Balancing the redox equation using half-reactions

Steps:

• Reduction half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (dichromate gains 6 electrons).
• Oxidation half-reaction: Zn → Zn²⁺ + 2e⁻ (zinc loses 2 electrons).
• Multiply oxidation by 3 to match electrons: 3Zn → 3Zn²⁺ + 6e⁻.
• Combine half-reactions: Cr₂O₇²⁻ + 3Zn + 14H⁺ → 2Cr³⁺ + 3Zn²⁺ + 7H₂O.

Why C is correct:

• It balances atoms, charges, and electrons (6 total transferred) per the half-reaction method.

Why the others are wrong:

• A: Only 1 Zn provides 2e⁻, but reduction requires 6e⁻; unbalanced electrons.
• B: 1 Zn cannot produce 3Zn²⁺; atoms unbalanced.
• D: 2Cr₂O₇²⁻ requires 12e⁻, but 3Zn provide only 6e⁻; electrons unbalanced.

Final answer: C