Potassium Permanganate (KMnO4)

Potassium Permanganate — Preparation and Structure

Potassium permanganate ($\text{KMnO}_4$) is a deep purple crystalline solid and one of the most widely used oxidising agents. Manganese is in the +7 oxidation state.

It is prepared from the mineral pyrolusite ($\text{MnO}_2$) in two steps:

1. Fuse $\text{MnO}_2$ with KOH and an oxidising agent (air or KNO$_3$) to get green potassium manganate ($\text{K}_2\text{MnO}_4$, Mn +6): $$2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}$$
2. Oxidise the manganate (electrolytically or with Cl$_2$) to permanganate: $$3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$$

The permanganate ion ($\text{MnO}_4^-$) is tetrahedral. Its intense purple colour arises not from a d-d transition (Mn$^{7+}$ is $d^0$) but from a charge-transfer transition (electron movement from oxygen to manganese).

Oxidising Action in Acidic Medium

$\text{KMnO}_4$ is a powerful oxidising agent, and its behaviour depends on the medium. In acidic solution it is strongest, with the half-reaction ($E^\circ = +1.51$ V):

$$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

Note: 5 electrons gained; Mn goes +7 → +2 (purple → almost colourless/pale pink).

Standard oxidations by acidified $\text{KMnO}_4$:

• Iron(II) to iron(III): $$5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$$
• Oxalate (C$_2$O$_4^{2-}$) to CO$_2$: $$5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2$$
• Iodide to iodine: $$10\text{I}^- + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{I}_2$$

Key Point: In acid, permanganate gains 5 electrons (Mn$^{7+}$ → Mn$^{2+}$), so its equivalent mass is (molar mass)/5 = 158/5 = 31.6 — essential for permanganometric titrations.

Oxidising Action in Neutral/Alkaline Media

The number of electrons changes with the medium:

• Neutral or faintly alkaline medium (Mn$^{7+}$ → Mn$^{4+}$, 3 electrons): $$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$$ (e.g. oxidation of iodide to iodate, $\text{I}^- \rightarrow \text{IO}_3^-$. )
• Strongly alkaline medium (Mn$^{7+}$ → Mn$^{6+}$, 1 electron): $$\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}$$

Summary of electrons gained: acidic → 5 (Mn$^{2+}$); neutral → 3 (MnO$_2$); strongly alkaline → 1 (MnO$_4^{2-}$).

[NEET Important] The electron count by medium (5 / 3 / 1) is a guaranteed exam point. In acid the colour goes deep purple → colourless; in neutral it gives a brown MnO$_2$ precipitate.

Solved Examples

Example 1: Oxidation state of Mn in permanganate

What is the oxidation state of manganese in $\text{MnO}_4^-$?

Solution: $x + 4(-2) = -1 \Rightarrow x = +7$. Manganese is in the +7 state.

Example 2: Electrons gained in acid

How many electrons does $\text{MnO}_4^-$ gain in acidic medium?

Solution: From $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, it gains 5 electrons (Mn +7 → +2).

Example 3: Equivalent mass of KMnO4 in acid

The molar mass of $\text{KMnO}_4$ is 158 g mol$^{-1}$. Find its equivalent mass in acidic medium.

Solution: It gains 5 electrons in acid, so equivalent mass $= \dfrac{158}{5} = 31.6$ g equiv$^{-1}$.

Example 4: Reaction with oxalate

Write the ionic equation for the oxidation of oxalate by acidified permanganate.

Solution: $5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2$.

Example 5: Why MnO4- is coloured though Mn is d0

The Mn in $\text{MnO}_4^-$ is $d^0$, yet permanganate is intensely purple. Explain.

Solution: With $d^0$ there can be no d-d transition. The colour comes from a charge-transfer transition — an electron is momentarily transferred from oxygen (ligand) to manganese — which absorbs strongly in the visible region.

Example 6: Electrons in neutral medium

How many electrons does $\text{MnO}_4^-$ gain in neutral/faintly alkaline medium, and what is the product?

Solution: 3 electrons (Mn +7 → +4), giving MnO$_2$: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$.

Example 7: Reaction with Fe2+

Write the equation for the oxidation of Fe$^{2+}$ by acidified $\text{KMnO}_4$.

Solution: $5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$.

Example 8: Electrons in strongly alkaline medium

How many electrons does $\text{MnO}_4^-$ gain in strongly alkaline medium?

Solution: 1 electron (Mn +7 → +6): $\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}$ (green manganate).

Example 9: Colour change in acidic titration

What colour change marks the end point when acidified $\text{KMnO}_4$ is used as a self-indicator?

Solution: During titration the purple $\text{MnO}_4^-$ is decolourised (reduced to nearly colourless Mn$^{2+}$). At the end point one excess drop turns the solution permanent pink — so KMnO$_4$ is its own indicator.

Example 10: Oxidation of iodide in acid

Write the equation for the oxidation of iodide by acidified permanganate.

Solution: $10\text{I}^- + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{I}_2$.