Balancing Redox Equations

1. Balancing Redox Equations The Molecular Method Ms. Reid

2. Assign oxidation numbers to all atoms in the equation. F2 + H2O → HF + O3 ` 0 +1 -2 +1 -1 0 F goes from 0 to -1 (reduced). O goes from -2 to 0 (oxidized).

3. Write the half-reaction with the reduced or oxidized atoms only as they appear in the equation. F2 + H2O → HF + O3 R: F2→ F O: O → O3 0 -1 -2 0 F on the left is F2 so write that down and on the right is HF so write F. O on the left is H2O so write O and on the right is O3 so write O3. In other words, write only the element involved along with its subscript.

4. Balance the atoms being reduced or oxidized using coefficients R: F2→ F O: O → O3 R: F2→ 2 F O: 3 O → O3 0 -1 -2 0 0 -1 -2 0

5. Balance the charge by adding electrons to the most positive side. R: F2→ 2 F O: 3 O → O3 R: F2 + 2e-→ 2F O: 3O → O3+ 6e- 0 -1 -2 0 (net charge of 0 on left and -2 on right) (net charge of 6- on left and 0 on right)

6. Get the same number of electrons in each half-reaction R: F2 + 2e-→ 2F O: 3O → O3+ 6e- R: 3x(F2 + 2e- → 2F) O: 3O → O3+ 6e- (Multiplying the reduction half-reaction by a factor of 3 gives 6 electrons in each half-reaction.)

7. Transfer the coefficients to the main equation. 3 F2 + 3 H2O → 6 HF + O3 R: 3F2 + 6e- → 6 F O: 3 O → O3+ 6e-

8. Balance any remaining atoms by inspection. • Not necessary in this case. 3 F2 + 3 H2O → 6 HF + O3

9. Now one that’s not quite so easy. Sb + HNO3→ Sb2O5 + NO2 + H2O

10. Assign oxidation numbers to all atoms in the equation. Sb + HNO3→ Sb2O5 + NO2 + H2O 0 +1 +5 --2 +5 -2 +4 -2 +1 -2 Sb goes from 0 to +5 (oxidized). N goes from +5 to +4 (reduced).

11. Write the half-reaction with the reduced or oxidized atoms only as they appear in the equation. Sb + HNO3→ Sb2O5 + NO2 + H2O O: Sb → Sb2 R: N → N 0 +5 +5 +4 Sb is in Sb2O5 on the right so write down Sb2.

12. Balance the atoms being reduced or oxidized using coefficients O: Sb → Sb2 R: N → N O: 2 Sb → Sb2 R: N → N 0 +5 +5 +4

13. Balance the charge by adding electrons to the most positive side. O: 2 Sb → Sb2 R: N → N 0 +5 +5 +4 O: 2 Sb → Sb2+ 10e- R: N + 1e- → N 0 +5 +5 +4 (net charge of 0 on left and +10 on right) (net charge of +5 on left and +4 on right)

14. Get the same number of electrons in each half-reaction O: 2 Sb → Sb2+ 10e- R: 10x(N + 1e- → N) 0 +5 +5 +4 (Now there are 10 electrons in each half-reaction.)

15. Transfer the coefficients to the main equation. O: 2 Sb → Sb2+ 10e- R: 10N + 10e- → 10N 2Sb + 10 HNO3→ Sb2O5 + 10 NO2 + H2O

16. Balance any remaining atoms by inspection. 2Sb + 10 HNO3→ Sb2O5 + 10 NO2 + 5 H2O