To balance this redox equation in an acidic solution, follow these steps:

1. Break the equation into two half-reactions:
Half-reaction 1: MnO4⁻ → Mn²⁺ (reduction)
Half-reaction 2: H2O2 → O2 (oxidation)

2. Balance the atoms other than oxygen and hydrogen:
Half-reaction 1: MnO4⁻ → Mn²⁺ (already balanced)
Half-reaction 2: 2H2O2 → O2 (balance the hydrogen atoms by adding 4H2O on the right side)

The half-reactions become:
Half-reaction 1: MnO4⁻ → Mn²⁺
Half-reaction 2: 2H2O2 → O2 + 4H2O

3. Balance the oxygen atoms by adding H2O to the side that needs more oxygen:
Half-reaction 1: MnO4⁻ + 4H2O → Mn²⁺ (already balanced)
Half-reaction 2: 2H2O2 → O2 + 4H2O (already balanced)

The half-reactions become:
Half-reaction 1: MnO4⁻ + 4H2O → Mn²⁺
Half-reaction 2: 2H2O2 → O2 + 4H2O

4. Balance the hydrogen atoms by adding H+ ions:
Half-reaction 1: MnO4⁻ + 8H+ + 4H2O → Mn²⁺
Half-reaction 2: 2H2O2 → O2 + 4H2O

5. Balance the charges by adding electrons:
Half-reaction 1: MnO4⁻ + 8H+ + 4e⁻ → Mn²⁺
Half-reaction 2: 2H2O2 → O2 + 4H2O + 4e⁻

6. Multiply each half-reaction to equalize the number of electrons:
Half-reaction 1: 2MnO4⁻ + 16H+ + 8e⁻ → 2Mn²⁺ + 8H2O
Half-reaction 2: 4H2O2 → 2O2 + 8H2O + 8e⁻

7. Add the two half-reactions together:
2MnO4⁻ + 16H+ + 8e⁻ + 4H2O2 → 2Mn²⁺ + 8H2O + 2O2 + 8H2O + 8e⁻

8. Simplify the equation:
2MnO4⁻ + 16H+ + 4H2O2 → 2Mn²⁺ + 10H2O + 2O2

Therefore, the balanced redox equation in an acidic solution is:
2MnO4⁻ + 16H+ + 4H2O2 → 2Mn²⁺ + 10H2O + 2O2

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