Potassium permanganate

A purple speck and a brown stain

A few crystals of potassium permanganate turn a glass of water deep purple. The same crystals leave a brown stain on skin. Why purple in water and brown on skin?

The answer is in the oxidation numbers. In water, manganese is $+7$ in the purple $\text{MnO}_4^-$ ion. On skin, the permanganate oxidises something and is reduced itself, to brown manganese dioxide, where manganese is $+4$. One ion, different products. Now we see how the product depends on the conditions.

Making potassium permanganate

The ore is pyrolusite, $\text{MnO}_2$. The making has two stages.

Stage 1. Fuse manganese dioxide with potassium hydroxide in air. Potassium nitrate can replace the air. Green potassium manganate forms:

Manganese goes from $+4$ to $+6$ and loses 2 electrons. Two manganese atoms lose 4, and one $\text{O}_2$ takes 4.

Stage 2. Oxidise the manganate ion, $\text{MnO}_4^{2-}$, to permanganate, $\text{MnO}_4^-$. There are two ways.

• Electrolysis in alkali. At the anode, $\text{MnO}_4^{2-} \rightarrow \text{MnO}_4^- + e^-$, and the green solution turns purple. At the cathode, hydrogen is given off. The overall change is $2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O} \rightarrow 2\text{KMnO}_4 + 2\text{KOH} + \text{H}_2$. Every manganese atom ends up as permanganate.
• Acid, or carbon dioxide. The manganate disproportionates. $3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$. Of every three manganese atoms, two become permanganate and one is lost as brown $\text{MnO}_2$. With carbon dioxide the equation is $3\text{K}_2\text{MnO}_4 + 2\text{CO}_2 \rightarrow 2\text{KMnO}_4 + \text{MnO}_2 + 2\text{K}_2\text{CO}_3$.

Electrolysis wastes nothing, so it is used in industry. A third method oxidises manganese(II) with peroxodisulfate, with silver ions as a catalyst: $2\text{Mn}^{2+} + 5\text{S}_2\text{O}_8^{2-} + 8\text{H}_2\text{O} \rightarrow 2\text{MnO}_4^- + 10\text{SO}_4^{2-} + 16\text{H}^+$.

The ion and its colour

Potassium permanganate forms dark purple crystals that look almost black. It dissolves in water to give a deep purple solution. On heating to about $513\ \text{K}$ it gives off oxygen:

The permanganate ion is a tetrahedron. Manganese is $+7$, so it has no d electrons: the ion has no unpaired electron and is diamagnetic. Its colour is charge transfer, as in dichromate.

The manganate ion is also a tetrahedron, but manganese is $+6$ with one d electron. So it has one unpaired electron. The spin-only formula gives the magnetic moment:

Here $n = 1$. Manganate is paramagnetic, and its solution is green.

Three media, three products

The product of reduction depends on how much $\text{H}^+$ is present.

• In acid: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, with $E^\circ = +1.51\ \text{V}$. Purple goes to almost colourless. Manganese falls by 5.
• In neutral or weakly alkaline solution: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$. Purple goes to a brown solid. Manganese falls by 3.
• In strongly alkaline solution: $\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}$. Purple goes to green. Manganese falls by 1.

So the n-factor of $\text{KMnO}_4$ is 5, 3 or 1. The equivalent mass is the molar mass divided by the n-factor, so it changes too.

Practise with new numbers

What it does in each medium

In acid. The potential of $+1.51\ \text{V}$ makes permanganate a very strong oxidising agent. You balanced its reactions with iron(II) and oxalate in Class 11. Three more:

• With hydrogen sulfide: $2\text{MnO}_4^- + 5\text{H}_2\text{S} + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{S} + 8\text{H}_2\text{O}$.
• With nitrite: $2\text{MnO}_4^- + 5\text{NO}_2^- + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{NO}_3^- + 3\text{H}_2\text{O}$.
• With chloride: $2\text{MnO}_4^- + 16\text{H}^+ + 10\text{Cl}^- \rightarrow 2\text{Mn}^{2+} + 5\text{Cl}_2 + 8\text{H}_2\text{O}$.

The third one is why hydrochloric acid is not used in a permanganate titration. The permanganate is strong enough to oxidise chloride, which dichromate is not. Count the electrons in the nitrite reaction. Each nitrite goes from $+3$ to $+5$ and loses 2. Five of them lose 10, and two permanganate ions take $2 \times 5 = 10$.

In neutral or weakly alkaline solution. Iodide is oxidised all the way to iodate:

Iodine goes from $-1$ to $+5$ and loses 6 electrons. Two permanganate ions take $2 \times 3 = 6$. The same medium is used in the test for a double bond in organic chemistry. Cold, dilute, alkaline permanganate is Baeyer's reagent. The purple disappears and brown $\text{MnO}_2$ appears when an alkene is present.

In strongly alkaline solution. Sulfite is oxidised, and the purple turns green:

With hot, acidic permanganate in organic chemistry, a side chain on a benzene ring is oxidised to a carboxylic acid group. Toluene gives benzoic acid:

The carbon of the methyl group goes from $-3$ to $+3$, so each toluene molecule loses 6 electrons. Five toluene molecules lose 30, and six permanganate ions take $6 \times 5 = 30$.

With concentrated sulfuric acid, potassium permanganate gives manganese(VII) oxide, the acidic oxide met earlier: $2\text{KMnO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Mn}_2\text{O}_7 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O}$. It is a green oil that explodes. Never add concentrated acid to the crystals.

Try it in Python

The program finds the oxidation number of manganese in four species. Then it counts the electrons that permanganate gains in each medium, and the equivalent mass.

Common mistakes

• Using 5 electrons in every medium. The n-factor is 5 in acid, 3 in neutral solution and 1 in strong alkali.
• Swapping purple and green. Permanganate is purple and manganate is green.
• Calling $\text{Mn}^{2+}$ purple. It is very pale pink, almost colourless in a dilute solution.
• Using hydrochloric acid for a permanganate titration. Permanganate oxidises chloride to chlorine, so the titre is too large.
• Leaving the electrons unmatched. Match them first. Five nitrite ions give 10 electrons, and two permanganate ions take 10. Then write the ratio.

Remember these

• $2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}$, then the manganate is oxidised to permanganate, best by electrolysis in alkali.
• Heated, $2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$.
• Permanganate is $d^0$ and diamagnetic. Manganate is $d^1$ with $\mu = 1.73\ \text{BM}$.
• Acid gives $\text{Mn}^{2+}$ (5 electrons). Neutral gives $\text{MnO}_2$ (3). Strong alkali gives $\text{MnO}_4^{2-}$ (1).
• Equivalent mass of $\text{KMnO}_4$ is $158/5$, $158/3$ or $158/1$ in $\text{g/eq}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why do crystals of potassium permanganate leave a brown stain on skin?

2. Why is electrolytic oxidation of manganate better than using acid?

3. How many unpaired electrons does the manganate ion have, and what is its spin-only magnetic moment?

4. What is the n-factor of $\text{KMnO}_4$ when it turns to $\text{MnO}_4^{2-}$ in strong alkali?

5. Purple permanganate turns green when sulfite and alkali are added. Which ion is green, and what has manganese done?