiple of the electrons is 6 so 

Oxidation:    2 x (Al --> Al 3+ + 3e-)

Reduction:    3 x ( 2e- + Cu2+ --> Cu)

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Recombine the reactions                                             6e- + 2 Al  + 3 Cu2+--> 2 Al 3+ + 3Cu + 6e-

The electrons must cancel.                                                   2 Al  + 3 Cu2+--> 2 Al 3+ + 3Cu

Atoms and charges must be conserved. 

 

AP Balancing Redox Reactions (Acidic Conditions)

 

Given          MnO4- + I- --> I2 + Mn2+ (acidic)

Step 1 Half Reactions

MnO4-  -->   Mn2+

 I-   -->     I2

Lets balance the reduction one first

for every Oxygen add a water on the other side

For every hydrogen add a H+ to the other side

Balance the imbalance of charge with electrons (+7 vs. +2)

 

MnO4-  -->   Mn2+ + 4H2O

8H+ + MnO4-  -->   Mn2+ + 4H2O

5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O 

Now for the oxidation

Balance the atoms

Balance the imbalance of charge with electrons (-2 vs. 0)

 I-   -->     I2

2I-   -->     I2

2I-   -->     I2 + 2e-

Step 2 Balance the electrons by finding the common multiple and multiply the half reactions accordingly.  Common Multiple here is 10.

2( 5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O )

5( 2I-   -->     I2 + 2e- )

Step 3 Check electrons, atoms and charge. Clean it up

10e- + 16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 
 

16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O 

Basic Conditions

 

Balancing redox reactions under Basic Conditions

Given          Cr(OH)3 + ClO3-    -->  CrO42- + Cl-   (basic)

Step 1 Half Reactions 

Lets balance the  reduction one first

for every Oxygen add a water on the other side

For every hydrogen add a H+ to the other side

Each H+  will react with an  OH- on both sides

H+ and OH-  make water

cancel the waters

 Balance the imbalance of charge with electrons (-1 vs. -7)

ClO3- --> Cl- 

 

ClO3- --> Cl- + 3H2O 

 

6H+ + ClO3- --> Cl- + 3H2O  

 

6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH- 

 

6H2O + ClO3- --> Cl- + 3H2O + 6 OH- 

 

3H2O + ClO3- --> Cl- + 6 OH- 

 

6e- + 3H2O + ClO3- --> Cl-  + 6 OH- 

Now for the oxidation

for every Oxygen add a water on the other side

For every hydrogen add a H+ to the other side

Each H+  will react with an  OH- on both sides

H+ and OH-  make water

cancel the waters

Balance the imbalance of chagre with electrons (-2 vs. 0)

Cr(OH)3 --> CrO42-

H2O  + Cr(OH)3 --> CrO42-

H2O  + Cr(OH)3 --> CrO42- + 5H+

5 OH-  + H2O  + Cr(OH)3 --> CrO42- + 5H++ 5OH- 

5 OH-  + H2O  + Cr(OH)3 --> CrO42- + 5H2O

5 OH-   + Cr(OH)3 --> CrO42- + 4H2O

5 OH-  + Cr(OH)3 --> CrO42- + 4H2O + 3e- 

Step 2 Balance the electrons by finding the common multiple and multiply the half reactions accordingly.  Common Multiple here is 6.

1(6e- + 3H2O + ClO3- --> Cl-  + 6OH- )

2(5 OH-   + Cr(OH)3 --> CrO42- + 4H2O + 3e- )

Step 3 Check electrons, atoms and charge then clean it up.

6e- + 3H2O + ClO3- + 10 OH-  + 2Cr(OH)3 -->Cl-  + 6OH- + 2CrO42- + 8H2O + 6e- 

 ClO3- + 4 OH-  + 2Cr(OH)3 -->Cl-  + 2CrO42- + 5H2O 

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