Aqueous electrolysis of potassium manganate $K_{2}Mn{O}_4$in basic solution is used for the production of tens of thousands of tons of potassium permanganate $KMn{O}_4$annually.

(a) Write a balanced equation for the electrolysis reaction. (Water is reduced also.)

(b) How many moles of $Mn{O}_4$can be formed if$12$A flows through a tank of aqueous $Mn{O_4}^{2-}$for $96$h? $1$

The production of tens of thousands of tons of potassium permanganate $K_{2}Mn{O}_4$annually takes place annually by the aqueous electrolysis of it in the basic solution.

The equation for the electrolysis reaction is:

$Mn{{O_4}^{2-}}_{\left(aq\right)}+H_2{O}_{\left(l\right)}\to Mn{{O_4}^{-}}_{\left(aq\right)}+H_{2\left(g\right)}\left[basic\right]$

The equation can be balanced using the half-reaction method along with electrons as:

$$\begin{gathered} Mn{O_4}^{2-}\to Mn{O_4}^{-}+e^{-} \\ 2H_{2}O+2e^{-}\to H_2+2O{H}^{-} \end{gathered}$$

The charges to moles of electrons is determined by dividing the charge by Faraday’s constant,

$C=mol{e}^{-}\times F$

It corresponds to the amount of electricity being carried by one mole of electrons using the process of electrolysis.

(a)The electrolysis reaction is balanced further by combining and canceling the charges.

The chemical equation will be:

$2Mn{O_{4\left(aq\right)}}^{2-}+2H_2{O}_{\left(l\right)\to }2Mn{{O_4}^{-}}_{\left(aq\right)}+H_{2\left(g\right)}+2O{H^{-}}_{\left(aq\right)}$

(b) The number of moles formed can be determined as:

$$\begin{gathered} 12A=12\frac{C}{8} \\ 12\frac{C}{s}\times 96hr\times \frac{60min}{1hr}\times \frac{60sec}{1min}=4.1472\times {10}^{6}C \end{gathered}$$

Determining the moles,

$$\begin{gathered} mol{e}^{-}=\frac{C}{F} \\ =\frac{4.1472\times {10}^{6}C}{96500C/mol{E}^{-}} \\ =42.98mol{e}^{-} \end{gathered}$$

Converting the moles to that of $Mn{O_4}^{2-}$,

$$\begin{gathered} 1mol{e}^{-}=1molMn{O_4}^2\backslash \mathrm{hfill} \\ 42.98mol{e}^{-}=42.98molMn{O_4}^2 \end{gathered}$$