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)

(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