KMnO4 (purple) is reduced to K2MnO4 (green) by SO2-3 in basic medium. 1 mole of KMnO4 is reduced by

Question

KMnO4 (purple) is reduced to K2MnO4 (green) by SO2-3 in basic medium. 1 mole of KMnO4 is reduced by (A) 1 mole of SO2-3 (B) 2 moles of SO32- (C) 1.5 mole of SO32- (D) 0.5 mole of SO32-

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/d428b25b5e047878dd74ef2b260da58b-.jpeg" /> Multiply by change in oxidation number 2MO4- + SO32- → SO2-4 + 2MnO2-4 2MnO4- ≡ 1SO32- 1MnO4- ≡0.5SO32-

Q. $K M n O_{4}$ (purple) is reduced to $K_{2} M n O_{4}$ (green) by $S O_{3}^{2 -}$ in basic medium. 1 mole of $K M n O_{4}$ is reduced by

1 mole of $S O_{3}^{2 -}$

2 moles of $S O_{3}^{2 -}$

1.5 mole of $S O_{3}^{2 -}$

0.5 mole of $S O_{3}^{2 -}$

Multiply by change in oxidation number
$2 M O_{4}^{-} + S O_{3}^{2 -} \to S O_{4}^{2 -} + 2 M n O_{4}^{2 -}$
$2 M n O_{4}^{-} \equiv 1 S O_{3}^{2 -}$
$1 M n O_{4}^{-} \equiv 0.5 S O_{3}^{2 -}$

Q. KMnO4​ (purple) is reduced to K2​MnO4​ (green) by SO32−​ in basic medium. 1 mole of KMnO4​ is reduced by

1 mole of SO32−​

2 moles of SO32−​

1.5 mole of SO32−​

0.5 mole of SO32−​