d- and f-Block Elements · Lesson 9 of 12
Permanganate Reactions in Different Media
“Derive medium-dependent reduction half-reactions, balance oxidations and calculate reacting amounts.”
• Choose the appropriate Mn(VII) reduction product from the stated medium and reaction. • Balance acidic permanganate reactions with several reductants. • Balance neutral or mildly alkaline reactions producing MnO₂. • Use electron stoichiometry in quantitative problems and explain kinetic effects.
The medium is part of the reaction
Purple permanganate may become nearly colourless Mn²⁺ in acid, brown MnO₂ in neutral or mildly alkaline conditions, or green manganate in strongly alkaline conditions with appropriate reductants. These products contain different manganese oxidation states and require different electron counts. Consequently there is no universal “permanganate factor” independent of the reaction. Specify the actual final state before calculating a mole ratio.
| Reduction product | Balanced half-reaction | Electrons per Mn |
|---|---|---|
| Mn²⁺, commonly in sufficiently acidic reactions | MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O | 5 |
| MnO₂, commonly neutral or mildly alkaline | MnO₄⁻+2H₂O+3e⁻→MnO₂+4OH⁻ | 3 |
| MnO₄²⁻, favoured in strongly alkaline suitable reactions | MnO₄⁻+e⁻→MnO₄²⁻ | 1 |
The acidic representation of the dioxide half is MnO₄⁻+4H⁺+3e⁻→MnO₂+2H₂O. Adding OH⁻ to both sides and cancelling water gives the basic form. The standard potentials about +1.52 V for Mn²⁺, +1.69 V for MnO₂ in its acidic convention and +0.56 V for manganate refer to precisely written half-reactions. They must not be interpreted as measurements at one identical neutral pH. Acid concentration changes the potential when H⁺ participates.
An electron-balancing method that scales
Problem
Derive the full ionic equation and explain every coefficient.
- 1.Mn changes +7 to +2 and gains five electrons. Write MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O after balancing O,H and charge.
- 2.Fe²⁺→Fe³⁺+e⁻ supplies one electron, so use five Fe²⁺ per MnO₄⁻.
- 3.Add to obtain MnO₄⁻+8H⁺+5Fe²⁺→Mn²⁺+4H₂O+5Fe³⁺. The charge is +17 on either side.
- 4.Potassium and sulfate counterions can be omitted from the net ionic equation if they remain spectators. The five-electron ratio belongs to reduction to Mn²⁺, not to every possible medium.
| Reductant and change | Balanced reaction in acid |
|---|---|
| I⁻ to I₂ | 2MnO₄⁻+16H⁺+10I⁻→2Mn²⁺+8H₂O+5I₂ |
| Oxalate to CO₂ | 2MnO₄⁻+16H⁺+5C₂O₄²⁻→2Mn²⁺+8H₂O+10CO₂ |
| H₂S to S | 2MnO₄⁻+6H⁺+5H₂S→2Mn²⁺+8H₂O+5S |
| Sulfite to sulfate | 2MnO₄⁻+6H⁺+5SO₃²⁻→2Mn²⁺+3H₂O+5SO₄²⁻ |
| SO₂ to sulfate | 2MnO₄⁻+2H₂O+5SO₂→2Mn²⁺+5SO₄²⁻+4H⁺ |
| Nitrite to nitrate | 2MnO₄⁻+6H⁺+5NO₂⁻→2Mn²⁺+5NO₃⁻+3H₂O |
Oxalate loses two electrons per C₂O₄²⁻: each carbon changes +3 to +4. For oxalic acid, the same chemistry can be written 2MnO₄⁻+6H⁺+5H₂C₂O₄→2Mn²⁺+8H₂O+10CO₂. The acid requirement differs because the neutral acid supplies its own hydrogen atoms. Likewise SO₂ and SO₃²⁻ representations are chemically linked but have different H⁺/H₂O coefficients. Balance the species that is actually written.
Problem
Why is the permanganate:oxalate ratio 2:5, and why may the reaction be warmed?
- 1.Reduction of two Mn(VII) to Mn(II) consumes ten electrons. Oxidation of five oxalates to ten CO₂ releases ten electrons, so the ratio is 2:5.
- 2.A favourable thermodynamic balance does not guarantee a fast initial reaction. Acidic oxalate oxidation is commonly warmed near 333 K in the described setting to obtain a suitable rate.
- 3.Mn²⁺ formed during reaction can accelerate the pathway, giving autocatalytic behaviour. The amount ratio is unchanged by warming or catalysis.
Neutral and mildly alkaline reactions form dioxide
| Oxidation | Balanced net reaction |
|---|---|
| Iodide to iodate | 2MnO₄⁻+H₂O+I⁻→2MnO₂+2OH⁻+IO₃⁻ |
| Thiosulfate to sulfate | 8MnO₄⁻+3S₂O₃²⁻+H₂O→8MnO₂+6SO₄²⁻+2OH⁻ |
| Mn²⁺ to MnO₂ | 2MnO₄⁻+3Mn²⁺+2H₂O→5MnO₂+4H⁺ |
Iodide changes from −1 to +5 in iodate, releasing six electrons. Two permanganates becoming MnO₂ consume six, explaining the 2:1 ratio. In thiosulfate the two sulfur atoms are not equivalent, but their total oxidation-state sum rises from +4 per S₂O₃²⁻ to +12 in two sulfates: eight electrons are released per thiosulfate. Three thiosulfates supply the twenty-four electrons needed by eight permanganates becoming dioxide.
The last equation combines Mn(VII) and Mn(II) to produce an intermediate state Mn(IV); it is comproportionation, the opposite pattern to disproportionation. Its formation of H⁺ also shows that reaction can change the medium unless buffering or other conditions compensate. Zinc sulfate or zinc oxide is reported to promote this oxidation in the described system; such kinetic assistance does not change the balanced manganese ratio.
Quantitative chemistry begins with the product
Problem
25.00 mL Fe²⁺ solution needs 18.60 mL of 0.02000 mol L⁻¹ permanganate in sufficient acid. Find [Fe²⁺].
- 1.Permanganate amount=0.02000 mol L⁻¹×0.01860 L=3.720×10⁻⁴ mol.
- 2.Use five Fe²⁺ per permanganate: n(Fe²⁺)=1.860×10⁻³ mol.
- 3.Divide by the sample volume, 0.02500 L: concentration=0.07440 mol L⁻¹. The digits follow the stated measured precision.
- 4.The calculation assumes the measured oxidant was consumed only by Fe²⁺ and reduced to Mn²⁺. A side reductant or an MnO₂ pathway would invalidate this simple interpretation.
Permanganate’s own colour can indicate an endpoint when a faint persistent excess appears, but its working solution needs standardisation rather than assuming every weighed sample gives an exact primary-standard concentration. Hydrochloric acid is unsuitable for the stated titration because chloride can be oxidised to chlorine, consuming oxidant. Nitric acid can introduce its own oxidising chemistry; the reaction medium must support the intended net equation. These considerations explain analytical selectivity, not just a memorised acid choice.
Oxidising applications include chemical analysis, organic synthesis and some bleaching or decolourising processes. The same substance need not react at the same rate with every possible reductant. Thermodynamics predicts that some reactions, including oxidation of water in a strongly acidic comparison, can be favourable while the unassisted pathway is slow. The observed behaviour requires both potential and kinetics.
Quick Quiz
Quiz
How many electrons are accepted when MnO₄⁻ becomes MnO₂?
Explanation: Mn changes +7 to +4.
In sufficient acid, how many Fe²⁺ react per MnO₄⁻?
Explanation: Five one-electron Fe oxidations supply the five electrons for Mn(II) formation.
Why is chloride-containing acid problematic in the stated titration?
Explanation: A competing oxidation changes the measured amount attributed to the analyte.
What is Mn(VII)+Mn(II)→Mn(IV) an example of?
Explanation: Two different initial states converge to an intermediate state.
What happens to the oxalate:permanganate mole ratio on warming the same reaction?
Explanation: Temperature can alter rate; the specified balanced reaction retains its electron ratio.
Practice Problems
- Balance acidic permanganate oxidation of sulfite.
- Derive the acidic equation for oxalic acid rather than oxalate.
- Calculate oxalate amount reacting with 15.00 mL of 0.02000 M permanganate in acid.
- Why does neutral iodide oxidation use two permanganates per iodide?
- Find the thiosulfate:permanganate ratio when sulfate and MnO₂ are products.
- A calculation uses a factor of five although brown MnO₂ forms. Explain the error.
- Balance acid oxidation of H₂S to elemental sulfur.
Oxidation is SO₃²⁻+H₂O→SO₄²⁻+2H⁺+2e⁻. Multiply by five and the Mn(VII)→Mn(II) half by two. Cancelling electrons, water and H⁺ gives 2MnO₄⁻+6H⁺+5SO₃²⁻→2Mn²⁺+3H₂O+5SO₄²⁻. Charge is −6 on each side.
Key Takeaways
• Permanganate can accept one, three or five electrons depending on its reduction product. • Medium and reductant determine which balanced reaction is appropriate. • Half-reactions prevent errors in H⁺, OH⁻, water and electron coefficients. • Acidic Fe²⁺ oxidation uses a 5:1 iron-to-permanganate ratio. • Oxalate oxidation gives CO₂ and can show temperature dependence and autocatalysis. • Neutral or mildly alkaline oxidations often yield MnO₂. • A side reaction can invalidate an otherwise correct titration calculation. • Potential and reaction rate provide different information.