Which of the following is /are correct about redox reactio?
`MnO_(4)^(ϴ)+H^(ϴ)toMn^(+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 an 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_(2)O_(3)^(2-)` ions are oxidised to `HSO_(4)^(-)`

Answer

D

Explanation

a) `5e^(-)+MnO_(4)^(-)rarrMn^(2+)(n=5)`
`2S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)(n=2/2=1)`
Eq of `MnO_(4)^(-)-=` Eq of `S_2O_(3)^(-)`
`5xx` moles of `MnO_(4)^(-)-=1xx` moles of `S_2O_(3)^(2-)`
`:.1` mol of `MnO_(4)^(-)-=1xx` moles of `S_(2)O_3^(2-)`
`:.1 ` mol of `S_(2)O_(3)^(2-)=5` mol of `MnO_4^(-)`
b) pH change from 4 to 10 (acidic to strongly basic )
`e^(-)+MnO_(4)^(-)rarrMnO_(4)^(2-)(n=1)`
`S_(2)O_3^(-2)rarr2SO_(4)^(-2)+8e^(-)(n=8)`
Eq of `MnO_4^(-) =" Eq of " S_(2)O_3^(2-)`
`:.1` mol of `S_2O_(3)^(2-)=1/8 ` mol of `MnO_4^(-)`
Hence with change of pH from 4 to 10, will change the stiochiometry of reaction and also changes the product.
c) pH change 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