NEET (UG) 2023 — Question 76

Write the unbalanced equation: $\ce{Cr2O7^{2-} + SO3^{2-} -> Cr^{3+} + SO4^{2-}}$. Identify half-reactions: Reduction: $\ce{Cr2O7^{2-} + 6e- -> 2Cr^{3+}}$; Oxidation: $\ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e-}$. Balance electrons: Multiply oxidation half by 3: $\ce{3SO3^{2-} + 3H2O -> 3SO4^{2-} + 6H+ + 6e-}$. Add half-reactions: $\ce{Cr2O7^{2-} + 3SO3^{2-} + 3H2O + 6e- -> 2Cr^{3+} + 3SO4^{2-} + 6H+ + 6e-}$. Cancel electrons: $\ce{Cr2O7^{2-} + 3SO3^{2-} + 3H2O -> 2Cr^{3+} + 3SO4^{2-} + 6H+}$. Balance oxygen: Left has 7+9+3=19 O, right has 12 O from $\ce{SO4^{2-}}$ plus 3 O from $\ce{H2O}$? Wait, the equation from step 4 already includes $\ce{H2O}$ and $\ce{H+}$. Let's redo: After adding, we have $\ce{Cr2O7^{2-} + 3SO3^{2-} + 3H2O -> 2Cr^{3+} + 3SO4^{2-} + 6H+}$. Count O: left 7+9+3=19, right 12 from $\ce{SO4^{2-}}$ and 3 from $\ce{H2O}$? Actually right has no $\ce{H2O}$ in this equation. Correct re-balance: The net ionic equation in acidic medium: $\ce{Cr2O7^{2-} + 3SO3^{2-} + 8H+ -> 2Cr^{3+} + 3SO4^{2-} + 4H2O}$. Check atoms: Left: Cr=2, O=7+9=16, H=8, S=3; Right: Cr=2, S=3, O=12+4=16, H=8. Balanced. Thus coefficients: $a=1, b=3, c=8$.