Which of the following is/are correct about the redox reation?
`MnO_(4)^(-) +S_(2)O_(3)^(2-) +H^(o+) rarr Mn^(2+) +S_(4)O_(6)^(2-)`

Options

A. 1 mol of `S_(2)O_(3)^(2-)` is oxidised by 8 mol of `MnO_(4)^(-)`
B. The above redox reaction with the change of pH from 4 to 10 will have and effect on the stiochiometry of the reaction
C. Change of pH from 4 to 7 will change the nature of the product
D. At `pH = 7, S_(20O_(3)^(2-)` ions are oxidised to `HSO_(4)^(-)`

Answer

B, C, D

Explanation

(a) `5e^(-) + MnO_(4)^(-) rarr Mn^(2+) (n= 5)`
`2S_(2)O_(3)^(-) rarr S_(4)O_(6)^(2-) +2e^(-) (n = (2)/(2) = 1)`
Eq of `MnO_(4)^(-) -=` Eq of `S_(2)O_(3)^(-)`
`5xx `moles of `MnO_(4)^(-) -= 1 xx` moles of `S_(2)O_(3)^(2-)`
`:. 1` mol of `S_(2)O_(3)^(2-) = 5` mol of `MnO_(4)^(-)`
(b) pH changes from 4 to 10 (acidic to strongly basix)
`e^(-) + MnO_(4)^(-) rarr MnO_(4)^(2-) (n = 1)`
`S_(2)O_(3)^(2-) rarr 2SO_(4)^(2-) +8e^(-) (n = 8)`
Eq of `MnO_(4)^(-) =` Eq of `S_(2)O_(3)^(-)`
`:. 1` mol of `S_(2)O_(3)^(2-) = (1)/(8)` mol of `MnO_(4)^(-)`
Hence with change of pH from 4 to 10, will change the stiochiometry of reaction and alo changes the product.
(c) pH changes from 4 to 7 (acidic to neutral medium)
`3e^(+) +MnO_(4)^(-) rarr MnO_(2) (n = 3)`
`S_(2)O_(3)^(2-) rarr 2HSO_(4)^(-) +8e^(-) (n = 8)`
Hence it will also effect the stoichiometry of reaction and nature of product.
(d) At `pH = 7, S_(2)O_(3)^(2-)` is oxidised to `HSO_(4)^(-)` ion