Arkansas State University Department of Chemistry and Physics

1.   1.   Bi(OH)₃   +     SnO₂^2-    --------->     SnO₃^2-   +  Bi    (basic solution)

Step 1. Break into half-reactions:

     Bi(OH)₃    ------->    Bi
      SnO₂^2-     ------->    SnO₃^2-

Step 2. Balance atoms other than H and O

     Bi(OH)₃    ------->    Bi
      SnO₂^2-     ------->    SnO₃^2-

Step 3. Balance O by adding H₂O

                  Bi(OH)₃    ------->    Bi  +  3 H₂O
    H₂O  +   SnO₂^2-     ------->    SnO₃^2-

Step 4. Balance H by adding H⁺

       3 H⁺   +   Bi(OH)₃    ------->    Bi  +  3 H₂O
        H₂O  +   SnO₂^2-     ------->    SnO₃^2-  +    2H⁺

Step 5. Balance charge by adding electron(s)

3 e^-   +  3 H⁺   +   Bi(OH)₃    ------->    Bi  +  3 H₂O
            H₂O  +   SnO₂^2-     ------->    SnO₃^2-  +   2 H⁺  +  2 e^-

Step 6. Electrons lost = electrons gained

 (3 e^- + 3 H⁺   +   Bi(OH)₃    ------->    Bi  +  3  H₂O) x 2
 (           H₂O  +   SnO₂^2-     ------->    SnO₃^2-  +   2 H⁺  +  2 e^- ) x 3

Gives:
 6 e^- +  6 H⁺ + 2 Bi(OH)₃     ------->   2 Bi     +   6 H₂O
        3 H₂O  +   3 SnO₂^2-     ------->  3 SnO₃^2-  +    6H⁺  +   6 e^-

Step 7. Cancel like terms and add half reactions

 2 Bi(OH)₃    +     3 SnO₂^2-       ------->   2Bi   +  3 SnO₃^2-  +  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
 

Problem 2 answer
Problem 3 answer