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The following reaction takes place in an acidic solution. MnO4–(aq) + Cl–(aq) --> Mn2+ + Cl2(g) (unbalanced).Do Each Step Individually**: Write the half reactions. Balance the equations for atoms (except O and H). Balance the equations for atoms O and H using H2O and H+. Balance the charge in the half reactions. Multiply each half reaction by the proper number to balance charges in the reaction. Add the equations and simplify to get a balanced equation.

Sagot :

Answer

2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O

Explanation

What is given:

The following reaction takes place in an acidic solution:

MnO₄⁻(aq) + Cl⁻(aq) → Mn²⁺(aq) + Cl₂(g) (unbalanced).

What to find:

To identify the half-reactions and balance the redox reaction.

Step-by-step solution:

Step 1: Identify the half-reactions.

The oxidation half-reaction is: Cl⁻(aq) → Cl₂(g)

The reduction half-reaction is: MnO₄⁻(aq) → Mn²⁺(aq)

Step 2: Balance the equations for atoms (except O and H).

Oxidation: 2Cl⁻(aq) → Cl₂(g)

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Step 3: Balance the equations for atoms O and H using H₂O and H⁺.

Oxidation: 2Cl⁻(aq) → Cl₂(g)

Reduction: MnO₄⁻(aq) + 8H⁺ → Mn²⁺(aq) + 4H₂O

Step 4: Balance the charge in the half-reactions.

Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻

Reduction: MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺(aq) + 4H₂O

Step 5: Multiply each half-reaction by the proper number to balance charges in the reaction.

Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ x 5

Reduction: MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺(aq) + 4H₂O x 2

Oxidation: 10Cl⁻(aq) → 5Cl₂(g) + 10e⁻

Reduction: 2MnO₄⁻(aq) + 16H⁺ + 10e⁻ → 2Mn²⁺(aq) + 8H₂O

Step 6: Add the equations and simplify to get a balanced equation.

2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ + 10e⁻ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O + 10e⁻

Simplifying the equation, we have:

2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O

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