Balancing Redox Reactions (Solutions)
- Page ID
- 11049
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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-