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Chemistry LibreTexts

Balancing Redox Solutions

1.  

Bi(OH)3   +     SnO22-    --------->     SnO32-   +  Bi    (basic solution)

Step 1. Break into half-reactions:

     Bi(OH)3    ------->    Bi
      SnO22-     ------->    SnO32-

Step 2. Balance atoms other than H and O

     Bi(OH)3    ------->    Bi
      SnO22-     ------->    SnO32-

Step 3. Balance O by adding H2O

                  Bi(OH)3    ------->    Bi  +  3 H2O
    H2O  +   SnO22-     ------->    SnO32-

Step 4. Balance H by adding H+

       3 H+   +   Bi(OH)3    ------->    Bi  +  3 H2O
        H2O  +   SnO22-     ------->    SnO32+    2H+

Step 5. Balance charge by adding electron(s)

3 e-   +  3 H+   +   Bi(OH)3    ------->    Bi  +  3 H2O
            H2O  +   SnO22-     ------->    SnO32+   2H+  +  2 e-

Step 6. Electrons lost = electrons gained

 (3 e- + 3 H+   +   Bi(OH)3    ------->    Bi  +  3  H2O) x 2
 (           H2O  +   SnO22-     ------->    SnO32-  +   2H+  +  2 e- ) x 3

Gives:
 6 e- +  6 H+ + 2 Bi(OH)3     ------->   2 Bi     +   6 H2O
        3 H2O  +   3 SnO22-     ------->  3 SnO32+    6H+  +   6 e-

Step 7. Cancel like terms and add half reactions

 2 Bi(OH)3    +     3 SnO22-       ------->   2Bi   +  3 SnO32+  3 H2O

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

No H+, finished

2.

   S2O32-   +   I2   -------->      I-   +   S4O62-    (acidic solution)

Step 1.  Break into half-reactions:

      S2O32-    ------->    S4O62-
              I2    ------->     I-

Step 2.  Balance atoms other than H and O

   2 S2O32-    ------->     S4O62-
              I2    ------->     2 I-

Step 3. Balance O by adding H2O (already balanced)

      2  S2O32-   ------->     S4O62-
                  I2    ------->      2 I-

Step 4. Balance H by adding H(No H's)

   2 S2O32-    ------->     S4O62-
               I2    ------->      2 I-

Step 5. Balance charge by adding electron(s)

   2  S2O32-    ------->     S4O62-     + 2 e-
  2 e-   +  I2    ------->      2 I-

Step 6. Electrons lost = electrons gained  (2 e- lost, 2 e- gained)

   2  S2O32-    ------->    S4O62-     + 2 e-
  2 e- + I2    ------->      2 I-

Step 7. Cancel like terms and add half reactions

      2  S2O32-      +  I2           ------->    S4O62-     +   2 I-

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-