On oxidation of S2O32 - by MnO4- in neutral aqueous medium, the oxidation number of S would change from:

Question

On oxidation of S2O32 - by MnO4- in neutral aqueous medium, the oxidation number of S would change from: (A) +6 to -2 (B) -2 to +2 (C) +2 to +6 (D) +4 to +6

Tardigrade Admin · Accepted Answer

In neutral aqueous medium, thiosulphate is oxidised almost quantitatively to sulphate:

8 Mn O_{4}^{-} + 3 S_{2} O_{3}^{2 -} + H_{2} O \to 8 MnO_{2} + 6 SO_{4}^{2 -} + 2 OH^{-}

In

S_{2} O_{3}^{2 -}

,average oxidation number of Sulphur =

\frac{+ 6 - 2}{2} = + 2

Q. On oxidation of S2​O32−​ by MnO4−​ in neutral aqueous medium, the oxidation number of S would change from:

+6 to -2

-2 to +2

+2 to +6

+4 to +6

In neutral aqueous medium, thiosulphate is oxidised almost quantitatively to sulphate: 8MnO4−​+3S2​O32−​+H2​O→8MnO2​+6SO42−​+2OH− In S2​O32−​ ,average oxidation number of Sulphur = 2+6−2​=+2

Q. On oxidation of $S_{2} O_{3}^{2 -}$ by $M n O_{4}^{-}$ in neutral aqueous medium, the oxidation number of S would change from:

In neutral aqueous medium, thiosulphate is oxidised almost quantitatively to sulphate: $8 Mn O_{4}^{-} + 3 S_{2} O_{3}^{2 -} + H_{2} O \to 8 MnO_{2} + 6 SO_{4}^{2 -} + 2 OH^{-}$
In $S_{2} O_{3}^{2 -}$ ,average oxidation number of Sulphur = $\frac{+ 6 - 2}{2} = + 2$