Question

The charge used for the oxidation of one mole ${\mathrm{Mn}}_3{\mathrm{O}}_4$ into ${{\mathrm{MnO}}_4}^{2-}$ in presence of alkaline medium is

Answer 1

Step 1: Calculating the oxidation state of Mn

We know that the loss of electrons is referred to as oxidation.

Let us assume the oxidation state of Manganese be $\mathrm{x}$.

$\therefore$The oxidation state of Manganese in ${\mathrm{Mn}}_3{\mathrm{O}}_4$is

$$\begin{gathered} 3\mathrm{x}+(4\times -2)=0 \\ \Rightarrow 3\mathrm{x}=8 \\ \Rightarrow \mathrm{x}=\frac{8}{3} \end{gathered}$$

$\therefore$Oxidation state of Manganese in ${{\mathrm{MnO}}_4}^{2-}$is

$$\begin{gathered} \mathrm{x}+(4\times -2)=-2 \\ \Rightarrow \mathrm{x}=-2+8 \\ \Rightarrow \mathrm{x}=+6 \end{gathered}$$

Step 2: Calculating the charge required

Thus total charge required for the oxidation of 1 mole of ${\mathrm{Mn}}_3{\mathrm{O}}_4$ into ${{\mathrm{MnO}}_4}^{2-}$is:

$$\begin{gathered} ∆=\left|\left(3\times 6\right)-\left(\frac{8}{3}\times 3\right)\right| \\ ∆=\left|18-8\right| \\ ∆=10 \end{gathered}$$

Therefore, the complete balanced reaction will be:

${\mathrm{Mn}}_3{\mathrm{O}}_4+16{\mathrm{OH}}^{-}\stackrel{\mathrm{Alkaline}\mathrm{medium}}{\to }3{{\mathrm{MnO}}_4}^{2-}+8{\mathrm{H}}_2\mathrm{O}+10{\mathrm{e}}^{-}$

Therefore, the charge used for the oxidation of one mole ${\mathrm{Mn}}_3{\mathrm{O}}_4$ into ${{\mathrm{MnO}}_4}^{2-}$ in presence of an alkaline medium is $10\mathrm{F}$.