Answer:
14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 e⁻ → 2 Cr⁺³ + 7 H₂O(l)
Step-by-step explanation:
Let's consider the half-reaction for the reduction of dichromate ion Cr₂O₇²⁻ to chromium ion Cr⁺³ in acidic aqueous solution.
Cr₂O₇²⁻(aq) → Cr⁺³
Step 1: Perform the mass balance adding H₂O(l) and H⁺(aq) where appropriate.
14 H⁺(aq) + Cr₂O₇²⁻(aq) → 2 Cr⁺³ + 7 H₂O(l)
Step 2: Perform the charge balance adding electrons where appropriate.
14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 e⁻ → 2 Cr⁺³ + 7 H₂O(l)