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Complete and balance the following redox reaction in acidic solution. be sure to include the proper phases for all species within the reaction. ClO 3



(aq)+I 2

( s)→ClO 2

( g)+IO 3

(aq)

User Relm
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2 Answers

5 votes

Final answer:

To balance the given redox reaction in acidic solution, follow the steps of identifying the half-reactions, balancing the atoms, balancing the charges, balancing the electrons, and combining the half-reactions. The balanced redox reaction is 2ClO3- (aq) + 12H+ (aq) + I2 (s) + 6H2O (l) → 2ClO2 (g) + 2IO3- (aq) + 6H+ (aq).

Step-by-step explanation:

To balance the redox reaction in acidic solution, follow these steps:

  1. Identify the oxidation and reduction half-reactions. In this case, the oxidation half-reaction is: ClO3−(aq) → ClO2(g) (reducing agent) and the reduction half-reaction is: I2(s) → IO3−(aq) (oxidizing agent).
  2. Balance the atoms in each half-reaction by adding the necessary coefficients. For the oxidation half-reaction, you need to balance the number of Cl (chlorine) atoms, while for the reduction half-reaction, you need to balance the number of I (iodine) atoms. The balanced half-reactions are: ClO3−(aq) + 6H+(aq) → ClO2(g) + 3H2O(l) (oxidation) and I2(s) + 6H2O(l) → 2IO3−(aq) + 12H+(aq) (reduction).
  3. Balance the charges in each half-reaction using H+ ions. For the oxidation half-reaction, add 6H+ ions to the left side, and for the reduction half-reaction, add 12H+ ions to the right side. The balanced half-reactions with charges are: ClO3−(aq) + 6H+(aq) → ClO2(g) + 3H2O(l) + 6e− (oxidation) and I2(s) + 6H2O(l) + 12e− → 2IO3−(aq) + 12H+(aq) (reduction).
  4. Balance the number of electrons in each half-reaction by multiplying one or both of the half-reactions by integers. In this case, we can multiply the oxidation half-reaction by 2 to equalize the number of electrons in both reactions. The adjusted half-reactions are: 2ClO3−(aq) + 12H+(aq) → 2ClO2(g) + 6H2O(l) + 12e− (oxidation) and I2(s) + 6H2O(l) + 12e− → 2IO3−(aq) + 12H+(aq) (reduction).
  5. Add the two half-reactions together and combine any like terms. The final balanced redox reaction in acidic solution is: 2ClO3−(aq) + 12H+(aq) + I2(s) + 6H2O(l) → 2ClO2(g) + 2IO3−(aq) + 6H+(aq).

User Amichaud
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7.9k points
6 votes

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)

User Glacasa
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