Potassium Permanganate: Preparation & Reactions

CHEImportant Compounds

Potassium permanganate is the standard strong oxidant whose product depends on the medium. Exam focus: preparation from pyrolusite, the acidic/neutral/alkaline products, balanced ionic oxidations, and the autocatalysis of the oxalic-acid reaction.

Acidic-medium reduction (5 electrons)

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \qquad E^\circ = +1.51\ \text{V}

Mn goes

+7 \rightarrow +2

; the strongest, most-used mode of

KMnO_4

Neutral/alkaline reductions

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \qquad MnO_4^- + e^- \rightarrow MnO_4^{2-}

neutral/faintly alkaline gives

MnO_2

(

+4

, 3 electrons); strongly alkaline gives

MnO_4^{2-}

(

+6

, 1 electron).

Oxidation of oxalate in acid

2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O

warm to about

333

K; autocatalysed by the

Mn^{2+}

formed.

Preparation from pyrolusite $MnO_2$ : fuse with $KOH$ and an oxidant ( $KNO_3$ or air) to give green $K_2MnO_4$ , then oxidise green manganate to purple permanganate, electrolytically or with $Cl_2$ or $O_3$ (in acid it also disproportionates: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ ).
Acidic medium ( $5e^-$ , $Mn$ goes $+7 \rightarrow +2$ ): $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ , $E^\circ = +1.51$ V.
Neutral or faintly alkaline ( $3e^-$ , $+7 \rightarrow +4$ ): $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$ ; strongly alkaline ( $1e^-$ , $+7 \rightarrow +6$ ): $MnO_4^- + e^- \rightarrow MnO_4^{2-}$ .
Oxidation of oxalic acid (warm to about $333$ K): $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$ .
Oxidation of iron(II): $MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$ ; of $SO_2$ : $2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$ .
The $KMnO_4$ –oxalic acid reaction is slow at first but speeds up because the $Mn^{2+}$ produced acts as a catalyst (autocatalysis); warming to about $333$ K also accelerates it.
$KMnO_4$ is a dark-purple crystalline solid; the intense colour is a charge-transfer band (O $\rightarrow$ Mn) since $Mn$ is $d^0$ in $MnO_4^-$ .
Structure: the $MnO_4^-$ ion is tetrahedral with $Mn$ in the $+7$ state and four equivalent Mn–O bonds with partial double-bond character.
Oxidation of iodide in acid: $2MnO_4^- + 16H^+ + 10I^- \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$ ; the purple colour is discharged as $Mn^{2+}$ forms.
$KMnO_4$ is widely used as a strong oxidant in volumetric analysis (permanganometry); it is self-indicating because the first excess drop gives a permanent pink colour.
Because the product (and the number of electrons) changes with pH, always identify the medium before writing the half-reaction: $5e^-$ in acid, $3e^-$ in neutral, $1e^-$ in strong alkali.

Where the marks go

Using the acidic 5-electron product ( $Mn^{2+}$ ) when the medium is neutral or alkaline; in neutral medium the product is $MnO_2$ ( $+4$ ), in strong alkali it is $MnO_4^{2-}$ ( $+6$ ).
Forgetting that $KMnO_4$ is self-indicating in titrations and adding a separate indicator unnecessarily.
Ascribing the purple colour to $d\text{-}d$ transitions; $Mn$ is $d^0$ in $MnO_4^-$ , so the colour is charge-transfer.
Mis-balancing the oxalate reaction — the correct stoichiometry is 2 $MnO_4^-$ : 5 $C_2O_4^{2-}$ : 16 $H^+$ .
Saying the oxalic-acid reaction is catalysed by added acid; the autocatalyst is the $Mn^{2+}$ produced (warming also helps), not $H^+$ .

How the board asks it

Conversionmedium-dependent reductions ( $5e^-$ acid, $3e^-$ neutral, $1e^-$ alkali)
Write the balanced ionic equation for the oxidation of $Fe^{2+}$ by acidified $KMnO_4$ , and state the change in oxidation state of manganese.
Give reasonsautocatalysis by $Mn^{2+}$
Account for the fact that the reaction between $KMnO_4$ and oxalic acid is slow initially but becomes rapid after some time, even at constant temperature.
Conversionpreparation from pyrolusite
How is $KMnO_4$ obtained from pyrolusite $MnO_2$ ? Give the equations for the conversion of $MnO_2$ to the green manganate and its subsequent oxidation to permanganate.
Numericalpermanganometry and the acidic half-reaction
$25.0\ \text{mL}$ of an oxalic acid solution required $20.0\ \text{mL}$ of $0.02\ \text{M}$ $KMnO_4$ for complete oxidation in acidic medium. Calculate the molarity of the oxalic acid solution.
Structure / namingtetrahedral $MnO_4^-$ ion
Draw the structure of the permanganate ion $MnO_4^-$ , indicating the oxidation state of $Mn$ and the nature of the four $Mn$ – $O$ bonds.
Give reasonscharge-transfer colour and $d^0$ configuration
Give a reason: the intense purple colour of the $MnO_4^-$ ion is not due to a $d$ – $d$ transition.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.