Răspuns:
3H Cl + HNO3 → N O Cl+ CI 2+ 2 H 2 O
2 Cl -I - 2 e - → 2 Cl 0 (oxidare)
N V + 2 e - → N III (reducere)
2 Na O H + Cl 2 → Na CI+ Na CI O + H 2 O
Cl 0 - 1 e - → Cl I (oxidare)
Cl 0 + e - → Cl -I(reducere)
PbO 2+ 4 H Cl→ Pb Cl 2 + Cl 2 + 2 H 2 O
Pb IV + 2 e - → Pb II (reducere)
2 Cl -I - 2 e - → 2 Cl 0 (oxidare)
