Question

Question #fa0cb

Answer 1

Oxidation half-reaction: H₂C₂O₄ + 2H⁺ → 2CO₂ + 4H⁺ + 2e⁻
Reduction half-reaction: CrO₄²⁻ + 8H⁺ + 3e⁻ → Cr³⁺ + 4H₂O
Balanced equation: 3H₂C₂O₄ + 2K₂CrO₄ + 10HCl → 6CO₂ +2CrCl₃ + 4KCl + 8H₂O

Here's how you do get the answers by the ion electron method.

Step 1. Write the net ionic equation

Omit all spectator ions (K⁺ and Cl⁻). Also omit H⁺, OH⁻, and H₂O (they come in automatically during the balancing procedure).

H₂C₂O₄ + CrO₄²⁻ → Cr³⁺+ CO₂

Step 2. Split into half-reactions
H₂C₂O₄ → CO₂
CrO₄²⁻ → Cr³⁺

Step 3. Balance atoms other than H and O
H₂C₂O₄ → 2CO₂
CrO₄²⁻ → Cr³⁺

Step 4. Balance O
H₂C₂O₄ → 2CO₂
CrO₄²⁻ → Cr³⁺ + 4H₂O

Step 5. Balance H
H₂C₂O₄ → 2CO₂ + 2H⁺
CrO₄²⁻ + 8H⁺ → Cr³⁺ + 4H₂O

Step 6. Balance charge
H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻
CrO₄²⁻ + 8H⁺ + 3e⁻ → Cr³⁺ + 4H₂O

Step 7. Equalize electrons transferred
3×[H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻]
2×[CrO₄²⁻ + 8H⁺ + 3e⁻ → Cr³⁺ + 4H₂O]

Step 8. Add half-reactions
3H₂C₂O₄ + 2CrO₄²⁻ + 10H⁺ → 6CO₂ + 2Cr³⁺ + 8H₂O

Step 9. Re-insert the spectator ions
3H₂C₂O₄ + 2K₂CrO₄ + 10HCl → 6CO₂ + 2CrCl₃ + 4KCl + 8H₂O