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KMnO4 can be prepared from K2MnO4 as per...

`KMnO_4` can be prepared from `K_2MnO_4` as per the reaction `3MnO_4^(2-)+2H_2O

A

`SO_2`

B

`CO_2`

C

KOH

D

HCl

Text Solution

Verified by Experts

`SO_2` and HCl are reducing agents. If these two gases are passed through the solution, they will reduce `MnO_4^(2-)` ions. On the other hand, `CO_2` is neither a reducing agent, nor an oxidising agent. When `CO_2` is passed through the solution, it forms `H_2CO_3` , which then reacts with `OH^-` ions produced in the given reaction. This causes the equilibrium to shift to the right, thereby making the reaction go to completion.
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