Answer:
5 H₂O₂ + 6 H⁺ + 2 MnO₄⁻ → 5 O₂ + 2 Mn²⁺ + 8 H₂O
Step-by-step explanation:
In order to balance a redox reaction, we will use the ion-electron method.
Step 1: Identify both half-reactions
Oxidation: H₂O₂ → O₂
Reduction: MnO₄⁻ → Mn²⁺
Step 2: Perform the mass balance, adding H⁺ and H₂O where necessary.
H₂O₂ → O₂ + 2 H⁺
8 H⁺ + MnO₄⁻ → Mn²⁺ + 4 H₂O
Step 3: Perform the electrical balance, adding electrons where necessary.
H₂O₂ → O₂ + 2 H⁺ + 2 e⁻
8 H⁺ + MnO₄⁻ + 5 e⁻ → Mn²⁺ + 4 H₂O
Step 4: Multiply both half-reactions by numbers that secure that the number of electrons gained and lost are equal.
5 × (H₂O₂ → O₂ + 2 H⁺ + 2 e⁻)
2 × (8 H⁺ + MnO₄⁻ + 5 e⁻ → Mn²⁺ + 4 H₂O)
Step 5: Add both half-reactions and cancel what is repeated in both sides.
5 H₂O₂ + 16 H⁺ + 2 MnO₄⁻ → 5 O₂ + 10 H⁺ + 2 Mn²⁺ + 8 H₂O
5 H₂O₂ + 6 H⁺ + 2 MnO₄⁻ → 5 O₂ + 2 Mn²⁺ + 8 H₂O