Self-indicator titration: KMnO4 (purple) → Mn²⁺ (colourless) in acidic medium (dilute H2SO4). Half-reaction: MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O. Oxalate: C2O4²⁻ → 2CO2 + 2e⁻. Heat to 60–70 °C to initiate the auto-catalysed reaction. End-point: first persistent pink colour.
-- NCERT Chemistry Lab Manual Class 12, Unit 6, p. 40Titrimetric Redox
8 MCQs1 revision card9-step worked example
Source: NCERT Practical and Analytical ChemistryPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 27 Sep 2026
Titrimetric Redox, explained for NEET
The trap that costs marks: In KMnO₄ vs oxalic acid titrations, aspirants frequently overshoot the end-point — they see a pink flash, swirl, watch it fade, and add more titrant. That extra drop converts a correct concordant reading into a discordant one, and the calculated molarity drifts high.
The reaction. In acidic medium (dilute H₂SO₄), permanganate oxidises oxalic acid:
2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → 2 MnSO₄ + K₂SO₄ + 10 CO₂ + 8 H₂O
KMnO₄ is self-indicating: purple MnO₄⁻ reduces to colourless Mn²⁺. The end-point is the first permanent light pink colour, given by a slight excess of permanganate once all the oxalate is used up (NCERT Chemistry Lab Manual Class 12, Unit 6, pages 41–43).
Normality approach. The equivalent factor (n-factor) of KMnO₄ in acidic medium is 5 (Mn goes from +7 to +2). For oxalic acid, n-factor is 2 (C goes from +3 to +4, two carbon atoms). The normality equation applies:
N₁V₁ = N₂V₂
where subscripts 1 and 2 refer to KMnO₄ and oxalic acid respectively (or vice versa — consistency matters).
Why heating matters. The reaction is slow at room temperature. The flask is heated to 60–70 °C before titration begins. Overheating (>70 °C) decomposes oxalic acid, giving falsely low titre values.
Watch-out: A transient pink that vanishes on swirling is NOT the end-point. Only a persistent colour change (≥30 s) signals equivalence. Overshooting even by one drop inflates the recorded KMnO₄ volume. When KMnO₄ is being standardised against oxalic acid, that volume is in the denominator, so the calculated normality of KMnO₄ comes out LOWER than its true value; if instead the oxalic acid were the unknown, the same overshoot would make ITS calculated normality too HIGH.
Can you answer these Titrimetric Redox MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the titration of oxalic acid with KMnO₄ in acidic medium, what is the n-factor (equivalent factor) of KMnO₄?
Show answer and why every option is right or wrong
Answer: A. A is correct. Mn in KMnO₄ is reduced from +7 to +2 in acidic medium, a change of 5 electrons per Mn atom, so the n-factor is 5.
Why B is wrong: B (n-factor = 3) — no standard redox change of Mn gives 3; this conflates with Cr₂O₇²⁻ (Cr: +6→+3 = 3) or phosphoric acid's basicity.
Why C is wrong: C (n-factor = 2) — this is the n-factor of oxalic acid (each molecule loses 2 electrons total per carbon × 2 carbons... actually oxalic acid n-factor is 2 because C goes +3→+4 for each of 2 C atoms). Confusing the two reagents' n-factors is the root error.
Why D is wrong: D (n-factor = 7) — this is the oxidation state of Mn in KMnO₄, not the change in oxidation state. The n-factor equals the change (7−2 = 5), not the initial state.
MCQ 2Easy RecallPractice
Why is no separate indicator needed in the titration of oxalic acid against KMnO₄?
Show answer and why every option is right or wrong
Answer: B. B is correct. KMnO₄ (purple) is reduced to Mn²⁺ (nearly colourless) until all oxalic acid is consumed. The first excess drop of KMnO₄ imparts a persistent pale pink — this IS the end-point signal, making it self-indicating (NCERT Chemistry Lab Manual Class 12, Unit 6, page 41).
Why A is wrong: A — oxalic acid being colourless does not help detect the end-point; detection requires a colour appearing, not disappearing.
Why C is wrong: C — CO₂ evolution occurs throughout the reaction, not specifically at the end-point, so it cannot signal equivalence.
Why D is wrong: D — MnSO₄ (Mn²⁺) is very pale pink/colourless in dilute solution, not brown. Brown MnO₂ forms only in neutral/basic medium.
MCQ 3Easy RecallPractice
The balanced equation for the reaction between KMnO₄ and oxalic acid in acidic medium is: 2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → products. What is the molar ratio of KMnO₄ to H₂C₂O₄?
Show answer and why every option is right or wrong
Answer: D. D is correct. From the balanced equation, 2 mol KMnO₄ reacts with 5 mol H₂C₂O₄, giving the ratio 2 : 5.
Why A is wrong: A — 1 : 5 would require only 1 mol KMnO₄ for 5 mol oxalic acid; the balanced equation explicitly shows coefficient 2 for KMnO₄.
Why B is wrong: B — 1 : 2 does not satisfy electron balance: 5 electrons gained per Mn × 2 Mn = 10 electrons; oxalic acid loses 2 electrons per molecule, requiring 5 molecules to balance.
Why C is wrong: C — 5 : 2 inverts the ratio (KMnO₄ : oxalic acid should be 2 : 5, not 5 : 2). Swapping reactant positions is a common careless error.
MCQ 4Direct ApplicationPractice
During KMnO₄ vs oxalic acid titration, the solution is heated to 60–70 °C before adding KMnO₄. What happens if the temperature exceeds 70 °C?
Show answer and why every option is right or wrong
Answer: B. B is correct. Above 70 °C, oxalic acid (H₂C₂O₄) thermally decomposes, reducing the amount available to react with KMnO₄. Less KMnO₄ is needed to reach end-point, so the recorded titre is lower than the true value.
Why A is wrong: A — KMnO₄ can decompose to MnO₂ in strongly alkaline or neutral medium on prolonged heating, but in acidic titration conditions at 70–80 °C the primary issue is oxalic acid decomposition, not KMnO₄ decomposition.
Why C is wrong: C — concentrated H₂SO₄ has a boiling point of 337 °C; dilute H₂SO₄ will not evaporate significantly at 70–80 °C. The medium remains acidic.
Why D is wrong: D — while the reaction does speed up, the practical error is not 'too fast to see colour' but rather loss of analyte (oxalic acid) by decomposition, which systematically biases the titre downward.
MCQ 5Direct ApplicationPractice
A student standardises a KMnO₄ solution against standard 0.1 N oxalic acid, with the KMnO₄ in the burette. The student overshoots the end-point by one drop. How does this affect the calculated normality of the KMnO₄?
Show answer and why every option is right or wrong
Answer: B. B is correct. At the equivalence point N(KMnO₄) × V(KMnO₄) = N(oxalic) × V(oxalic), so N(KMnO₄) = [N(oxalic) × V(oxalic)] / V(KMnO₄). Overshooting records a V(KMnO₄) larger than the true equivalence volume, and because it is in the denominator the calculated N(KMnO₄) comes out lower than the true value. (trap: end-point overshoot)
Why A is wrong: A is wrong because it reasons 'more titrant used, so a stronger solution detected'. The formula runs the other way: N(KMnO₄) = N(oxalic) × V(oxalic) / V(KMnO₄), and V(KMnO₄) is in the denominator, so recording too large a volume makes the calculated normality too SMALL.
Why C is wrong: C is wrong because one drop, about 0.05 mL, is a 0.25% error on a 20 mL titre — comparable to the whole reading error of the burette — and, unlike a reading error, it always pushes the result the same way. That is a systematic error, not noise.
Why D is wrong: D is wrong because the brown precipitate MnO₂ forms when the solution is not acidic enough. In a properly acidified titration, excess permanganate simply stays as pink MnO₄⁻.
MCQ 6Direct ApplicationPractice
In KMnO₄ titration with oxalic acid, dilute H₂SO₄ is used to provide acidic medium. Why is HCl NOT used instead?
Show answer and why every option is right or wrong
Answer: C. C is correct. Cl⁻ ions from HCl are oxidised by KMnO₄ (a strong oxidiser) to Cl₂. This side reaction consumes additional KMnO₄ beyond what is needed for oxalic acid alone, making the recorded titre volume erroneously high.
Why A is wrong: A — HCl does not react with or decompose oxalic acid. Both are acids; no redox or decomposition occurs between them.
Why B is wrong: B — MnCl₂ is highly soluble in water. No precipitation occurs. This option confuses MnCl₂ with insoluble compounds like AgCl.
Why D is wrong: D — pH has minimal effect on KMnO₄'s purple-to-colourless transition visibility. The real issue is the parasitic oxidation of Cl⁻, not visibility of the colour change.
MCQ 7CalculationPractice
A 1.96 g sample of impure oxalic acid (H₂C₂O₄·2H₂O, molar mass 126 g/mol) is dissolved to make 100 mL of solution. A 10.0 mL portion of this solution requires 22.4 mL of 0.025 M KMnO₄ for complete oxidation in acidic medium. What is the percentage purity of the oxalic acid sample?
Show answer and why every option is right or wrong
Answer: C. C is correct. Moles KMnO₄ reacted = 0.0224 L × 0.025 mol/L = 5.6 × 10⁻⁴ mol. From the balanced equation, 2 mol KMnO₄ ≡ 5 mol H₂C₂O₄, so moles of oxalic acid in the 10.0 mL portion = 5.6 × 10⁻⁴ × (5/2) = 1.4 × 10⁻³ mol. Scaling to the full 100 mL (×10) gives 1.4 × 10⁻² mol, i.e. 1.4 × 10⁻² × 126 g/mol = 1.764 g of pure H₂C₂O₄·2H₂O. Percentage purity = 1.764/1.96 × 100 = 90.0%.
Why A is wrong: A — 64.3% comes from using the anhydrous molar mass (~90 g/mol) instead of the dihydrate's 126 g/mol: 1.4 × 10⁻² mol × 90 g/mol = 1.26 g, and 1.26/1.96 × 100 = 64.3%, which understates the mass of pure acid found and so understates the purity.
Why B is wrong: B — 80.0% can result from an arithmetic slip while chaining the mole-ratio conversion and the ×10 dilution scale-up, giving a value below the correct 90.0% without a single clean cause.
Why D is wrong: D — 100% wrongly assumes the entire sample is pure oxalic acid, which defeats the purpose of the titration: the whole point of the calculation is that the titre gives LESS acid than the full sample mass would imply if it were 100% pure.
MCQ 8Concept TrapPractice
In an acidic-medium KMnO₄ titration, a student observes that the initial drops of KMnO₄ decolourise very slowly, but after some oxalic acid has reacted, subsequent drops decolourise almost instantly. What explains this acceleration?
Show answer and why every option is right or wrong
Answer: A. A is correct. The Mn²⁺ ions produced during the reduction of MnO₄⁻ catalyse the reaction between permanganate and oxalic acid (autocatalysis). This is why the first few drops react slowly (no catalyst present) but later drops react rapidly (Mn²⁺ accumulates).
Why B is wrong: B — while the reaction is mildly exothermic, the solution is already pre-heated to 60–70 °C. The small additional heat from the reaction does not explain the dramatic rate acceleration observed after the first few drops.
Why C is wrong: C — oxalic acid concentration DECREASES as it is consumed by KMnO₄. CO₂ escaping does not increase the concentration of the remaining acid — it is a product leaving the system.
Why D is wrong: D — the amount of water produced is negligible relative to the total solution volume (~50 mL). H₂SO₄ concentration change is insignificant and cannot explain the observed rate increase.
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Titrimetric Redox: quick recall before you leave
How do you solve a Titrimetric Redox question? A worked example
Pattern: NEET pattern: practical bundle (redox titration calculation)
- 1
Given
• Volume of oxalic acid solution (V₂) = 25.0 mL• Normality of oxalic acid (N₂) = 0.05 N (prepared from 0.63 g H₂C₂O₄·2H₂O in 200 mL)• Volume of KMnO₄ at end-point (V₁) = 18.5 mL• Medium: dilute H₂SO₄ (acidic)
- 2
Required
Find the normality and molarity of the KMnO₄ solution.
- 3
Concept
At the equivalence point, equivalents of oxidant = equivalents of reductant. In acidic medium, KMnO₄ has n-factor = 5 (Mn: +7 → +2). Oxalic acid has n-factor = 2 (each C: +3 → +4, two C atoms).
- 4
Formula
N₁V₁ = N₂V₂
Molarity = Normality / n-factor - 5
Substitution
N₁ × 18.5 = 0.05 × 25.0
- 6
Calculation
N₁ = (0.05 × 25.0) / 18.5
N₁ = 1.25 / 18.5
N₁ = 0.0676 N
Molarity of KMnO₄ = N₁ / n-factor = 0.0676 / 5 = 0.01351 M
Note on exact values: The stoichiometric coefficients (2, 5) and n-factors (5, 2) are exact integers and do not limit significant figures. The result is limited by the 3 significant figures in the measured volumes and normality. - 7
Final answer
Normality of KMnO₄ = 6.76 × 10⁻² N
Molarity of KMnO₄ = 1.35 × 10⁻² M
(Scientific notation used to avoid ambiguity in trailing zeros — per Rule 2.) - 8
Common trap
End-point overshoot (trap: end point overshoot): If the student recorded 19.0 mL instead of 18.5 mL by overshooting, the calculated N₁ would be (1.25/19.0) = 0.0658 N — about 2.7% lower than the true value. In NEET MCQs, the "overshoot" distractor typically gives the next-lower option.
- 9
Similar NEET-style question
"20.0 mL of an acidified oxalic acid solution requires 16.0 mL of 0.02 M KMnO₄ for complete oxidation. Find the molarity of the oxalic acid solution."
(Approach: Convert M(KMnO₄) to normality using n=5, apply N₁V₁ = N₂V₂, then convert back to molarity of oxalic acid using n=2.)
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What to remember before solving Titrimetric Redox questions
Which Titrimetric Redox formulas do you need for NEET?
Normality equation in titration
Equivalents of acid = equivalents of base at end-point. Or for redox: equivalents of oxidant = equivalents of reductant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N | normality | eq/L |
| V | volume | mL or L |
Valid when
- Same titration end-point
- Equivalent factors known
Where do students lose marks on Titrimetric Redox?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Continuing to add titrant past the first persistent colour change because the colour seemed to fade after a swirl.
When it triggers
Question describes 'colour faded after swirling' or 'persistent colour' — distinguishes transient vs end-point.
How to avoid
End-point = first PERSISTENT colour change (lasts ≥30 s). Transient fades back to original on swirling.
Root cause: rushed under time pressure
Correction
Slow drop-wise addition near the end-point; first persistent colour change is the end-point. Re-do if overshot.
More in Practical and Analytical Chemistry: 4 exam traps and mistakes · 1 formula · 1 question pattern from its other lessons.
Titrimetric Redox questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys.
NEET 2026
10.10 M
20.20 M
30.25 M
40.15 M
NTA Answer: Option 1(final)
All 7 past-paper questions from Practical and Analytical Chemistry →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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