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

Options

A. `SO_2`
B. `CO_2`
C. KOH
D. HCl

Explanation

`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.