Question

$KMnO_{4}$ (purple) is reduced to $K_{2}MnO_{4}$ (green) by $SO^{2-}_{3}$ in basic medium. 1 mole of $KMnO_{4}$ is reduced by

• A. 1 mole of $SO^{2-}_{3}$
• B. 2 moles of $SO_{3}^{2-}$
• C. 1.5 mole of $SO_{3}^{2-}$
• D. 0.5 mole of $SO_{3}^{2-}$

Answer 1

Answer: (D)

Multiply by change in oxidation number
$2MO_{4}^{-} + SO_{3}^{2-} \to SO^{2-}_{4} + 2MnO^{2-}_{4}$
$2MnO_{4}^{-} \equiv 1SO_{3}^{2-}$
$1MnO_{4}^{-} \equiv0.5SO_{3}^{2-}$