Q.Permanganate(VII) ion, MnO4–, in basic medium, oxidises iodide ion (I–) to produce molecular iodine (I2) and manganese dioxide (MnO2). Write a balanced ionic equation to represent this reaction.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Redox Titration
Redox Titration: The Intuition First
Imagine you have a dark room and you want to know exactly how much water is in a bucket. You can't see the water level directly. But you have a measuring cup of ink — and you know that each drop of ink turns a fixed amount of water completely black. You add ink drop by drop, stirring, until the water just turns black. The number of drops tells you exactly how much water was there.
Redox titration works on the same principle — except instead of ink and water, we use an oxidising agent and a reducing agent. One of them is the "unknown" (the water), the other is the "known solution" (the ink). They react with each other in a fixed, predictable ratio. We add the known solution until the reaction is just complete, and that tells us the amount of the unknown.
The Precise Statement
Redox titration is a volumetric analysis technique where a solution of unknown concentration (the analyte) is reacted with a standard solution of known concentration (the titrant) in a redox reaction — one substance gets oxidised, the other gets reduced — until the equivalence point is reached. The volume of titrant used allows calculation of the unknown concentration.
The key difference from acid-base titration: here, electrons are transferred, not protons.
How It Actually Works
You have a flask containing the analyte — say, a solution of ferrous ions (Fe2+). You don't know its concentration. You fill a burette with a standard solution of potassium permanganate (KMnO4), which is a strong oxidising agent. You know its concentration exactly.
You add the permanganate drop by drop. Each drop reacts with Fe2+:
MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O
The purple permanganate gets consumed as it reacts. As long as any Fe2+ remains, the purple colour disappears. The moment all Fe2+ is used up, the next drop of permanganate stays purple — the solution turns pink. That's your end point.
In this case, the titrant itself acts as the indicator — no separate indicator needed. This is called a self-indicating titration. Not all redox titrations are self-indicating; some need a separate redox indicator (like starch for iodine titrations).
The Core Idea in One Sentence
You measure the volume of a known oxidising (or reducing) agent needed to completely react with an unknown reducing (or oxidising) agent, and from that volume you calculate the unknown concentration.
The Calculation (Simple Version)
Suppose you titrate 25.0 mL of Fe2+ solution with 0.0200 M KMnO4. You use 15.0 mL of permanganate to reach the end point.
From the balanced equation: 1 mole MnO4− reacts with 5 moles Fe2+.
Moles of KMnO4 used = 0.0200×0.0150=3.00×10−4 mol
Moles of Fe2+ present = 5×3.00×10−4=1.50×10−3 mol
Concentration of Fe2+ = 0.02501.50×10−3=0.0600 M
Canalyte=Vanalyten×Mtitrant×Vtitrant …
Concept: Redox reaction stoichiometry — balancing by the ion-electron (half-reaction) method in basic medium, with iodide as the specific reducing agent.
Step 1: Write the half-reactions.
Oxidation: I−(aq)→I2(s)
Reduction: MnO4−(aq)→MnO2(s)
Step 2: Balance each half-reaction (atoms, then O with H₂O and H with H⁺, then convert to basic medium with OH⁻, then charge with electrons).
Oxidation: 2I−(aq)→I2(s)+2e−
Reduction: MnO4−(aq)+2H2O(l)+3e−→MnO2(s)+4OH−(aq)
Step 3: Equalise electrons (LCM of 2 and 3 is 6) and add. …
In basic solution, permanganate (MnO4−) oxidises iodide (I−) to iodine (I2) and is itself reduced to manganese dioxide (MnO2). Using the half-reaction method, the balanced ionic equation is
6I−+2MnO4−+4H2O→3I2+2MnO2+8OH−
The Concept: Permanganate in Basic Medium
Permanganate is a powerful oxidising agent, but its reduction product depends critically on the pH of the solution. In acidic medium, it goes all the way down to Mn2+ (colourless). In basic medium, the reduction stops at MnO2, a dark brown solid, because Mn2+ is unstable in base and would immediately precipitate and oxidise further.
A common mistake is to assume permanganate always reduces to Mn2+. In basic solution, the product is MnO2, not Mn2+ — the colour change is from purple (MnO4−) to brown (MnO2), not to colourless.
Step-by-Step: The Half-Reaction (Ion-Electron) Method
1. Write the skeletal ionic equation.
MnO4−(aq)+I−(aq)→MnO2(s)+I2(s)
2. Split into the two half-reactions.
Oxidation half: I−(aq)→I2(s)
Reduction half: MnO4−(aq)→MnO2(s)
3. Balance atoms other than O and H.
Oxidation half needs 2 iodide ions to give 1 I2:
2I−(aq)→I2(s)
Reduction half already has 1 Mn on each side.
4. Balance O and H — first as if in acidic medium, then convert to basic.
For the reduction half, balance O by adding 2H2O to the right:
MnO4−(aq)→MnO2(s)+2H2O(l)
Balance H by adding 4H+ to the left:
MnO4−(aq)+4H+(aq)→MnO2(s)+2H2O(l)
Convert to basic medium: add 4OH− to both sides, combine H++OH− into H2O on the left, then cancel 2H2O common to both sides:
MnO4−(aq)+2H2O(l)→MnO2(s)+4OH−(aq)
5. Balance charge by adding electrons.
Oxidation half: left charge −2, right charge 0 — add 2e− to the right:
2I−(aq)→I2(s)+2e−
Reduction half: left charge −1, right charge −4 — add 3e− to the left: …
- AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQQ.In acidic medium, potassium permanganate oxidizes H2O2 to O2 as per the equation given below 2MnO4−+6H++5H2O2→2Mn2++8H2O+5O2 100 mL of 0.02 M KMnO4 oxidises 10 mL of X vol H2O2 completely. The value of X approximately is (A) 2.8 (B) 5.6 (C) 7.2 (D) 10.0
›Reveal solutionSolution
This is a redox titration combined with the "volume strength" definition of H2O2; the value of X works out to 5.6.
Concept and Intuition
The given equation fixes the mole ratio between MnO4− and H2O2 (5 mol H2O2 per 2 mol MnO4−). "X volume" H2O2 means one litre of that solution liberates X litres of O2 gas (at STP) on complete decomposition (2H2O2→2H2O+O2); since 1 mole of H2O2 yields 0.5 mole O2 (i.e. 11.2 L at STP per mole H2O2), the molarity of an "X volume" solution is M=X/11.2.
Step-by-Step Solution
- Moles of KMnO4 used =0.100L×0.02mol/L=0.002 mol.
- From the balanced equation, 2MnO4− reacts with 5H2O2, so moles H2O2 reacted =25×0.002=0.005 mol. …
- AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQQ.Which of the following is not correct regarding K2Cr2O7? (A) The angle Cr−O−Cr in it is 118° (B) It oxidises Sn2+ to Sn4+ in acidic medium (C) It is used as a primary standard in volumetric analysis (D) It is an orange colored solid
›Reveal solutionSolution
Everything about K2Cr2O7 is true except the bond-angle value quoted — the real Cr−O−Cr bridge angle is 126°, not 118°.
Concept and Intuition
Dichromate ion structure: two tetrahedral CrO4 units fused through a shared bridging oxygen (like Cr2O72− analogous to pyrophosphate/pyrosulfate structures). The Cr–O(bridge)–Cr angle in this bent bridge is a specific, commonly quoted structural fact (126°), distinguishing it from linear (180°) or tetrahedral (109.5°) angles.
Step-by-Step Solution
- (A) Structural fact: Cr−O−Cr angle in Cr2O72− is 126° — the statement claims 118°, which is wrong.
- (B) Cr2O72− is a strong oxidiser in acidic medium: Cr2O72−+14H++6e−→2Cr3++7H2O; it oxidises Sn2+→Sn4+ — true.
- (C) K2Cr2O7 can be obtained in a very pure state, is stable, and is a standard primary standard in redox titrations — true.
- (D) K2Cr2O7 crystals are orange-red coloured — true. …
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.x mL of 0.05 M KMnO4 solution is required to oxidise completely 1.52 g of FeSO4 in acidic medium. The value of x is (At. wt: Fe = 56 u, S = 32 u, O = 16 u) (A) 40 (B) 20 (C) 30 (D) 50
›Reveal solutionSolution
The 1:5 KMnO4:FeSO4 stoichiometry in acidic medium gives x=40 mL.
Concept and Intuition
In acidic medium, MnO4− is reduced from Mn(+7) to Mn(+2) — a 5-electron gain — while each Fe2+ is oxidized to Fe3+, losing 1 electron. Balancing electrons means 1 mole of MnO4− reacts with exactly 5 moles of Fe2+. This fixed mole ratio is the whole basis of the classic permanganate titration.
Step-by-Step Solution
- Balanced half-reactions combine to: MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O.
- Molar mass of FeSO4=56+32+4(16)=152 g/mol.
- Moles of FeSO4=152 g/mol1.52 g=0.01 mol.
- From stoichiometry, moles of KMnO4 needed =50.01=0.002 mol. …
- AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQQ.The volume (in mL) of 10 volume H2O2 solution required to completely react with 200 mL of 0.4 M KMnO4 solution in acidic medium is (A) 112 (B) 336 (C) 224 (D) 448
›Reveal solutionSolution
This tests converting "volume strength" of H₂O₂ to molarity and then using stoichiometric equivalence with KMnO₄ in acidic redox titration.
Concept and Intuition
"10 volume" H₂O₂ means 1 L of that solution liberates 10 L of O₂ gas (measured at STP) on complete decomposition. Since 2H2O2→2H2O+O2, 1 mole of H₂O₂ produces 0.5 mole O₂ = 11.2 L of O₂ at STP. So a solution of molarity M has volume strength =11.2M. In the redox reaction with acidified KMnO₄, H₂O₂ acts as the reducing agent (getting oxidized to O₂), and KMnO₄ is reduced Mn7+→Mn2+ (gain of 5 electrons per KMnO₄).
Step-by-Step Solution
- Molarity of H₂O₂: M=11.2volume strength=11.210=0.893 M.
- Balanced acidic redox equation: 2KMnO4+5H2O2+3H2SO4→K2SO4+2MnSO4+8H2O+5O2.
- Moles of KMnO4 = 0.4 M×0.200 L=0.08 mol. …
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.Which one of the following acts as autocatalyst during titration of KMnO4 and oxalic acid in presence of dilute H2SO4 ? (A) H2SO4 (B) KMnO4 (C) H2C2O4 (D) MnSO4
›Reveal solutionSolution
This tests the classic autocatalysis example: Mn2+ ions, generated as the reaction proceeds, catalyse the same reaction — a product acting as its own catalyst.
Concept and Intuition
Autocatalysis occurs when a product of a reaction catalyses the reaction itself, so the reaction rate increases as the reaction proceeds (rather than steadily decreasing as reactants are consumed). The classic textbook example is precisely the KMnO4–oxalic acid redox titration.
Step-by-Step Solution
- The overall redox reaction: 2MnO4−+5H2C2O4+6H+→2Mn2++10CO2+8H2O.
- Initially, before any Mn2+ has formed, the reaction between MnO4− and oxalic acid is slow (the first few drops of KMnO4 decolourise slowly).
- As Mn2+ ions accumulate (as MnSO4 in this sulphuric acid medium), they catalyse the further reduction of MnO4−, and the reaction speeds up markedly. …
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.100 mL of aqueous solution of 0.05 M Cu2+ is added to 1 L of 0.1 M KI solution. The resultant solution was titrated with 0.01 M Na2S2O3 solution using starch indicator till blue color disappeared. What is the volume (in mL) of Na2S2O3 used? (A) 2000 (B) 1000 (C) 500 (D) 250
›Reveal solutionSolution
This is the classic iodometric estimation of Cu(II): Cu2+ oxidizes excess iodide to iodine, and the liberated iodine is titrated with thiosulfate; working through the stoichiometry gives 500 mL of thiosulfate needed.
Concept and Intuition
When Cu2+ is added to excess iodide, it is reduced to insoluble CuI while iodide is oxidized to iodine: 2Cu2++4I−→2CuI↓+I2. The liberated I2 is then estimated by titrating against standard sodium thiosulfate using starch as the indicator (the blue starch-iodine complex disappears at the endpoint): I2+2S2O32−→2I−+S4O62−. Note that KI here is in large excess (0.1 mol vs. 0.005 mol Cu2+ needing only 0.01 mol I−), so iodide is not limiting and all the Cu2+ reacts.
Step-by-Step Solution
- Moles of Cu2+: 0.1L×0.05mol/L=5×10−3mol.
- Check KI is in excess: reaction needs 2×5×10−3=0.01 mol I−; available KI =1L×0.1M=0.1 mol, far more than needed, so Cu2+ is the limiting reagent.
- From 2Cu2+→1I2: moles of I2 formed =25×10−3=2.5×10−3mol. …
- AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQQ.100 mL of 0.1 M Fe2+ solution was titrated with 601 M Cr2O72− solution in acid medium. What is the volume (in L) of Cr2O72− solution consumed? (A) 100 (B) 10 (C) 1 (D) 0.1
›Reveal solutionSolution
Using the 6:1 Fe²⁺-to-dichromate stoichiometry, the dichromate volume needed is 0.1 L.
Concept and Intuition
In acidic medium, dichromate oxidises Fe²⁺ to Fe³⁺ while itself being reduced from Cr(+6) to Cr(+3): Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2O. Each Cr2O72− ion accepts 6 electrons (2 Cr atoms × 3 electrons each), matching the 6 electrons donated by 6 Fe2+ ions (each losing 1 electron).
Step-by-Step Solution
- Moles of Fe2+: 0.1 L×0.1 mol/L=0.01 mol.
- Stoichiometry requires 6 mol Fe2+ per 1 mol Cr2O72−, so moles of Cr2O72− needed =0.01/6 mol. …
- AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQQ.Match List I with List II. List I (Medium): A. Acidic medium B. Neutral medium C. Strongly basic medium List II (equivalent weight of KMnO4):(i) 158.0(ii) 31.6(iii) 52.7 (A) A → i ; B → ii ; C → iii (B) A → ii ; B → iii ; C → i (C) A → i ; B → iii ; C → ii (D) A → ii ; B → i ; C → iii
›Reveal solutionSolution
KMnO4's equivalent weight depends on how many electrons Mn gains, which differs by medium: acidic (5e⁻, 31.6), neutral (3e⁻, 52.7), strongly basic (1e⁻, 158).
Concept and Intuition
Equivalent weight of an oxidizing agent equals its molar mass divided by the number of electrons it accepts per formula unit in a given reaction. KMnO4's reduction product — and hence the electron count — depends on the medium's acidity/basicity.
Step-by-Step Solution
- In acidic medium: MnO4−+8H++5e−→Mn2++4H2O, so n=5. Eq. wt =158/5=31.6 — matches (ii).
- In neutral (or faintly alkaline) medium: MnO4−+2H2O+3e−→MnO2+4OH−, so n=3. Eq. wt =158/3=52.67≈52.7 — matches (iii).
- In strongly alkaline medium: MnO4−+e−→MnO42−, so n=1. Eq. wt =158/1=158 — matches (i). …
- AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQQ.Which among the following species acts as a self-indicator? (A) H2O2 (B) I− (C) Cr2O72− (D) MnO4−
›Reveal solutionSolution
Permanganate ion is the classic self-indicator, since its own deep purple colour disappears sharply at the equivalence point of a redox titration.
Concept and Intuition
A self-indicator is a titrant (or reactant) whose own colour change signals the endpoint, without needing to add a separate indicator dye. Potassium permanganate solutions are intensely violet/purple due to MnO4−; when reduced to nearly colourless Mn2+ (in acidic medium) at the endpoint, the disappearance of the purple colour (or the first faint persisting pink) is used to detect completion.
Step-by-Step Solution
- H2O2 can act as an oxidizing or reducing agent but is not intensely coloured — not a self-indicator.
- I− forms a colour with starch as an added indicator, but it isn't a striking self-indicator on its own. …
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