Redox with Permanganate in Acid

Introduction

Permanganate is a useful oxidising reagent, but its manganese product depends on conditions. In the acidic examples on this page, MnO₄⁻ becomes Mn²⁺. Manganese falls from +7 to +2 and each permanganate ion accepts five electrons. That five-electron fact explains stoichiometric ratios with different reductants and provides a quick check on any proposed equation.

Core explanation

Assign manganese's oxidation state in MnO₄⁻: Mn + 4(−2) = −1, so Mn = +7. The reduction half in acid is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The eight protons and four waters balance H and O; five electrons balance charge. On the left, −1 + 8 − 5 = +2, equal to the product charge. No free electron remains after a compatible oxidation half is added.

With Fe²⁺ as reductant, Fe²⁺ → Fe³⁺ + e⁻ supplies one electron per iron ion. Five Fe²⁺ ions therefore supply the five electrons accepted by one MnO₄⁻. The net equation is MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. The ratio MnO₄⁻:Fe²⁺ is 1:5. If 0.001 mol permanganate reacts completely under these assumptions, it corresponds to 0.005 mol Fe²⁺, provided no competing reductant consumes it.

Oxalate gives a different ratio. The oxidation half is C₂O₄²⁻ → 2CO₂ + 2e⁻. In oxalate, each carbon is +3: 2C + 4(−2) = −2 gives 2C = +6. Each carbon becomes +4 in CO₂, so two carbon atoms release two electrons in total. The least common multiple of five and two is ten. Multiply the permanganate half by two and oxalate half by five, then combine: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.

Audit this oxalate equation. Mn 2, C 10, O 28 and H 16 occur on both sides. Left charge is 2(−1) + 5(−2) + 16(+1) = +4; right is 2(+2) = +4. Two permanganate ions accept ten electrons, and five oxalate ions release ten. The 2:5 permanganate-to-oxalate ratio differs from the 1:5 ratio with Fe²⁺ because an oxalate ion releases two electrons rather than one.

Do not transfer the Mn²⁺ half-reaction uncritically to neutral or basic medium. Permanganate may instead produce MnO₂ under other stated conditions, with manganese falling by three units and a different stoichiometric ratio. Likewise, concentrated or unusual conditions may involve other chemistry. The visible colour of a solution can help monitor a reaction, but the balanced equation must be based on specified species, pH and electron conservation rather than colour alone.

In titration, permanganate's strong colour can mark the endpoint under suitable conditions. The analytical calculation uses moles, not colour intensity: convert delivered volume and concentration to moles of MnO₄⁻, then use the balanced coefficient ratio to calculate reductant moles. Excess acid is required for the stated Mn²⁺ product. Experimental details affect practical accuracy, but the stoichiometric ratio comes from the net ionic equation.

Step-by-step reasoning

1. Confirm that the reaction is in acid and Mn²⁺ is the intended manganese product. 2. Write the five-electron permanganate reduction half-reaction. 3. Write and balance the reductant's oxidation half-reaction. 4. Find a common electron count, multiply complete halves and combine. 5. Audit atoms and charge before using coefficients for mole calculations.

Visual explanation

Draw Mn at +7 at the top of a five-step staircase ending at +2. Place five electron symbols along the descent. Beside it, draw five one-step Fe²⁺ → Fe³⁺ arrows or two five-step-per-pair oxalate columns to show how different reductants supply the same total ten or five electrons.

Real-world analogy

A device requiring five identical units of input can be supplied by five one-unit packs or by matching several two-unit packs to a common total of ten. Permanganate accepts five electrons per ion in this acidic reaction, while Fe²⁺ and oxalate supply different numbers per formula unit. The analogy concerns numerical matching, not a literal battery inside each ion.

Real-world example

Acidic permanganate can be used in volumetric analysis of iron(II) or oxalate-containing samples. A measured permanganate amount is translated to sample amount only after choosing the correct balanced reaction. Using the Fe²⁺ 1:5 ratio for oxalate would produce a major systematic error.

Why?

Why does acidic permanganate accept five electrons? Manganese's formal oxidation number changes from +7 in MnO₄⁻ to +2 in Mn²⁺. The five-unit decrease requires five electron charges in the reduction half-reaction, after oxygen and hydrogen are balanced with water and H⁺.

Common misconception

“One permanganate ion always reacts with five molecules of any reductant.” The number five is electrons accepted, not a universal reductant coefficient. Fe²⁺ supplies one electron per ion, whereas oxalate supplies two per ion, giving different mole ratios.

Worked example

How many moles of Fe²⁺ react with 0.0020 mol MnO₄⁻ in the stated acidic reaction? Use MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. The coefficient ratio is five moles Fe²⁺ per one mole MnO₄⁻. Therefore n(Fe²⁺) = 5 × 0.0020 = 0.010 mol. The acid must be adequate and other reductants absent for this calculation to describe the sample.

Quick check

1. How many electrons does one MnO₄⁻ ion accept when it becomes Mn²⁺ in acid? Answer: Five, because Mn falls from oxidation state +7 to +2.

Exam focus

State the acid medium and manganese product before using the five-electron factor. Derive the partner's electron count from its oxidation states. Include charge and atom audits for the final ionic equation.

Advanced insight

The acidic permanganate half-reaction is a formal net transformation. The pathway by which a particular reductant reacts can involve intermediates and may have a rate sensitive to temperature or catalysts. Stoichiometric electron accounting remains valid for the complete reaction when the assumed products are correct.

Summary

In the stated acidic reaction, MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Each permanganate accepts five electrons. That yields a 1:5 ratio with Fe²⁺ and a 2:5 ratio with oxalate, with the final ratio determined by the reductant's own electron loss.

Practice questions

1. What is manganese's oxidation state in MnO₄⁻? Answer: +7, since Mn + 4(−2) = −1. 2. What is the MnO₄⁻:C₂O₄²⁻ ratio for oxalate oxidation to CO₂ in acid? Answer: 2:5, because two permanganate ions accept ten electrons and five oxalate ions release ten. 3. Why is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O inappropriate as a universal permanganate half-reaction? Answer: The manganese product depends on medium and conditions; the five-electron Mn²⁺ result is for the specified acidic case.