Balancing Redox Reactions in Acid

Introduction

• Redox Reaction: Involves both reduction and oxidation.
• Oxidation States: Assign them to determine what is oxidized and what is reduced.

Assigning Oxidation States

• Dichromate Anion (Cr₂O₇²⁻):
  • Oxygen: -2
  • Total for 7 Oxygens: -14
  • Anion Total Charge: -2
  • Chromium Total: +12 (two Cr = +6 each)
• Chloride Anion (Cl⁻):
  • Oxidation State: -1
• Chromium Ion (Cr³⁺):
  • Oxidation State: +3
• Chlorine (Cl₂):
  • Oxidation State: 0

Determining Oxidation and Reduction

• Chlorine Oxidation: Cl⁻ (-1) to Cl₂ (0)
  • Increase in oxidation state = Oxidized.
• Chromium Reduction: Cr in Cr₂O₇²⁻ (+6) to Cr³⁺ (+3)
  • Decrease in oxidation state = Reduced.

Steps to Balance the Reaction

1. Write Half Reactions

   • Oxidation: Cl⁻ → Cl₂
   • Reduction: Cr₂O₇²⁻ → Cr³⁺

2. Balance Atoms (except O and H)

   • Oxidation: 2Cl⁻ → Cl₂
   • Reduction: Cr₂O₇²⁻ → 2Cr³⁺

3. Balance Oxygens (by adding H₂O)

   • Reduction: Cr₂O₇²⁻ + 7H₂O → 2Cr³⁺

4. Balance Hydrogens (by adding H⁺)

   • Reduction: 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

5. Balance Charges (by adding electrons)

   • Oxidation: 2Cl⁻ → Cl₂ + 2e⁻
   • Reduction: 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

6. Make Electrons Equal

   • Multiply oxidation by 3: 6Cl⁻ → 3Cl₂ + 6e⁻
   • Reduction remains unchanged.

7. Combine Half Reactions

   • Final Reaction: 6Cl⁻ + 14H⁺ + Cr₂O₇²⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

Verification

• Atoms:
  • Chlorine: 6 on both sides
  • Hydrogen: 14 on both sides
  • Chromium: 2 on both sides
  • Oxygen: 7 on both sides
• Charge Balance:
  • Reactants: +6 (14 positive - 8 negative)
  • Products: +6 (2 x 3 positive)