Cân bằng các phản ứng oxi hóa – khử sau
Giải thích
a) H+1Cl−1 + Pb+4O−22 → Pb+2Cl−12 + Cl02 + H+12O−2
Pb+4 +2e → Pb +2Cl−1 → Cl0 +1e
1 × 2 × Pb+4 +2e → Pb +2Cl−1 → Cl0 +1e
⇒ Pb+4 + 2Cl−1 → Pb+2 + 2Cl0
4HCl + PbO2 → PbCl2 + Cl2 + 2H2O
b) K+1Mn+7O−24 + H+1Cl−1 → K+1Cl−1 + Mn+2Cl−12 + Cl02 + H+12O−2
Mn+7 +5e → Mn+2Cl−1 → Cl0 +1e
1 × 5 ×Mn+7 +5e → Mn+2Cl−1 → Cl0 +1e
⇒ Mn+7 + 5Cl−1 → Mn+2 + 5Cl0
⇒ 2Mn+7 + 10Cl−1 → 2Mn+2 + 10Cl0
2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O