Răspuns:
3 C + 4 H N O 3 → 3 C O 2 + 4 N O + 2 H 2 O
3 C 0 - 12 e - → 3 C IV (oxidare)
4 N V + 12 e - → 4 N II (reducere)
6 K N O 3 + 10 Fe → 5 Fe 2 O 3 + 3K 2 O + 3 N 2
6 N V + 30 e - → 6 N 0 (reducere)
10 Fe 0 - 30 e - → 10 Fe III (oxidare)
6 H I + 2 H NO3 → 3 I2 + 2 NO + 4H2O
6 I -I - 6 e - → 6 I 0 (oxidare)
2 N V + 6 e - → 2 N II (reducere)
5FeCl2 + KMnO4 + 8HCl → 5 FeCl3 + MnCl2 + KCl + 4H2O
Mn VII + 5 e - → Mn II (reducere)
5 Fe II - 5 e - → 5 Fe III (oxidare)