How to Balance Redox Equations?

How do you balance the following redox equations by the ion-electron method?

• (Use the lowest possible coefficients. Omit states-of-matter in your answer.) (a) Mn2+ + H2O2 → MnO2 + H2O (in basic solution) (b) Bi(OH)3 + SnO22- → Bi + SnO32- (in basic solution) (c) Cr2O72- + C2O42- → Cr3+ + CO2 (in acidic solution) (d) Cl ‾ + ClO3‾ → Cl2 + ClO2 (in acidic solution)

• Answer:

  (a) Mn2+ + 4OH- ---> MnO2 + 2H2O + 2e- H2O2 + 2e- ----> 2OH- Mn2+ + 2OH- + H2O2 ---> MnO2 + 2H2O (b) Bi(OH)3 + 3e- ---> Bi + 3 OH- SnO2^-2 + 2OH- ------> SnO3^-2 + H2O + 2e- 2 Bi(OH)3 + 3 SnO2^-2 -----> 2 Bi + 3 SnO3^-2 + 3H2O (c) Cr2O7^-2 + 14 H+ + 6 e- ---> 2 Cr^3+ + 7 H2O C2O4^-2 -----> 2 CO2 + 2e- 3 C2O4^-2 + Cr2O7^-2 + 14 H+ ---> 6 CO2 + 2 Cr^3+ + 7 H2O (d) 2Cl- ----> Cl2 +2e- ClO3^- + 2H+ +1e- ---> ClO2 + H2O 2Cl- + 2ClO3^- + 4H+ ---> Cl2 + 2ClO2 + 2H2O