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Balancing Redox Solutions

  • Page ID
    11049
  • 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-