Arkansas State University
Department of Chemistry
and Physics
Answer - Problem 3
Balancing Redox Reactions

3.   MnO4-   +    I-   -------->   MnO2     +    I2   (basic solution)

Step 1. Break into half-reactions:

  MnO4-    ------->    MnO2
          I-     ------->     I2

Step 2. Balance atoms other than H and O

  MnO4-    ------->    MnO2
      2  I-     ------->     I2

Step 3. Balance O by adding H2O

  MnO4-    ------->   MnO2   +   2 H2O
      2  I-     ------->     I2

Step 4. Balance H by adding H+

 4 H+ +  MnO4-    ------->    MnO2    +   2 H2O
                  2  I-     ------->     I2

Step 5. Balance charge by adding electron(s)

3 e- +  4 H+ +  MnO4-    ------->   MnO2       + 2 H2O
                               2  I-     ------->     I2   +    2 e-

Step 6. Electrons lost = electrons gained

 (3 e- +  4 H+ +  MnO4-    ------->    MnO2       + 2 H2O) x 2
 (                             2  I-     ------->     I2   +    2 e- ) x 3

Gives:
 6 e- + 8 H+ + 2 MnO4-    ------->   2 MnO2       + 4 H2O
          6 I-      ------->     3  I2   +   6 e-

Step 7. Cancel like terms and add half reactions

  8 H+ + 2 MnO4-   +   6 I-       ------->   2 MnO2       + 4 H2O +  3 I2

Step 8. Since it is a basic solution, add OH-  for each H+ (add OH- to both sides of eqn)

8 OH-   +    8 H + 2 MnO4-   +   6 I-       ------->   2 MnO2    + 4 H2O +  3  I2   +   8 OH-

9. Combine H+ with OH- to form H2O

8 OH-   +    8 H+  + 2 MnO4-   +   6 I-       ------->   2 MnO2   +  4 H2O +   3 I2   +   8 OH-
Gives:
      8 H2O   +    2 MnO4-   +   6 I-       ------->   2 MnO2   +   4 H2O +  3 I2   +   8 OH-

10. Cancel like terms:

   4 H2O   +    2 MnO4-   +   6 I-       ------->   2 MnO2    +  3  I2   +   8 OH-
 

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