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InterviewSolution
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Which of the following is/are correct about the redox reaction? `MnO_(4)^(ɵ)+S_(2)O_(3)^(2-)+H^(o+)toMn^(+2)+S_(4)O_(6)^(2-)`A. 1 " mol of "`S_(2)O_(3)^(2-)` is oxidised by 8 " mol of "`MnO_(4)^(ɵ)`B. The above redox reaction with the change of pH from 4 to 10 will have an effect on the stoichiometry of the reaction.C. Change of pH form 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)^(ɵ)` |
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Answer» Correct Answer - B::C::D (a). `5e^(-)+MnO_(4)^(ɵ)toMn^(2+)(n=5)` `2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-)(n=(2)/(2)=1)` " Eq of "`MnO_(4)^(ɵ)-=" Eq of "S_(2)O_(3)^(2-)` `5xx` moles of `MnO_(4)^(ɵ)-=1xx` moles of `S_(2)O_(3)^(2-)` `therefore` 1 " mol of "`S_(2)O_(3)^(2-)=5 " mol of "MnO_(4)^(ɵ)` (b). pH changes from 4 to 10 (acidic to strongly basic) `e^(-)+MnO_(4)^(ɵ)toMnO_(4)^(2-)(n=1)` `S_(2)O_(3)^(2-)to2SO_(4)^(2-)+8e^(-)(n=8)` " Eq of "`MnO_(4)^(ɵ)=" Eq of "S_(2)O_(3)^(2-)` `therefore` 1 " mol of "`S_(2)O_(3)^(2-)=(1)/(8)" mol of "MnO_(4)^(ɵ)` Hence with change of pH from 4 to 10, will change the stoichiometry of reaction and also changes the product. (c). pH changes from 4 to 7 (acidic to neutral medium) `3e^(-)+MnO_(4)^(ɵ)toMnO_(2)(n=3)` `S_(2)O_(3)^(2-)to2HSO_(4)^(ɵ)+8e^(-)(n=8)` Hence it will also effect the stoichiometry of reaction and natural of product. (d). `At pH=7,S_(2)O_(3)^(2-)` is oxidised to `HSO_(4)^(ɵ)` ion. |
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