Potassium permanganate

Deep purple crystals that dissolve to disinfect a wound, decolourise a flask of oxalic acid, and stain your fingers for a week — potassium permanganate, $\text{KMnO}_4$ , is chemistry's favourite demonstration of manganese pushed to its highest oxidation state, $+7$ , doing real oxidising work. Within the d- and f-Block Elements unit it is one of only two named compounds (alongside $\text{K}_2\text{Cr}_2\text{O}_7$ ) that the ISC syllabus asks you to know in full detail: its tetrahedral $\text{MnO}_4^-$ structure, its two-stage manufacture from pyrolusite ore, and — the part exam papers love — how its oxidising behaviour and reduction product change with the medium it reacts in. This article compresses that syllabus scope into one tight story: why $\text{MnO}_4^-$ is tetrahedral, how pyrolusite becomes permanganate through a green manganate intermediate, why acidic, neutral and alkaline media each stop the reduction at a different manganese oxidation state, and how that acidic-medium reduction underlies the redox titrations you will be asked to balance and calculate.

The core insight

$\text{KMnO}_4$ is manganese at its maximum $+7$ oxidation state acting as a strong oxidising agent whose reduction product — colourless $\text{Mn}^{2+}$ , brown $\text{MnO}_2$ , or green $\text{MnO}_4^{2-}$ — depends entirely on whether the medium is acidic, neutral/faintly alkaline, or strongly alkaline.

The core idea

Structure and shape. In the permanganate ion $\text{MnO}_4^-$ , manganese sits at oxidation state $+7$ , bonded to four oxygen atoms. The ion is tetrahedral ( $sp^3$ hybridisation at Mn), and resonance delocalises the single negative charge equally over all four oxygens, making the four Mn–O bonds identical in length and strength.

Extraction from pyrolusite ( $\text{MnO}_2$ ). Manufacture is a two-stage oxidation that climbs the manganese oxidation ladder from $+4$ to $+6$ to $+7$ . First, pyrolusite is fused with $\text{KOH}$ in the presence of air or an oxidant such as $\text{KNO}_3$ , oxidising Mn( $+4$ ) to green potassium manganate, $\text{K}_2\text{MnO}_4$ (Mn at $+6$ ):

2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \xrightarrow{\Delta} 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}

Second, the green manganate is oxidised further — by $\text{Cl}_2$ or $\text{O}_3$ , or electrolytically at the anode — to purple permanganate:

2\text{K}_2\text{MnO}_4 + \text{Cl}_2 \rightarrow 2\text{KMnO}_4 + 2\text{KCl}

Skipping this intermediate manganate stage, or muddling which oxidation state belongs to which colour, is the single commonest slip students make when asked to write out the manufacture.

Oxidising nature — the reduction product depends on the medium. In acidic medium (dilute $\text{H}_2\text{SO}_4$ ), $\text{MnO}_4^-$ is reduced all the way to colourless $\text{Mn}^{2+}$ , a five-electron change:

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

where: the eight $\text{H}^+$ come from the acidified medium, the five electrons are what a single $\text{MnO}_4^-$ ion gains as Mn falls from $+7$ to $+2$ , and the four $\text{H}_2\text{O}$ balance the oxygens released. In neutral or faintly alkaline medium there is not enough $\text{H}^+$ available to drive the reduction that far, so $\text{MnO}_4^-$ stops at $\text{MnO}_2$ (a brown precipitate, Mn at $+4$ ), a smaller three-electron change: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$ . In strongly alkaline medium the reduction is gentler still, stopping at green manganate, $\text{MnO}_4^{2-}$ (Mn at $+6$ ), a one-electron change. The pattern is simple once seen: more available $\text{H}^+$ pushes Mn further down the oxidation ladder.

Figure

Three reduction pathways of permanganate showing the manganese oxidation state falling further as the medium becomes more acidic, from plus six in strongly alkaline to plus four in neutral to plus two in acidic medium

Derivation ADeriving the mole ratio between KMnO4 and oxalic acid in an acidic titration

The goal is to convert two separately balanced half-reactions — one for permanganate being reduced, one for oxalate being oxidised — into a single whole-number mole ratio of $2:5$ that lets a titration reading be turned directly into moles of oxalic acid, without ever writing electrons in the final equation.

Step 1. Write the reduction half-reaction for $\text{MnO}_4^-$ in acidic medium, already established above: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ . Each permanganate ion accepts exactly five electrons as manganese falls from $+7$ to $+2$ .

Step 2. Write the oxidation half-reaction for oxalate, $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$ . Each carbon rises from $+3$ to $+4$ , and since there are two carbons per oxalate ion, each oxalate ion releases exactly two electrons.

Step 3. Equalise electrons transferred, since electrons lost must equal electrons gained. The least common multiple of $5$ and $2$ is $10$ , so multiply the Mn half-reaction by $2$ (giving $10e^-$ gained) and the oxalate half-reaction by $5$ (giving $10e^-$ lost). This fixes the coefficients at $2\ \text{MnO}_4^-$ to $5\ \text{C}_2\text{O}_4^{2-}$ .

Step 4. Add the two scaled half-reactions and cancel the $10e^-$ on each side, along with any spectator terms, to reach the full balanced ionic equation:

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

Step 5. Read off the mole ratio directly from the coefficients: every $2$ moles of $\text{KMnO}_4$ that react consume exactly $5$ moles of oxalic acid. This $2:5$ ratio is exactly what a titration calculation needs — once you know moles of $\text{KMnO}_4$ delivered from the burette, multiply by $5/2$ to get moles of oxalic acid, with no need to track electrons at all after this point.

Worked example

Problem: In a titration, $25.0\ \text{mL}$ of $0.0200\ \text{M}$ $\text{KMnO}_4$ solution, acidified with dilute $\text{H}_2\text{SO}_4$ and warmed to about $60^\circ\text{C}$ , exactly reacts with a solution of oxalic acid. Find the mass of oxalic acid ( $(\text{COOH})_2\cdot 2\text{H}_2\text{O}$ , molar mass $126\ \text{g/mol}$ ) that reacted.

Step 1. Find moles of $\text{KMnO}_4$ used: moles $= \text{molarity} \times \text{volume} = 0.0200\ \text{mol/L} \times 0.0250\ \text{L} = 5.00\times10^{-4}\ \text{mol}$ .

Step 2. Apply the $2:5$ mole ratio derived above (2 mol $\text{KMnO}_4$ reacts with 5 mol oxalic acid), so moles of oxalic acid $= 5.00\times10^{-4} \times \dfrac{5}{2} = 1.25\times10^{-3}\ \text{mol}$ .

Step 3. Convert to mass: mass $=$ moles $\times$ molar mass $= 1.25\times10^{-3}\ \text{mol} \times 126\ \text{g/mol} = 0.158\ \text{g}$ .

Answer: $0.158\ \text{g}$ of oxalic acid dihydrate reacted, and $\text{KMnO}_4$ (as $\text{MnO}_4^-$ ) is the oxidising agent while oxalic acid is the reducing agent, confirmed by the colour change from purple to colourless at the endpoint.

Common traps

Acidifying a $\text{KMnO}_4$ titration with dilute HCl instead of dilute $\text{H}_2\text{SO}_4$ — $\text{KMnO}_4$ oxidises $\text{Cl}^-$ to $\text{Cl}_2$ , consuming extra permanganate and giving a falsely high titre; $\text{H}_2\text{SO}_4$ is inert to oxidation by $\text{KMnO}_4$ .
Quoting the wrong reduction product for the stated medium: $\text{Mn}^{2+}$ (colourless) in acid, $\text{MnO}_2$ (brown precipitate) in neutral/faintly alkaline, and $\text{MnO}_4^{2-}$ (green) in strongly alkaline medium — mixing these up is the most common exam slip on this topic.
Skipping the intermediate manganate stage when writing the manufacture, or reversing which colour belongs to which oxidation state ( $\text{MnO}_2$ at $+4$ is not green; $\text{K}_2\text{MnO}_4$ at $+6$ is).
Adding an external indicator to a $\text{KMnO}_4$ titration — it needs none, since it is its own self-indicator, with the intense purple colour vanishing at the endpoint.

Ask the tutor: ask the AI tutor to walk through balancing the $\text{KMnO}_4$ – $\text{FeSO}_4$ or $\text{KMnO}_4$ –KI equations the same way, or to set up another titration mass calculation for practice.

Why does the same KMnO4 give a different coloured product depending on the medium?

The reduction stops at different manganese oxidation states because the amount of available $\text{H}^+$ controls how far the reduction can go. Strongly acidic medium supplies plenty of $\text{H}^+$ to protonate the released oxygens fully, letting Mn fall all the way from $+7$ to $+2$ . With less or no $\text{H}^+$ available, as in neutral or alkaline medium, the reduction stalls earlier, at $+4$ ( $\text{MnO}_2$ ) or even $+6$ (green manganate) in strongly alkaline conditions.

Why is KMnO4 called a self-indicator in titrations?

Because the reagent itself changes colour as the reaction proceeds, with no separate indicator dye needed. As long as excess oxalic acid or $\text{Fe}^{2+}$ is present, every drop of $\text{KMnO}_4$ added is immediately reduced and decolourised. Once all the reducing agent is used up, the very next drop of $\text{KMnO}_4$ survives unreacted, tinting the solution a faint permanent pink — that is the endpoint, spotted directly from $\text{KMnO}_4$ 's own strong purple colour.

Do I need to memorise the full balanced equation with oxalic acid, or just the mole ratio?

For calculations, the $2:5$ mole ratio between $\text{KMnO}_4$ and oxalic acid is what you actually use, so memorising that number saves time. But examiners often ask you to derive or write the full balanced ionic equation from the two half-reactions, so you should be comfortable doing Steps 1 to 4 of the derivation above from scratch rather than only recalling the final ratio.

Potassium permanganate

The core idea

Worked example

Common traps

Before this

Where this leads