1. Bi(OH) 3 + SnO 2 2 - ---------> SnO 3 2 - + Bi (basic solution) Step 1. Break into half-reactions: Bi(OH) 3 -------> Bi SnO 2 2 - -------> SnO 3 2 - Step 2. Balance atoms other than H and O Bi(OH) 3 -------> Bi SnO 2 2 - -------> SnO 3 2 - Step 3. Balance O by adding H 2 O Bi(OH) 3 -------> Bi + 3 H 2 O H 2 O + SnO 2 2 - -------> SnO 3 2 - Step 4. Balance H by adding H + 3 H + + Bi(OH) 3 -------> Bi + 3 H 2 O H 2 O + SnO 2 2 - -------> SnO 3 2 - + 2H + Step 5. Balance charge by adding electron(s) 3 e - + 3 H + + Bi(OH) 3 -------> Bi + 3 H 2 O H 2 O + SnO 2 2 - -------> SnO 3 2 - + 2H + + 2 e - Step 6. Electrons lost = electrons gained (3 e - + 3 H + + Bi(OH) 3 -------> Bi + 3 H 2 O) x 2 ( H 2 O + SnO 2 2 - -------> SnO 3 2 - + 2H + + 2 e - ) x 3 Gives: 6 e - + 6 H + + 2 Bi(OH) 3 -------> 2 Bi + 6 H 2 O 3 H 2 O + 3 SnO 2 2 - -------> 3 SnO 3 2 - + 6H + + 6 e - Step 7. Cancel like terms and add half reactions 2 Bi(OH) 3 + 3 SnO 2 2 - -------> 2Bi + 3 SnO 3 2 - + 3 H 2 O Step 8. Since it is a basic solution, add OH - for each H + (add OH - to both sides of eqn) No H + , finished 2. S 2 O 3 2 - + I 2 --------> I - + S 4 O 6 2 - (acidic solution) Step 1. Break into half-reactions: S 2 O 3 2 - -------> S 4 O 6 2 - I 2 -------> I - Step 2. Balance atoms other than H and O 2 S 2 O 3 2 - -------> S 4 O 6 2 - I 2 -------> 2 I - Step 3. Balance O by adding H 2 O (already balanced) 2 S 2 O 3 2 - -------> S 4 O 6 2 - I 2 -------> 2 I - Step 4. Balance H by adding H + (No H's) 2 S 2 O 3 2 - -------> S 4 O 6 2 - I 2 -------> 2 I - Step 5. Balance charge by adding electron(s) 2 S 2 O 3 2 - -------> S 4 O 6 2 - + 2 e - 2 e - + I 2 -------> 2 I - Step 6. Electrons lost = electrons gained (2 e - lost, 2 e - gained) 2 S 2 O 3 2 - -------> S 4 O 6 2 - + 2 e - 2 e - + I 2 -------> 2 I - Step 7. Cancel like terms and add half reactions 2 S 2 O 3 2 - + I 2 -------> S 4 O 6 2 - + 2 I - 3 MnO 4 - + I - --------> MnO 2 + I 2 (basic solution) Step 1. Break into half-reactions: MnO 4 - -------> MnO 2 I - -------> I 2 Step 2. Balance atoms other than H and O MnO 4 - -------> MnO 2 2 I - -------> I 2 Step 3. Balance O by adding H 2 O MnO 4 - -------> MnO 2 + 2 H 2 O 2 I - -------> I 2 Step 4. Balance H by adding H + 4 H + + MnO 4 - -------> MnO 2 + 2 H 2 O 2 I - -------> I 2 Step 5. Balance charge by adding electron(s) 3 e - + 4 H + + MnO 4 - -------> MnO 2 + 2 H 2 O 2 I - -------> I 2 + 2 e - Step 6. Electrons lost = electrons gained (3 e - + 4 H + + MnO 4 - -------> MnO 2 + 2 H 2 O) x 2 ( 2 I - -------> I 2 + 2 e - ) x 3 Gives: 6 e - + 8 H + + 2 MnO 4 - -------> 2 MnO 2 + 4 H 2 O 6 I - -------> 3 I 2 + 6 e - Step 7. Cancel like terms and add half reactions 8 H + + 2 MnO 4 - + 6 I - -------> 2 MnO 2 + 4 H 2 O + 3 I 2 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 MnO 4 - + 6 I - -------> 2 MnO 2 + 4 H 2 O + 3 I 2 + 8 OH - 9. Combine H + with OH - to form H 2 O 8 OH - + 8 H + + 2 MnO 4 - + 6 I - -------> 2 MnO 2 + 4 H 2 O + 3 I 2 + 8 OH - Gives: 8 H 2 O + 2 MnO 4 - + 6 I - -------> 2 MnO 2 + 4 H 2 O + 3 I 2 + 8 OH - 10. Cancel like terms: 4 H 2 O + 2 MnO 4 - + 6 I - -------> 2 MnO 2 + 3 I 2 + 8 OH -