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-