Math Problem Statement

HCLO3->CLO2+O2+H20 complete with half reaction method only

Solution

To balance the given reaction using the half-reaction method, let's follow these steps:

Given reaction:

HClO3ClO2+O2+H2O\text{HClO}_3 \rightarrow \text{ClO}_2 + \text{O}_2 + \text{H}_2\text{O}

Step 1: Split the reaction into two half-reactions.

  • Oxidation half-reaction: The species that loses electrons.
  • Reduction half-reaction: The species that gains electrons.
  1. Oxidation Half-Reaction: Cl in HClO3Cl in ClO2\text{Cl in HClO}_3 \rightarrow \text{Cl in ClO}_2

    The oxidation state of chlorine in HClO3\text{HClO}_3 is +5+5, and in ClO2\text{ClO}_2 it is +4+4.

  2. Reduction Half-Reaction: O in HClO3O2\text{O in HClO}_3 \rightarrow \text{O}_2

    The oxidation state of oxygen in O2\text{O}_2 is 00, while in HClO3\text{HClO}_3 it is 2-2.

Step 2: Balance the atoms in each half-reaction.

  1. Oxidation Half-Reaction (for Cl): ClO3ClO2+e\text{ClO}_3^- \rightarrow \text{ClO}_2 + e^-

    • Balance oxygen by adding water (H2O\text{H}_2\text{O}): ClO3ClO2+H2O+e\text{ClO}_3^- \rightarrow \text{ClO}_2 + \text{H}_2\text{O} + e^-
    • Balance hydrogen by adding protons (H+\text{H}^+): ClO3+2H+ClO2+H2O+e\text{ClO}_3^- + 2\text{H}^+ \rightarrow \text{ClO}_2 + \text{H}_2\text{O} + e^-
  2. Reduction Half-Reaction (for O): O32O2\text{O}_3^{2-} \rightarrow \text{O}_2

    • Balance oxygen: 2O323O2+4e2\text{O}_3^{2-} \rightarrow 3\text{O}_2 + 4e^-
    • Balance the charges by adding electrons.

Step 3: Combine the half-reactions.

Make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. To do this, we need to multiply the half-reactions by appropriate coefficients.

  1. Oxidation Half-Reaction: 2ClO3+4H+2ClO2+2H2O+2e2\text{ClO}_3^- + 4\text{H}^+ \rightarrow 2\text{ClO}_2 + 2\text{H}_2\text{O} + 2e^-

  2. Reduction Half-Reaction: 2O32+4e3O22\text{O}_3^{2-} + 4e^- \rightarrow 3\text{O}_2

Step 4: Add the half-reactions.

2ClO3+4H++2O322ClO2+3O2+2H2O2\text{ClO}_3^- + 4\text{H}^+ + 2\text{O}_3^{2-} \rightarrow 2\text{ClO}_2 + 3\text{O}_2 + 2\text{H}_2\text{O}

Final Balanced Equation:

2HClO32ClO2+3O2+2H2O2\text{HClO}_3 \rightarrow 2\text{ClO}_2 + 3\text{O}_2 + 2\text{H}_2\text{O}

This is the balanced equation using the half-reaction method.

Do you need any further details, or have any questions?


Related Questions:

  1. What are the steps for balancing redox reactions in acidic and basic solutions?
  2. How do oxidation states help in identifying the oxidized and reduced species?
  3. What are some common examples of redox reactions in everyday life?
  4. Can you explain the role of electrons in the half-reaction method?
  5. How would the procedure differ in a basic medium?

Tip: When balancing redox reactions, always verify the atom count and charge balance to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Chemistry
Redox Reactions
Half-Reaction Method

Formulas

Oxidation half-reaction: ClO3^- -> ClO2 + H2O + e^-
Reduction half-reaction: O3^{2-} -> O2

Theorems

Conservation of Mass
Conservation of Charge

Suitable Grade Level

High School (Grades 10-12)