Question

How many electrons are involved in the oxidation of Potassium Permangnate $\left({\mathrm{KMnO}}_4\right)$ in the basic medium?

Answer 1

1. Potassium Permanganate $\left({\mathrm{KMnO}}_4\right)$ in a strongly alkaline solution oxidises and takes up one of its electrons.
2. Reaction involved:
3. $\underset{\mathrm{Mangnate}\mathrm{ion}}{{{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)}+\underset{\mathrm{electron}}{{\mathrm{e}}^{-}}\to \underset{\mathrm{Mangnate}\mathrm{Ion}}{{{\mathrm{MnO}}_4}^{2-}\left(\mathrm{aq}\right)}$
4. Oxidation State of ${\mathrm{KMnO}}_4$, as oxygen has -2 negative charge and there have 4 oxygen atoms, the total charge in the compound is -1 so oxidation state of ${\mathrm{KMnO}}_4$ will be: $\mathrm{x}-8=-1,\mathrm{x}=7$, so oxidation state of Magnate ion is 7.
5. After the oxidation, it takes up one electron so the oxidation state after taking one electron is:$\mathrm{x}-8=-2,\mathrm{x}=6$, So oxidation state after oxidation is 7.
6. Change in Oxidation Number= $7-6=+1$
7. Potassium permanganate takes only one electron, thus it is a very much weaker oxidizing agent in a basic medium.

Therefore, one electron is involved in Potassium Permanganate $\left({\mathrm{KMnO}}_4\right)$ is one electron.