Question

Question: A good method of preparing pure oxygen on a small scale is the decomposition of${\mathbf{\text{KMnO}}}_{\mathbf{\text{4}}}$in a vacuum above $\mathbf{215}\mathbf{°}\mathit{C}$:

${\mathbf{\text{2KMnO}}}_{\mathbf{\text{4}}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{K}}}_{\mathbf{\text{2}}}{\mathbf{\text{MnO}}}_{\mathbf{\text{4}}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

Assign an oxidation number to each atom and verify that the total number of electrons lost is equal to the total number gained.

Short answer

The oxidation number of each atom is;

$\text{2}\stackrel{+1}{\text{K}}\stackrel{+7}{\text{Mn}}{\stackrel{-8}{\text{O}}}_{\text{4}}\left(s\right)\to {\stackrel{+2}{\text{K}}}_{\text{2}}\stackrel{+6}{\text{Mn}}{\stackrel{-8}{\text{O}}}_{\text{4}}\left(s\right)+\stackrel{+4}{\text{Mn}}{\stackrel{-2}{\text{O}}}_{\text{2}}\left(s\right)+{\stackrel{0}{\text{O}}}_{\text{2}}\left(g\right)$

As seen, the total number of electrons lost equals the total number gained.

Step-by-step solution

The oxidation state.

The oxidation state is the electron number that is lost or accepted by an atom during the formation of a chemical bond.

The decomposition of potassium permanganate.

The decomposition reaction of potassium permanganate is of great significance because of the production of pure oxygen gas.

Potassium permanganate is also an important chemical that is widely used in the treatment of well water for removing hydrogen sulfide and iron.

The oxidation number of each atom.

The balanced reaction goes as;

$\text{2}\stackrel{+1}{\text{K}}\stackrel{+7}{\text{Mn}}{\stackrel{-8}{\text{O}}}_{\text{4}}\left(s\right)\to {\stackrel{+2}{\text{K}}}_{\text{2}}\stackrel{+6}{\text{Mn}}{\stackrel{-8}{\text{O}}}_{\text{4}}\left(s\right)+\stackrel{+4}{\text{Mn}}{\stackrel{-2}{\text{O}}}_{\text{2}}\left(s\right)+{\stackrel{0}{\text{O}}}_{\text{2}}\left(g\right)$

For,

$$\begin{gathered} {\text{KMnO}}_{\text{4}} \\ +1+x+4\left(-2\right)=0 \\ x=8-1 \\ x=+7 \end{gathered}$$

The charge on manganese is +7.

In

$$\begin{gathered} {\text{K}}_{\text{2}}{\text{MnO}}_{\text{4}} \\ 2\left(+1\right)+x+4(-2)=0 \\ x=8-2 \\ x=+6 \end{gathered}$$

Here, the charge on manganese is +6.

$$\begin{gathered} {\text{MnO}}_{\text{2}} \\ x+2(-2)=0 \\ x=+4 \end{gathered}$$

Here, the charge on the manganese ion is +4.

In all the molecules, the total number of electrons lost equals the total number gained.