Balancing Redox Reactions

Balancing redox reactions by the ion-electron method requires equalising the electrons lost and gained between the two half-reactions. In the reduction half-reaction, chromium goes from +6 in Cr_2O_7^2- to +3 in Cr^3+; since there are two chromium atoms, each gaining 3 electrons, the dichromate ion accepts a total of 6 electrons. In the oxidation half-reaction, each Fe^2+ loses one electron to become Fe^3+. To balance the 6 electrons accepted by one dichromate, exactly 6 moles of Fe^2+ must be oxidised. The balanced equation is Cr_2O_7^2- + 6Fe^2+ + 14H^+ → 2Cr^3+ + 6Fe^3+ + 7H_2O. Option 3 wrongly counts electrons for a single chromium atom. Option 4 has no electron-balance basis. Option 2 ignores the dimeric nature of dichromate. This stoichiometry underlies the classic dichromate titration of iron in volumetric analysis. Carefully relating the data to the governing principle ensures the reasoning remains valid even when the numbers or species in the question are changed. A common JEE pitfall is to ignore the role of ion-electron method, yet it is exactly this factor that distinguishes the correct answer from the tempting alternatives. Plausibility check: total charge on the left, (-2) + 6(+2) + 14(+1) = +24, equals the right side, 2(+3) + 6(+3) = +24, confirming the balance.