Answer:
Oxidizing agent: ClO₂
Reducing agent: H₂O₂
2 ClO₂(aq) + H₂O₂(aq) → 2 ClO₂⁻(aq) + O₂(g) + 2 H⁺(aq)
Explanation:
In order to balance a redox reaction, we will use the ion-electron method.
Step 1: Identify both half-reactions
Reduction: ClO₂(aq) → ClO₂⁻(aq)
Oxidation: H₂O₂(aq) → O₂(g)
ClO₂ is reduced, so it is the oxidizing agent, while H₂O₂ is oxidized so it is the reducing agent.
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
ClO₂(aq) → ClO₂⁻(aq)
H₂O₂(aq) → O₂(g) + 2 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
1 e⁻ + ClO₂(aq) → ClO₂⁻(aq)
H₂O₂(aq) → O₂(g) + 2 H⁺(aq) + 2 e⁻
Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are equal, and add them
2 × [1 e⁻ + ClO₂(aq) → ClO₂⁻(aq)]
1 × [H₂O₂(aq) → O₂(g) + 2 H⁺(aq) + 2 e⁻]
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2 ClO₂(aq) + H₂O₂(aq) → 2 ClO₂⁻(aq) + O₂(g) + 2 H⁺(aq)