Q129P - Chemistry: Molecular Nature Of Matter And Change

Question

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? $1$

Step-by-Step Solution

Answer

(a) The balanced chemical equation for the electrolysis reaction is 

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

(b) The number of moles formed if A flows through the tank of the solution for the time period of 96 h is 42.98 mol.

 

Step 1: Given information

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.

Step 2: The definitions of electrolysis reaction, half-reaction method and Faraday’s constant

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.

Step 3: The balanced chemical equation depicting the electrolysis reaction

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

The chemical equation will be:

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

Step 4: The number of moles of potassium manganate formed

(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,

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

Step 5: The moles in potassium manganate ions

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

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