KMnO(4) can be prepared from K(2)MnO(4) ...

`KMnO_(4)` can be prepared from `K_(2)MnO_(4)` as per the reaction :
`3MnO_(4)^(2-)+2H_(2)O hArr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)`
The reaction can go to completion by removing `OH^(-)` ions by adding -

`SO_(2)`

HCl

KOH

`CO_(2)`

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The correct Answer is:

HCl and `SO_(2)` act as reducing agent. Both can reduce the `MnO_(4)^(-)` which is formed in the given reaction. However `CO_(2)` is neither oxidant nor reductant. Moreover it gives an alkaline medium driving the reaction in the forward direction.

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`KMnO_(4)` can be prepared from `K_(2)MnO_(4)` as per the reaction :
`3MnO_(4)^(2-)+2H_(2)O hArr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)`
The reaction can go to completion by removing `OH^(-)` ions by adding -

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CHHAYA PUBLICATION-D - AND F - BLOCK ELEMENTS -ENTRANCE QUESTION BANK

`KMnO_(4)` can be prepared from `K_(2)MnO_(4)` as per the reaction : `3MnO_(4)^(2-)+2H_(2)O hArr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)` The reaction can go to completion by removing `OH^(-)` ions by adding -

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CHHAYA PUBLICATION-D - AND F - BLOCK ELEMENTS -ENTRANCE QUESTION BANK

`KMnO_(4)` can be prepared from `K_(2)MnO_(4)` as per the reaction :
`3MnO_(4)^(2-)+2H_(2)O hArr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)`
The reaction can go to completion by removing `OH^(-)` ions by adding -