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Which of the following is /are correct about redox reactio? MnO_(4)^(ϴ)+H^(ϴ)toMn^(+2)+S_(4)O_(6)^(-2) |
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Answer» 1mol of `S_(2)O_(3)^(2-)`is oxidised by 8 mol of `MnO_4^(-)` `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 |
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