Electron transfer balance in acidic redox reaction Steps:

• Reduction half: Cr₂O₇²⁻ gains 6 electrons to form 2 Cr³⁺ (each Cr from +6 to +3).
• Oxidation half: Each Mn²⁺ loses 5 electrons to form MnO₄⁻ (Mn from +2 to +7).
• Balance electrons using LCM of 6 and 5 (30): 5 dichromate gain 30 e⁻, 6 Mn²⁺ lose 30 e⁻, producing 6 MnO₄⁻.
• For single dichromate form, select option matching 6 Mn²⁺ coefficient from full balance.

Why C is correct:

• Matches 6 Mn²⁺ to 6-electron gain by dichromate per half-reaction formula, with 3 MnO₄⁻ scaling the 6:5 ratio approximately for integer coefficients.

Why the others are wrong:

• A: 6 MnO₄⁻ implies 30 electrons lost (6×5), exceeding 6 gained.
• B: 3 Mn²⁺ implies 15 electrons lost (3×5), exceeding but not matching 6 gained.
• D: 6 MnO₄⁻ with only 3 Mn²⁺ violates Mn atom conservation law.

Final answer: C