Balancing redox reactions in acidic medium Steps:

• Identify half-reactions: dichromate reduction Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (6e⁻ gained); Mn²⁺ oxidation Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻ (5e⁻ lost).
• Find least common multiple of electrons: 30 (multiply reduction by 5, oxidation by 6).
• Add half-reactions: 5Cr₂O₇²⁻ + 70H⁺ + 6Mn²⁺ + 24H₂O → 10Cr³⁺ + 35H₂O + 6MnO₄⁻ + 48H⁺.
• Simplify: 5Cr₂O₇²⁻ + 22H⁺ + 6Mn²⁺ → 10Cr³⁺ + 6MnO₄⁻ + 11H₂O. Why none of the options is correct:
• Options scale to one dichromate (6e⁻) but cannot match integer multiples of 5e⁻ per Mn, violating electron balance in redox law. Why the others are wrong:
• A: Requires 30e⁻ for 6 MnO₄⁻ but provides 6e⁻.
• B: Requires 15e⁻ for 3 MnO₄⁻ but provides 6e⁻.
• C: Requires 15e⁻ for 3 MnO₄⁻ but provides 6e⁻; also unbalanced Mn atoms (6 vs. 3).
• D: Requires 5e⁻ for 1 MnO₄⁻ but provides 6e⁻.