Final answer:
To balance the redox reaction ClO₃⁻(aq) + I₂(s) → ClO₂(g) + IO₃(aq) in acidic solution, follow the stepwise process. Divide the reaction into two half-reactions, balance the equations for atoms, add water molecules and H+ ions to balance oxygen and hydrogen atoms, balance charges with electrons, and multiply the half-reactions to balance the electrons transferred. The overall balanced equation is 6ClO₃⁻(aq) + 2I₂(s) + 6H₂O(l) → 5ClO₂(g) + 4IO₃−(aq) + 6H+(aq).
Step-by-step explanation:
To balance the redox reaction ClO₃⁻(aq) + I₂(s) → ClO₂(g) + IO₃(aq) in acidic solution, we need to follow a stepwise process.
Step 1: Divide the reaction into two half-reactions, one for oxidation and one for reduction.
Oxidation half-reaction: ClO₃⁻(aq) → ClO₂(g)
Reduction half-reaction: I₂(s) → IO₃(aq)
Step 2: Balance the equations for atoms other than oxygen and hydrogen.
Oxidation half-reaction: 6ClO₃⁻(aq) → 5ClO₂(g) + Cl−(aq)
Reduction half-reaction: I2(s) → 2IO₃−(aq)
Step 3: Balance the oxygen atoms by adding water molecules.
Oxidation half-reaction: 6ClO₃⁻(aq) → 5ClO₂(g) + Cl−(aq) + 3H₂O(l)
Reduction half-reaction: I2(s) + 6H₂O(l) → 2IO₃−(aq) + 12H+(aq)
Step 4: Balance the hydrogen atoms by adding H+ ions.
Oxidation half-reaction: 6ClO₃⁻(aq) + 3H₂O(l) → 5ClO₂(g) + Cl−(aq) + 6H+(aq)
Reduction half-reaction: I2(s) + 6H₂O(l) → 2IO₃−(aq) + 12H+(aq)
Step 5: Balance the charges by adding electrons.
Oxidation half-reaction: 6ClO₃⁻(aq) + 3H₂O(l) → 5ClO₂(g) + Cl−(aq) + 6H+(aq) + 6e−
Reduction half-reaction: I2(s) + 6H₂O(l) → 2IO₃⁻(aq) + 12H+(aq) + 12e−
Step 6: Multiply the half-reactions by appropriate factors to balance the number of electrons transferred.
Oxidation half-reaction: 6ClO₃⁻(aq) + 3H₂O(l) → 5ClO₂(g) + Cl−(aq) + 6H+(aq) + 6e−
Reduction half-reaction: 2I₂(s) + 12H₂O(l) → 4IO₃−(aq) + 24H+(aq) + 24e−
Step 7: Add the balanced half-reactions together and eliminate the electrons.
Overall balanced redox equation: 6ClO₃⁻(aq) + 2I₂(s) + 6H₂O(l) → 5ClO₂(g) + 4IO₃⁻(aq) + 6H+(aq)