Respuesta :

dsdrajlin

Answer:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

Explanation:

Step 1: Write the unbalanced redox reaction

Fe(s) +  Cl₂(g) → Fe³⁺(aq) + Cl⁻(aq)

Step 2: Identify both half-reactions

Reduction: Cl₂(g) → Cl⁻(aq)

Oxidation: Fe(s) → Fe³⁺(aq)

Step 3: Perform the mass balance

Cl₂(g) → 2 Cl⁻(aq)

Fe(s) → Fe³⁺(aq)

Step 4: Perform the charge balance

2 e⁻ + Cl₂(g) → 2 Cl⁻(aq)

Fe(s) → Fe³⁺(aq) + 3 e⁻

Step 5: Multiply both half-reactions by numbers that make that the number of electrons gained and lost are equal

3 × [2 e⁻ + Cl₂(g) → 2 Cl⁻(aq)]

2 × [Fe(s) → Fe³⁺(aq) + 3 e⁻]

Step 6: Add both half-reactions and cancel what is repeated in both sides

6 e⁻ + 3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq) + 6 e⁻

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

samueladesida43

The coefficients needed to balance the redox reaction is as follows: 3Cl₂(g) + 2Fe(s) → 6Cl⁻(aq) + 2Fe³⁺(aq)

  • The unbalanced redox reaction is given as: Fe(s) + Cl₂(g) → Fe³⁺(aq) + Cl⁻(aq)

  • The balanced half-reactions in these equation are as follows:
  1. Reduction: Cl₂(g) → 2Cl⁻
  2. Oxidation: Fe(s) → Fe³⁺

  • Next, we balance the charges in both half-reactions as follows:
  1. 2e⁻ + Cl₂(g) → 2Cl⁻
  2. Fe(s) → Fe³⁺ + 3e⁻

  • Next, we multiply both half-reactions by coefficients that make number of electrons gained and lost equal.

  1. 3 × [2e⁻ + Cl₂ → 2 Cl⁻]
  2. 2 × [Fe → Fe³⁺ + 3e⁻]

  • Overall, we have:

  • 6e⁻ + 3Cl₂(g) + 2Fe(s) → 6Cl⁻(aq) + 2Fe³⁺(aq) + 6e⁻

  • We cross out both electrons on both sides to have:

  • 3Cl₂(g) + 2Fe(s) → 6Cl⁻(aq) + 2Fe³⁺(aq)

Learn more at: https://brainly.com/question/13293425?referrer=searchResults

Otras preguntas