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.

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Which of the following statement is correct about the balanced equation given below? $\ce{Cr2O7^{2-}(aq) + 14H^+(aq) + 3S^{2-}(aq) -> 2Cr^{3+}(aq) + 3S(s) + 7H2O(l)}$ $(\text{aq} = \text{aqueous},\ s = \text{solid},\ l = \text{liquid})$ (A) $\ce{Cr2O7^{2-}}$ reduces the $\ce{S^{2-}}$ (B) oxidation number of Cr changes from $+7$ to $+3$ (C) oxidation number of S remains $-2$ (D) $\ce{Cr2O7^{2-}}$ oxidises the $\ce{S^{2-}}$

   ›Reveal solutionSolution

   The reaction is a redox process where dichromate oxidises sulfide to sulfur; the correct statement is that $\ce{Cr2O7^{2-}}$ oxidises $\ce{S^{2-}}$, so option (D) is correct.

   Concept & Intuition

   This is a classic redox reaction. To decide which statement is correct, we need to track oxidation numbers (the charge an atom would have if all bonds were ionic). The key idea: an increase in oxidation number = oxidation (loss of electrons); a decrease = reduction (gain of electrons). The species that causes oxidation is itself reduced — it’s the oxidising agent. Here, dichromate ($\ce{Cr2O7^{2-}}$) contains chromium in a high oxidation state, so it tends to gain electrons and get reduced, while sulfide ($\ce{S^{2-}}$) is electron-rich and tends to lose electrons (get oxidised). Let’s verify step by step.

   1. Find the oxidation number of Cr in $\ce{Cr2O7^{2-}}$

      Oxygen is almost always $-2$ (except in peroxides). Let Cr’s oxidation number be $x$.

      For $\ce{Cr2O7^{2-}}$: $2x + 7(-2) = -2$

      $2x - 14 = -2 \implies 2x = +12 \implies x = +6$

      So Cr is $+6$ in dichromate, not $+7$. This already eliminates option (B).

   2. Find the oxidation number of Cr in $\ce{Cr^{3+}}$

      Simple: it’s $+3$. So Cr goes from $+6$ to $+3$ — a decrease of 3 per Cr atom. That means each Cr gains 3 electrons; Cr is reduced. Therefore $\ce{Cr2O7^{2-}}$ acts as an oxidising agent.

   3. Find the oxidation number of S in $\ce{S^{2-}}$

      The ion $\ce{S^{2-}}$ has sulfur at $-2$ (since it’s a simple monatomic ion).

      In the product, sulfur appears as elemental sulfur $\ce{S(s)}$, where the oxidation number of an element in its standard state is $0$.

      So S goes from $-2$ to $0$ — an increase of 2. That means each S loses 2 electrons; S is oxidised. Therefore $\ce{S^{2-}}$ acts as a reducing agent.

   4. Interpret the statements

      • (A) “$\ce{Cr2O7^{2-}}$ reduces the $\ce{S^{2-}}$” — This would mean dichromate causes reduction of sulfide, but reduction means gaining electrons. Here sulfide loses electrons (is oxidised), so this is false. …

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2. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Which of the following oxidation reactions of KMnO$_4$ occur in acidic medium? I. Oxidation of oxalic acid II. Oxidation of iodide to iodate III. Precipitation of sulphur from hydrogen sulphide (A) I, II, III (B) II, III only (C) I, III only (D) I, II only

   ›Reveal solutionSolution

   KMnO$_4$ acts as a strong oxidizer in acidic medium, reducing to Mn$^{2+}$. Among the given reactions, oxalic acid oxidation and H$_2$S oxidation to sulfur occur in acidic medium, while iodide-to-iodate oxidation requires alkaline conditions. The correct set is I and III only.

   The key to this question lies in understanding how the medium (acidic, neutral, or alkaline) dictates the reduction product of KMnO$_4$ and, consequently, which reactions are feasible. In acidic medium, MnO$_4^-$ is reduced to Mn$^{2+}$ (colourless), gaining 5 electrons. In neutral or faintly alkaline medium, it reduces to MnO$_2$ (brown precipitate), gaining 3 electrons. In strongly alkaline medium, it reduces to MnO$_4^{2-}$ (green), gaining 1 electron. Each reaction must be compatible with the medium for the oxidation to proceed.

   Let’s examine each reaction one by one.

   1. Oxidation of oxalic acid (H$_2$C$_2$O$_4$) by KMnO$_4$

      This is a classic redox titration performed in acidic medium (dilute H$_2$SO$_4$). The half-reactions are:

      • Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
      • Oxidation: $\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^-$ The reaction requires H$^+$ ions, so it occurs in acidic medium. Reaction I is valid.

   2. Oxidation of iodide (I$^-$) to iodate (IO$_3^-$)

      This conversion involves a large change in oxidation state: I$^-$ (oxidation state -1) to IO$_3^-$ (oxidation state +5), a loss of 6 electrons per iodine atom. In acidic medium, KMnO$_4$ would oxidize I$^-$ to I$_2$ (not IO$_3^-$), because the strong acidic conditions favour the formation of iodine. To push the oxidation all the way to iodate, an alkaline medium is used, where MnO$_4^-$ reduces to MnO$_2$ or MnO$_4^{2-}$ and the reaction proceeds as:

      $\text{I}^- + 6\text{OH}^- \rightarrow \text{IO}_3^- + 3\text{H}_2\text{O} + 6e^-$

      This is not an acidic medium reaction. Reaction II does NOT occur in acidic medium. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.In acidic medium, potassium permanganate oxidizes $H_2O_2$ to $O_2$ as per the equation given below $2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$ 100 mL of 0.02 M $KMnO_4$ oxidises 10 mL of X vol $H_2O_2$ 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 $H_2O_2$; the value of X works out to 5.6.

   Concept and Intuition

   The given equation fixes the mole ratio between $MnO_4^-$ and $H_2O_2$ (5 mol $H_2O_2$ per 2 mol $MnO_4^-$). "X volume" $H_2O_2$ means one litre of that solution liberates X litres of $O_2$ gas (at STP) on complete decomposition ($2H_2O_2\to2H_2O+O_2$); since 1 mole of $H_2O_2$ yields 0.5 mole $O_2$ (i.e. 11.2 L at STP per mole $H_2O_2$), the molarity of an "X volume" solution is $M=X/11.2$.

   Step-by-Step Solution

   1. Moles of $KMnO_4$ used $= 0.100\,L \times 0.02\,mol/L = 0.002$ mol.
   2. From the balanced equation, $2\,MnO_4^-$ reacts with $5\,H_2O_2$, so moles $H_2O_2$ reacted $= \dfrac{5}{2}\times0.002 = 0.005$ mol. …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Which of the following is not correct regarding $K_2Cr_2O_7$? (A) The angle $Cr-O-Cr$ in it is $118°$ (B) It oxidises $Sn^{2+}$ to $Sn^{4+}$ 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 $K_2Cr_2O_7$ 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 $CrO_4$ units fused through a shared bridging oxygen (like $Cr_2O_7^{2-}$ 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

   1. (A) Structural fact: $Cr-O-Cr$ angle in $Cr_2O_7^{2-}$ is $126°$ — the statement claims $118°$, which is wrong.
   2. (B) $Cr_2O_7^{2-}$ is a strong oxidiser in acidic medium: $Cr_2O_7^{2-}+14H^++6e^- \rightarrow 2Cr^{3+}+7H_2O$; it oxidises $Sn^{2+}\rightarrow Sn^{4+}$ — true.
   3. (C) $K_2Cr_2O_7$ can be obtained in a very pure state, is stable, and is a standard primary standard in redox titrations — true.
   4. (D) $K_2Cr_2O_7$ crystals are orange-red coloured — true. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$x$ mL of 0.05 M $KMnO_4$ solution is required to oxidise completely 1.52 g of $FeSO_4$ 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 $KMnO_4$:$FeSO_4$ stoichiometry in acidic medium gives $x = 40$ mL.

   Concept and Intuition

   In acidic medium, $MnO_4^-$ is reduced from Mn(+7) to Mn(+2) — a 5-electron gain — while each $Fe^{2+}$ is oxidized to $Fe^{3+}$, losing 1 electron. Balancing electrons means 1 mole of $MnO_4^-$ reacts with exactly 5 moles of $Fe^{2+}$. This fixed mole ratio is the whole basis of the classic permanganate titration.

   Step-by-Step Solution

   1. Balanced half-reactions combine to: $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$.
   2. Molar mass of $FeSO_4 = 56 + 32 + 4(16) = 152$ g/mol.
   3. Moles of $FeSO_4 = \dfrac{1.52\text{ g}}{152\text{ g/mol}} = 0.01$ mol.
   4. From stoichiometry, moles of $KMnO_4$ needed $= \dfrac{0.01}{5} = 0.002$ mol. …

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6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The change in oxidation state of sulphur during the oxidation of thiosulphate ion in neutral alkaline solution by KMnO$_4$ is (A) from $+2$ to $+6$ (B) from $-2$ to $+2$ (C) from $+2$ to $+3$ (D) from $+4$ to $+6$

   ›Reveal solutionSolution

   The thiosulphate ion ($S_2O_3^{2-}$) is oxidized to the sulphate ion ($SO_4^{2-}$) by KMnO$_4$ in neutral/alkaline solution. The average oxidation state of sulphur changes from $+2$ in thiosulphate to $+6$ in sulphate. The correct option is (A).

   The oxidation state of an element in a compound or ion represents the hypothetical charge an atom would have if all bonds were ionic. It's a useful concept for tracking electron transfer in redox reactions. To determine the change in oxidation state, we first need to identify the initial and final compounds containing the element in question, and then calculate its oxidation state in each.

   In this problem, we are looking at the oxidation of the thiosulphate ion ($S_2O_3^{2-}$). When thiosulphate is oxidized by a strong oxidizing agent like potassium permanganate (KMnO$_4$) in a neutral or alkaline solution, it is converted into the sulphate ion ($SO_4^{2-}$). This is a standard reaction outcome under these conditions.

   Let's calculate the oxidation state of sulphur in both the reactant and the product.

   1. Identify the reactant and product containing sulphur.

      • The reactant is the thiosulphate ion, $S_2O_3^{2-}$.
      • The product, under these specific reaction conditions (oxidation by KMnO$_4$ in neutral/alkaline solution), is the sulphate ion, $SO_4^{2-}$.

   2. Calculate the oxidation state of sulphur in the thiosulphate ion ($S_2O_3^{2-}$).

      We assign standard oxidation states to oxygen, which is typically $-2$. The sum of the oxidation states of all atoms in an ion must equal the charge of the ion.

      Let the oxidation state of sulphur be $x$.

      There are two sulphur atoms and three oxygen atoms. The overall charge is $-2$.

   $2(x) + 3(-2) = -2$

   $2x - 6 = -2$

   $2x = -2 + 6$

   $2x = 4$

   $x = +2$

   So, the *average* oxidation state of sulphur in $S_2O_3^{2-}$ is $+2$. …
   

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7. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Which of the following change is not correct about the oxidizing property of KMnO$_4$ in acidic medium? (A) S$^{2-}$ $\rightarrow$ S (B) Mn$^{2+}$ $\rightarrow$ MnO$_2$ (C) SO$_3^{2-}$ $\rightarrow$ SO$_4^{2-}$ (D) C$_2$O$_4^{2-}$ $\rightarrow$ CO$_2$

   ›Reveal solutionSolution

   In acidic medium, KMnO$_4$ is reduced to Mn$^{2+}$, not to MnO$_2$. Option (B) shows Mn$^{2+}$ being oxidized to MnO$_2$, which is the opposite of what happens — so (B) is the incorrect change.

   The question tests your understanding of the redox behaviour of KMnO$_4$ in acidic medium. Potassium permanganate is a powerful oxidizing agent, and its reduction product depends on the pH of the solution. In acidic medium, the half-reaction is:

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

   So the manganese ends up as Mn$^{2+}$ (colourless), not as MnO$_2$ (brown precipitate, which forms in neutral or alkaline medium). The question asks which change is not correct — meaning which transformation does NOT actually happen when KMnO$_4$ oxidizes something in acid.

   Let’s examine each option.

   1. Option (A): S$^{2-}$ $\rightarrow$ S

      Sulphide ion is oxidized to elemental sulphur. In acidic medium, KMnO$_4$ can oxidize S$^{2-}$ to S (or further to SO$_4^{2-}$ depending on conditions, but S is a valid product). This change is correct.

   2. Option (B): Mn$^{2+}$ $\rightarrow$ MnO$_2$

      Here, Mn$^{2+}$ is being oxidized to MnO$_2$. But in acidic medium, KMnO$_4$ itself gets reduced to Mn$^{2+}$ — it does not oxidize Mn$^{2+}$ further. In fact, Mn$^{2+}$ is the final reduced form of manganese in acid. So this change is backwards: KMnO$_4$ cannot turn Mn$^{2+}$ into MnO$_2$ in acidic conditions. This is the incorrect change.

   3. Option (C): SO$_3^{2-}$ $\rightarrow$ SO$_4^{2-}$ …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The volume (in mL) of 10 volume $\mathrm{H_2O_2}$ solution required to completely react with 200 mL of 0.4 M $\mathrm{KMnO_4}$ 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 $2H_2O_2 \to 2H_2O + O_2$, 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 $Mn^{7+}\to Mn^{2+}$ (gain of 5 electrons per KMnO₄).

   Step-by-Step Solution

   1. Molarity of H₂O₂: $M = \dfrac{\text{volume strength}}{11.2} = \dfrac{10}{11.2} = 0.893\ M$.
   2. Balanced acidic redox equation: $2KMnO_4 + 5H_2O_2 + 3H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.
   3. Moles of $KMnO_4$ = $0.4\ M \times 0.200\ L = 0.08$ mol. …

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9. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Which one of the following acts as autocatalyst during titration of $\mathrm{KMnO_4}$ and oxalic acid in presence of dilute $\mathrm{H_2SO_4}$ ? (A) $\mathrm{H_2SO_4}$ (B) $\mathrm{KMnO_4}$ (C) $\mathrm{H_2C_2O_4}$ (D) $\mathrm{MnSO_4}$

   ›Reveal solutionSolution

   This tests the classic autocatalysis example: $\mathrm{Mn^{2+}}$ 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 $\mathrm{KMnO_4}$–oxalic acid redox titration.

   Step-by-Step Solution

   1. The overall redox reaction: $2\mathrm{MnO_4^-} + 5\mathrm{H_2C_2O_4} + 6\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 10\mathrm{CO_2} + 8\mathrm{H_2O}$.
   2. Initially, before any $\mathrm{Mn^{2+}}$ has formed, the reaction between $\mathrm{MnO_4^-}$ and oxalic acid is slow (the first few drops of $\mathrm{KMnO_4}$ decolourise slowly).
   3. As $\mathrm{Mn^{2+}}$ ions accumulate (as $\mathrm{MnSO_4}$ in this sulphuric acid medium), they catalyse the further reduction of $\mathrm{MnO_4^-}$, and the reaction speeds up markedly. …

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10. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.100 mL of aqueous solution of 0.05 M $\mathrm{Cu^{2+}}$ is added to 1 L of 0.1 M KI solution. The resultant solution was titrated with 0.01 M $\mathrm{Na_2S_2O_3}$ solution using starch indicator till blue color disappeared. What is the volume (in mL) of $\mathrm{Na_2S_2O_3}$ used? (A) 2000 (B) 1000 (C) 500 (D) 250

    ›Reveal solutionSolution

    This is the classic iodometric estimation of Cu(II): $\mathrm{Cu^{2+}}$ 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 $\mathrm{Cu^{2+}}$ is added to excess iodide, it is reduced to insoluble $\mathrm{CuI}$ while iodide is oxidized to iodine: $2\mathrm{Cu^{2+}} + 4\mathrm{I^-} \to 2\mathrm{CuI}\downarrow + \mathrm{I_2}$. The liberated $\mathrm{I_2}$ is then estimated by titrating against standard sodium thiosulfate using starch as the indicator (the blue starch-iodine complex disappears at the endpoint): $\mathrm{I_2} + 2\mathrm{S_2O_3^{2-}} \to 2\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$. Note that KI here is in large excess (0.1 mol vs. 0.005 mol Cu$^{2+}$ needing only 0.01 mol I$^-$), so iodide is not limiting and all the Cu$^{2+}$ reacts.

    Step-by-Step Solution

    1. Moles of $\mathrm{Cu^{2+}}$: $0.1\,\text{L} \times 0.05\,\text{mol/L} = 5\times10^{-3}\,\text{mol}$.
    2. Check KI is in excess: reaction needs $2\times5\times10^{-3}=0.01$ mol I$^-$; available KI $=1\,\text{L}\times0.1\,\text{M}=0.1$ mol, far more than needed, so Cu$^{2+}$ is the limiting reagent.
    3. From $2\mathrm{Cu^{2+}} \to 1\,\mathrm{I_2}$: moles of $\mathrm{I_2}$ formed $= \dfrac{5\times10^{-3}}{2} = 2.5\times10^{-3}\,\text{mol}$. …

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11. COMEDK 2025Set 2025-M1 markMCQ
    Q.The concentration and percentage purity of Oxalic acid can be determined by titration with $\mathrm{KMnO}_4$ in presence of dil. $\mathrm{H}_2 \mathrm{SO}_4$. Instead of dil. $\mathrm{H}_2 \mathrm{SO}_4$, dil HCl cannot be used because (A) HCl can also reduce $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$ (B) HCl can also provide $\mathrm{H}^{+}$ions in addition to $\mathrm{H}^{+}$ions from Oxalic acid. (C) HCl can also oxidise Oxalic acid to $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. (D) Oxalic acid oxidises HCl to $\mathrm{Cl}_2$.

    ›Reveal solutionSolution

    The key idea is that HCl is a stronger reducing agent than oxalic acid, so it competes with oxalic acid for the permanganate, leading to an overestimation of oxalic acid. The correct option is (A).

    Concept and Intuition

    In a redox titration, the titrant (here, $\mathrm{KMnO}_4$) must react only with the analyte (oxalic acid) for the result to be accurate. If the solvent or any added reagent also reacts with the titrant, the volume of titrant used will be larger than expected, giving a falsely high concentration of the analyte. Dilute $\mathrm{H}_2\mathrm{SO}_4$ is safe because sulfate ions are not easily oxidised by permanganate. But chloride ions ($\mathrm{Cl}^-$) from HCl are oxidised by $\mathrm{MnO}_4^-$ in acidic medium — this is a well-known side reaction. So using HCl introduces an extra consumption of $\mathrm{KMnO}_4$, ruining the titration.

    Step-by-step reasoning

    1. Identify the main reaction In the titration, oxalic acid ($\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4$) is oxidised by permanganate in acidic medium:

    $2\,\mathrm{MnO}_4^- + 5\,\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + 6\,\mathrm{H}^+ \rightarrow 2\,\mathrm{Mn}^{2+} + 10\,\mathrm{CO}_2 + 8\,\mathrm{H}_2\mathrm{O}$

    The $\mathrm{H}^+$ ions are supplied by the dilute $\mathrm{H}_2\mathrm{SO}_4$.

    2. What happens if we replace $\mathrm{H}_2\mathrm{SO}_4$ with HCl? HCl dissociates to give $\mathrm{H}^+$ and $\mathrm{Cl}^-$. The $\mathrm{H}^+$ are fine, but the $\mathrm{Cl}^-$ ions are also capable of being oxidised by $\mathrm{MnO}_4^-$:

    $2\,\mathrm{MnO}_4^- + 10\,\mathrm{Cl}^- + 16\,\mathrm{H}^+ \rightarrow 2\,\mathrm{Mn}^{2+} + 5\,\mathrm{Cl}_2 + 8\,\mathrm{H}_2\mathrm{O}$

    This reaction is well known and occurs readily in acidic solution.

    3. Consequence for the titration

       Because $\mathrm{Cl}^-$ reduces $\mathrm{MnO}_4^-$ to $\mathrm{Mn}^{2+}$, some of the permanganate is “wasted” on the chloride instead of reacting with oxalic acid. The burette reading will be higher than it should be, leading to an overestimate of the oxalic acid concentration. Therefore, the titration becomes invalid.

    4. Evaluate the options …

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.Match the following List - I Substance A $\mathrm{Na_2CO_3}$ B $\mathrm{KMnO_4} \mid \mathrm{H^+}$ C $\mathrm{K_2Cr_2O_7} \mid \mathrm{H^+}$ D $\mathrm{KMnO_4} \mid \mathrm{H_2O}$ List - II Equivalent weight I $\dfrac{M}{5}$ II $\dfrac{M}{3}$ III $\dfrac{M}{2}$ IV $\dfrac{M}{6}$ (M = Formula weight) (A) A – III; B – I; C – IV; D – II (B) A – III; B – IV; C – I; D – II (C) A – II; B – III; C – IV; D – I (D) A – IV; B – II; C – III; D – I

    ›Reveal solutionSolution

    The equivalent weight of a substance in a redox or acid‑base reaction depends on the number of electrons gained/lost or the net charge change per formula unit. For the given substances: Na₂CO₃ (acid‑base, n=2) → M/2; KMnO₄/H⁺ (n=5) → M/5; K₂Cr₂O₇/H⁺ (n=6) → M/6; KMnO₄/H₂O (neutral, n=3) → M/3. The correct matching is A–III, B–I, C–IV, D–II, which corresponds to option (A).

    ---

    Concept & Intuition

    Equivalent weight is defined as the formula weight (M) divided by the n‑factor — the number of moles of electrons transferred (for redox) or the number of moles of H⁺/OH⁻ exchanged (for acid‑base). The trick is to identify the change in oxidation state (or charge) per formula unit in the given medium. Each substance here behaves differently depending on whether the environment is acidic, neutral, or (for Na₂CO₃) purely acid‑base.

    ---

    Step‑by‑step reasoning

    1. A: Na₂CO₃ (sodium carbonate)

       • This is an acid‑base reaction, not redox. In water, CO₃²⁻ accepts two protons to become H₂CO₃ (or CO₂ + H₂O).
       • Each CO₃²⁻ ion reacts with 2 H⁺. Hence the n‑factor = 2.
       • Equivalent weight = M/2 → matches III.

    2. B: KMnO₄ in acidic medium (H⁺)

       • Mn in KMnO₄ has oxidation state +7. In acidic solution, it reduces to Mn²⁺ (oxidation state +2).
       • Change in oxidation number = 7 – 2 = 5 electrons gained per Mn atom.
       • n‑factor = 5 → Equivalent weight = M/5 → matches I.

    3. C: K₂Cr₂O₇ in acidic medium (H⁺)

       • Cr in K₂Cr₂O₇ is +6. In acid, it reduces to Cr³⁺ (oxidation state +3). …

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