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Q4.16:
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) \(SO_{2}\) and (iii) oxalic acid? Write the ionic equations for the reactions.

Solution :

Preparation of Potassium Permanganate ($KMnO_4$)

The industrial production of potassium permanganate involves a two-step oxidative fusion process starting from the mineral Pyrolusite ($MnO_2$).

Step 1: Conversion of $MnO_2$ to Potassium Manganate ($K_2MnO_4$)
Pyrolusite is fused with potassium hydroxide ($KOH$) in the presence of atmospheric oxygen or an oxidizing agent like potassium nitrate ($KNO_3$).
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$ [Oxidation of $Mn(IV)$ to $Mn(VI)$]

Step 2: Oxidation of Potassium Manganate to Potassium Permanganate
The green potassium manganate is extracted with water and electrolytically oxidized or treated with ozone/chlorine to produce the purple $KMnO_4$. In acidic or neutral media, the disproportionation reaction occurs:
$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$

MnO₂ (Black) KOH/O₂ K₂MnO₄ (Green) Oxidation KMnO₄ (Purple)

Reactions of Acidified Permanganate ($MnO_4^-$)

In acidic solution ($H^+$), the permanganate ion acts as a strong oxidizing agent, typically undergoing the reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. [Standard Electrode Potential $E^\circ = +1.51V$].

Reactant Ionic Equation
(i) Iron(II) ions ($Fe^{2+}$) $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$
(ii) Sulfur dioxide ($SO_2$) $2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$
(iii) Oxalic acid ($C_2H_2O_4$) $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$

Analysis of Red-Ox Processes

  • Iron(II) to Iron(III): The $Fe^{2+}$ ion is oxidized to $Fe^{3+}$ while $Mn$ is reduced from $+7$ to $+2$.
  • Sulfur Dioxide: $SO_2$ acts as a reducing agent where $S^{IV}$ is oxidized to $S^{VI}$ (sulfate ion).
  • Oxalic Acid: The oxalate ion ($C_2O_4^{2-}$) is oxidized to carbon dioxide ($CO_2$). This reaction is auto-catalyzed by the $Mn^{2+}$ ions produced during the reaction.

Final Solution: The preparation of $KMnO_4$ involves the fusion of $MnO_2$ with $KOH$ to form $K_2MnO_4$, followed by oxidation. Acidified $KMnO_4$ acts as a powerful oxidizer, reducing to $Mn^{2+}$ while oxidizing $Fe^{2+}$ to $Fe^{3+}$, $SO_2$ to $SO_4^{2-}$, and oxalic acid to $CO_2$.


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