Custom Search

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)

____________________________________

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