Cân bằng phương trình bằng phương pháp thăng bằng e: (a) KMnO4 → K2MnO4 + MnO2 + O2 (b) KClO3 → KCl+ O2

a)

\[\left. {\begin{array}{*{20}{c}}{1 \times }\\{1 \times }\end{array}} \right|\begin{array}{*{20}{c}}{2\mathop {Mn}\limits^{ + 7} + 4e \to \mathop {Mn}\limits^{ + 6} + \mathop {Mn}\limits^{ + 4} }\\{2\mathop O\limits^{ - 2} \to {{\mathop O\limits^0 }_2} + 4e}\end{array}\]

2KMnO₄ K₂MnO₄ + MnO₂ + O₂

b)

\[\left. {\begin{array}{*{20}{c}}{2 \times }\\{3 \times }\end{array}} \right|\begin{array}{*{20}{c}}{\mathop {Cl}\limits^{ + 5} + 6e \to \mathop {Cl}\limits^{ - 1} }\\{2\mathop O\limits^{ - 2} \to {{\mathop O\limits^0 }_2} + 4e}\end{array}\]

2KClO₃ 2KCl + 3O₂