Q.The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.

Showing the 12 most recent of 27 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Observe the following unbalanced equation $a\,S_8(s) + b\,OH^-(aq) \rightarrow c\,S^{2-}(aq) + d\,S_2O_3^{-2}(aq) + e\,H_2O(l)$ From the balanced equation, identify the correct relations from the following sets I. $a + b = c + d + e + 1$ II. $b = 2e + d$ III. $b - a = c + e + 1$ IV. $b + d = 2c + e - 1$ The correct answer is (A) I, II, III only (B) II, IV only (C) I, III only (D) I, IV only

   ›Reveal solutionSolution

   This tests balancing a disproportionation redox equation ($S_8$ in basic medium disproportionating to sulfide and thiosulfate) and then verifying algebraic relations among the balanced coefficients. Answer: I, III only.

   Concept and Intuition

   Sulfur ($S_8$, oxidation state 0) disproportionates in hot alkali into sulfide ion $S^{2-}$ (oxidation state $-2$, reduced) and thiosulfate $S_2O_3^{2-}$ (average oxidation state $+2$, oxidized) — a classic disproportionation reaction. Balancing such an equation requires matching sulfur atoms, oxygen atoms, hydrogen atoms, and total charge simultaneously, which pins down the unique smallest-integer set of coefficients; once found, any proposed algebraic relation among $a,b,c,d,e$ can simply be checked by substitution.

   Step-by-Step Solution

   1. Set up conservation equations for $a\,S_8+b\,OH^-\to c\,S^{2-}+d\,S_2O_3^{2-}+e\,H_2O$:
      • Sulfur: $8a=c+2d$
      • Oxygen: $b=3d+e$
      • Hydrogen: $b=2e$
      • Charge: $-b=-2c-2d \Rightarrow b=2c+2d$
   2. From H: $e=b/2$. Substitute into O: $b=3d+b/2\Rightarrow b/2=3d\Rightarrow b=6d$, so $e=3d$.
   3. From charge: $c=(b-2d)/2=(6d-2d)/2=2d$.
   4. From S: $8a=c+2d=2d+2d=4d\Rightarrow a=d/2$, i.e. $d=2a$.
   5. Take $a=1\Rightarrow d=2,\ b=6d=12,\ e=3d=6,\ c=2d=4$. Balanced: $S_8+12OH^-\to4S^{2-}+2S_2O_3^{2-}+6H_2O$.
   6. Verify: S: $8=4+4$ ✓; O: $12=2(3)+6=12$ ✓; H: $12=2(6)=12$ ✓; charge: $-12=-2(4)-2(2)=-12$ ✓.
   7. Check I: $a+b=1+12=13$; $c+d+e+1=4+2+6+1=13$ — equal, holds. …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Consider the unbalanced equation given below $a\ Cr(OH)_3 + b\ IO_3^- + c\ OH^- \rightarrow d\ CrO_4^{2-} + e\ I^- + f\ H_2O$ Identify the correct statements/equations given below I. $Cr(OH)_3$ is reduced to $CrO_4^{2-}$ II. $a+b+c=7$ III. $a+c=d+f$ IV. $IO_3^-$ acts as an oxidizing agent in this reaction The correct answer is (A) I, II, IV only (B) II, IV only (C) I, III only (D) I, II, III, IV

   ›Reveal solutionSolution

   This tests redox balancing by the ion-electron (half-reaction) method in basic medium, then checking derived arithmetic relations among the balanced coefficients.

   Concept and Intuition

   $Cr(OH)_3$ has $Cr^{3+}$; in $CrO_4^{2-}$, $Cr$ is $+6$ — so chromium is oxidized, losing 3 electrons. $IO_3^-$ has $I^{5+}$; in $I^-$, iodine is $-1$ — so iodine is reduced, gaining 6 electrons. Balancing means matching the electrons lost to electrons gained (2 Cr atoms per 1 I atom), then balancing O and H with $OH^-$/$H_2O$ since the medium is basic.

   Step-by-Step Solution

   1. Oxidation half-reaction: $Cr(OH)_3 + 5OH^- \to CrO_4^{2-} + 4H_2O + 3e^-$ (check: O: $3+5=8$ left, $4+4=8$ right; H: $3+5=8$ left, $8$ right; charge: $-5$ left, $-2-3=-5$ right ✓).
   2. Reduction half-reaction: $IO_3^- + 3H_2O + 6e^- \to I^- + 6OH^-$ (check: O: $3+3=6$ left, $6$ right; H: $6$ left, $6$ right; charge: $-1-6=-7$ left, $-1-6=-7$ right ✓).
   3. Multiply oxidation ×2 to match 6 electrons: $2Cr(OH)_3 + 10OH^- \to 2CrO_4^{2-} + 8H_2O + 6e^-$.
   4. Add to the reduction half-reaction and cancel $6e^-$ and common $OH^-/H_2O$: $2Cr(OH)_3 + IO_3^- + 4OH^- \to 2CrO_4^{2-} + I^- + 5H_2O$. So $a=2,\ b=1,\ c=4,\ d=2,\ e=1,\ f=5$.
   5. Check each statement:
      • I: "$Cr(OH)_3$ is reduced to $CrO_4^{2-}$" — false, it's oxidized ($+3 \to +6$).
      • II: $a+b+c = 2+1+4 = 7$ — true. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the molar masses of $Na_2S_2O_3$ and $I_2$ are $M_1$ and $M_2$ respectively, then the equivalent weights of $Na_2S_2O_3$ and $I_2$ in the following reaction are respectively $2Na_2S_2O_3 + I_2 \to 2NaI + Na_2S_4O_6$ (A) $M_1, M_2$ (B) $M_1, \dfrac{M_2}{2}$ (C) $2M_1, M_2$ (D) $\dfrac{M_1}{2}, M_2$

   ›Reveal solutionSolution

   This tests computing n-factors (and hence equivalent weights) from a redox half-reaction analysis of the iodine–thiosulfate reaction. The equivalent weights are $M_1$ (thiosulfate) and $M_2/2$ (iodine).

   Concept and Intuition

   Equivalent weight $= \dfrac{\text{molar mass}}{n\text{-factor}}$, where the n-factor is the number of electrons gained or lost per formula unit in the redox change. For thiosulfate/iodine, the n-factor is fixed by how the sulfur oxidation states change on going to tetrathionate — this is a classic, frequently tested case because the S atoms in $S_2O_3^{2-}$ and $S_4O_6^{2-}$ are not all in the same oxidation state (there's a S–S bond), so you must track average oxidation state carefully.

   Step-by-Step Solution

   1. In $S_2O_3^{2-}$, the 2 sulfur atoms have an average oxidation state of $+2$ (one S is $+5$-like, the other $-1$-like, structurally).
   2. In $S_4O_6^{2-}$ (tetrathionate), the 4 sulfur atoms have average oxidation state $+2.5$ (computed from $4S + 6(-2) = -2 \Rightarrow S = +2.5$).
   3. Two $S_2O_3^{2-}$ ions (4 S atoms total, avg $+2$) combine to form one $S_4O_6^{2-}$ (4 S atoms, avg $+2.5$): total oxidation state increase $= 4\times 0.5 = 2$ electrons lost, spread over 2 moles of $Na_2S_2O_3$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Observe the following unbalanced reactions $KO_2 \xrightarrow{HOH} X + Y\uparrow + KOH$ $KMnO_4 \xrightarrow[\text{basic medium}]{X} Y + Z + KOH + H_2O$ Y and Z are respectively (A) $O_2$, MnO (B) $O_2$, $MnO_2$ (C) $O_3$, $MnO_2$ (D) $O_3$, MnO

   ›Reveal solutionSolution

   $KO_2$ hydrolysis gives $H_2O_2$ (X) and $O_2$ (Y); $H_2O_2$ then reduces $KMnO_4$ in basic medium to $MnO_2$ (Z), also releasing $O_2$ (Y) — so Y, Z are $O_2$, $MnO_2$.

   Concept and Intuition

   Potassium superoxide ($KO_2$, containing the superoxide ion $O_2^-$) disproportionates on contact with water into hydrogen peroxide and oxygen gas. In basic/neutral medium, $KMnO_4$ is a milder oxidant than in acid — it is reduced only to $MnO_2$ (Mn +4, a brown solid) rather than all the way to $Mn^{2+}$, and $H_2O_2$ acting here as the reducing agent is itself oxidised to $O_2$.

   Step-by-Step Solution

   1. Reaction 1 — hydrolysis of superoxide: $2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2\uparrow$. This identifies $X = H_2O_2$ and $Y = O_2$.
   2. Reaction 2 — $H_2O_2$ (that's $X$) reacting with $KMnO_4$ in basic medium: $2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2KOH + 2H_2O$. …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.$KMnO_4$ oxidises iodide ions in both acidic and neutral media. The change in oxidation state of manganese in acidic, neutral media are x, y respectively. The sum of x and y is (A) 7 (B) 8 (C) 9 (D) 11

   ›Reveal solutionSolution

   $KMnO_4$ goes $Mn^{7+}\to Mn^{2+}$ in acid (change 5) and $Mn^{7+}\to Mn^{4+}$ in neutral medium (change 3); sum = 8.

   Concept and Intuition

   Permanganate's product of reduction depends strongly on the medium's pH: in strongly acidic conditions it is reduced all the way to the colourless $Mn^{2+}$ ion (5-electron change), while in neutral or mildly alkaline conditions it is only reduced to brown insoluble $MnO_2$ (3-electron change), because the more strongly reducing/protonating acidic environment stabilises the lower oxidation state product.

   Step-by-Step Solution

   1. Acidic medium: $MnO_4^- (Mn = +7) \rightarrow Mn^{2+} (Mn = +2)$; change in oxidation state $x = 7-2 = 5$.
   2. Neutral medium: $MnO_4^- (Mn=+7) \rightarrow MnO_2 (Mn=+4)$; change in oxidation state $y = 7-4 = 3$. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Observe the following unbalanced equation $aS_8 + b\ OH^-(aq) \rightarrow cS^{2-}(aq) + dS_2O_3^{2-}(aq) + eH_2O(l)$ In the balanced equation, the ratio of c and d is (A) $1:2$ (B) $2:1$ (C) $1:3$ (D) $3:1$

   ›Reveal solutionSolution

   This tests disproportionation redox balancing of $S_8$ in hot concentrated alkali. Sulfur (0) splits into $S^{2-}$ ($-2$, reduced) and $S_2O_3^{2-}$ (avg. $+2$ per S, oxidised); the mole ratio $c:d = 2:1$.

   Concept and Intuition

   When $S_8$ is boiled with concentrated NaOH/KOH, sulfur undergoes disproportionation — the same element is simultaneously oxidised and reduced because it starts at an intermediate oxidation state (0) that can go both up and down. Some sulfur atoms are reduced to sulfide, $S^{2-}$ ($-2$), while others are oxidised to thiosulfate, $S_2O_3^{2-}$, where the average oxidation state of S works out to $+2$.

   Step-by-Step Solution

   1. Assign oxidation states: $S$ in $S_8$ is $0$. In $S^{2-}$, S is $-2$. In $S_2O_3^{2-}$: total charge $= -2$, oxygens contribute $3\times(-2) = -6$, so $2x - 6 = -2 \Rightarrow x = +2$ (average per S atom).
   2. Electrons gained per $S^{2-}$ formed: $0 \to -2$, i.e. 2 electrons gained per S atom.
   3. Electrons lost per $S_2O_3^{2-}$ formed: each of its 2 S atoms goes from $0 \to +2$ (loses 2 e$^-$), so $2\times 2 = 4$ electrons lost per $S_2O_3^{2-}$.
   4. Let $a S_8 \to c\,S^{2-} + d\,S_2O_3^{2-}$. Sulfur atom balance: $8a = c + 2d$. Electron balance: $2c$ (gained) $= 4d$ (lost) $\Rightarrow c = 2d$. …

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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\mathrm{H_2O_2}$ with $\mathrm{KMnO_4}$ in acidic medium gives a manganese compound 'X' and in basic medium gives another manganese compound 'Y'. The oxidation state of manganese in X and Y, respectively are (A) +2, +4 (B) +4, +2 (C) +3, +4 (D) +4, +3

   ›Reveal solutionSolution

   This tests the well-known medium-dependence of $KMnO_4$'s reduction product: acidic medium takes Mn all the way from +7 to +2 ($Mn^{2+}$), while basic/neutral medium stops the reduction at +4 ($MnO_2$).

   Concept and Intuition

   $KMnO_4$ is a versatile oxidising agent whose reduction product depends strongly on the pH of the medium, because the mechanism (and the number of electrons transferred to reach a stable Mn species) differs:

   • In strongly ACIDIC medium, $MnO_4^-$ (Mn +7) is reduced by a 5-electron step directly to the pale-pink/colourless $Mn^{2+}$ (Mn +2).
   • In NEUTRAL or BASIC medium, $MnO_4^-$ is reduced by only a 3-electron step to the brown insoluble $MnO_2$ (Mn +4), since further reduction to $Mn^{2+}$ is not favoured under those conditions. This distinction is a classic inorganic-chemistry fact tested repeatedly in competitive exams.

   Step-by-Step Solution

   1. Identify the reaction context: $H_2O_2$ reduces $KMnO_4$, with $H_2O_2$ itself being oxidised to $O_2$.
   2. In ACIDIC medium: $2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$. Product X = $Mn^{2+}$, oxidation state $+2$.
   3. In BASIC medium: $2MnO_4^- + H_2O_2 \rightarrow 2MnO_2 + 2OH^- + O_2 + \ldots$ (brown $MnO_2$ precipitate forms). Product Y = $MnO_2$, oxidation state $+4$. …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.In acid medium, dichromate oxidizes sulphite to sulphate. In this redox reaction, the oxidation state of sulphur changes from $x$ to $y$. What is the value of $(x + y)$? (A) 8 (B) 6 (C) 10 (D) 12

   ›Reveal solutionSolution

   Sulphur's oxidation state rises from $+4$ (sulphite) to $+6$ (sulphate), so $x+y=4+6=10$.

   Concept and Intuition

   In a redox reaction, we track the oxidation state of the atom being oxidized (or reduced) using the standard rule that oxygen is $-2$ in these oxoanions, and the sum of oxidation states equals the ion's overall charge.

   Step-by-Step Solution

   1. Sulphite ion, $\mathrm{SO_3^{2-}}$: let S have oxidation state $x$. $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = 4$
   2. Sulphate ion, $\mathrm{SO_4^{2-}}$: let S have oxidation state $y$. $y + 4(-2) = -2 \implies y - 8 = -2 \implies y = 6$ …

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9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.In acid medium dichromate oxidizes sulphite to sulphate as shown below. $x\mathrm{Cr_2O_7^{2-}} + y\mathrm{SO_3^{2-}} + z\mathrm{H^+} \longrightarrow a\mathrm{Cr^{3+}} + b\mathrm{SO_4^{2-}} + c\mathrm{H_2O}$ Identify correct statements about this balanced equation I) Sum of $x$ and $y$ is 4 II) Sum of $a$ and $c$ is equals to $(3+b)$ III) Sum of $x, y$ and $z$ is 11 (A) I, II, III (B) I, II only (C) I, III only (D) II, III only

   ›Reveal solutionSolution

   The key idea is to balance the redox reaction using the ion-electron method (half-reactions) in acidic medium. The balanced equation is $\mathrm{Cr_2O_7^{2-}} + 3\mathrm{SO_3^{2-}} + 8\mathrm{H^+} \longrightarrow 2\mathrm{Cr^{3+}} + 3\mathrm{SO_4^{2-}} + 4\mathrm{H_2O}$, so $x=1$, $y=3$, $z=8$, $a=2$, $b=3$, $c=4$. Checking the statements: I (1+3=4) true, II (2+4=6 equals 3+3=6) true, III (1+3+8=12, not 11) false. Hence only I and II are correct, option (B).

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   Concept & Intuition: Redox Reaction Stoichiometry

   This is a classic redox reaction where dichromate ($\mathrm{Cr_2O_7^{2-}}$) is the oxidizing agent (it gets reduced) and sulfite ($\mathrm{SO_3^{2-}}$) is the reducing agent (it gets oxidized). In acidic medium, we balance each half-reaction separately — first for atoms (other than H and O), then for oxygen using water, then for hydrogen using $\mathrm{H^+}$, and finally for charge using electrons. Once both half-reactions are balanced, we multiply them by appropriate factors so that the number of electrons lost equals the number gained. Adding them gives the full balanced equation. The coefficients $x, y, z, a, b, c$ are then determined, and we can test each statement.

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   Step-by-step solution

   1. Write the reduction half-reaction (dichromate to chromium(III))
      • Dichromate ion: $\mathrm{Cr_2O_7^{2-}}$ contains two Cr atoms, each going from +6 to +3, so total gain of $2 \times 3 = 6$ electrons.
      • Balance oxygen: add 7 water molecules to the right.
      • Balance hydrogen: add 14 $\mathrm{H^+}$ to the left.
      • Balance charge: left side charge = $-2 + 14 = +12$; right side charge = $2 \times (+3) = +6$. To balance, add 6 electrons to the left.
      • Balanced reduction half-reaction:

   $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6e^- \longrightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}$

   2. Write the oxidation half-reaction (sulfite to sulfate)
      • Sulfite ion: $\mathrm{SO_3^{2-}}$ (S oxidation state +4) becomes sulfate $\mathrm{SO_4^{2-}}$ (S oxidation state +6), so each S loses 2 electrons.
      • Balance oxygen: add one water molecule to the left.
      • Balance hydrogen: add 2 $\mathrm{H^+}$ to the right.
      • Balance charge: left side charge = $-2$; right side charge = $-2 + 2 = 0$. To balance, add 2 electrons to the right.
      • Balanced oxidation half-reaction:

   $\mathrm{SO_3^{2-}} + \mathrm{H_2O} \longrightarrow \mathrm{SO_4^{2-}} + 2\mathrm{H^+} + 2e^-$

   3. Equalize electrons transferred
      • Reduction gains 6 electrons; oxidation loses 2 electrons per sulfite. To match, multiply the oxidation half-reaction by 3:

   $3\mathrm{SO_3^{2-}} + 3\mathrm{H_2O} \longrightarrow 3\mathrm{SO_4^{2-}} + 6\mathrm{H^+} + 6e^-$

   4. Add the half-reactions
      • Add the reduction and the multiplied oxidation half-reaction. Cancel the 6 electrons on both sides. …

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10. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.In acidic medium $\mathrm{MnO_4^-}$ oxidises $\mathrm{NO_2^-}$ to $\mathrm{NO_3^-}$. How many moles of $\mathrm{NO_2^-}$ are oxidised by 10 moles of $\mathrm{MnO_4^-}$? (A) $20$ (B) $10$ (C) $25$ (D) $15$

    ›Reveal solutionSolution

    This is an electron-balance (equivalent) redox stoichiometry question. 10 moles of $\mathrm{MnO_4^-}$ oxidise 25 moles of $\mathrm{NO_2^-}$.

    Concept and Intuition

    In acidic medium, $\mathrm{MnO_4^-}$ (Mn: +7) is reduced to $\mathrm{Mn^{2+}}$ (+2), a 5-electron gain per $\mathrm{MnO_4^-}$. $\mathrm{NO_2^-}$ (N: +3) is oxidised to $\mathrm{NO_3^-}$ (N: +5), a 2-electron loss per $\mathrm{NO_2^-}$. For the redox reaction to balance, total electrons lost must equal total electrons gained.

    Step-by-Step Solution

    1. Half-reactions: $\mathrm{MnO_4^-} + 8H^+ + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4H_2O$ (5 e⁻ per Mn).
    2. $\mathrm{NO_2^-} + H_2O \rightarrow \mathrm{NO_3^-} + 2H^+ + 2e^-$ (2 e⁻ per N). …

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11. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.In which of the following reactions of $\mathrm{H_2O_2}$, dioxygen is evolved? (only = only) I) With HOCl in acidic medium II) With permanganate in acidic medium III) With permanganate in basic medium (A) I & II only (B) I, II & III (C) II & III only (D) I & III only

    ›Reveal solutionSolution

    $\mathrm{H_2O_2}$ behaves as a reducing agent (getting oxidised to $O_2$) in reactions with strong oxidants like HOCl and permanganate, whether the medium is acidic or basic — so all three listed reactions evolve dioxygen.

    Concept and Intuition

    Hydrogen peroxide can act as either an oxidising agent (itself reduced to $H_2O$) or a reducing agent (itself oxidised to $O_2$), depending on the other reactant. Against strong oxidants such as acidified/basic permanganate or hypochlorous acid, $\mathrm{H_2O_2}$ is the one being oxidised, and dioxygen gas is released as the product.

    Step-by-Step Solution

    1. I) $\mathrm{H_2O_2 + HOCl \rightarrow H_3O^+ + Cl^- + O_2}$ — $H_2O_2$ reduces HOCl, releasing $O_2$. Evolves $O_2$.
    2. II) In acidic medium: $2\mathrm{MnO_4^-}+5\mathrm{H_2O_2}+6H^+\rightarrow 2\mathrm{Mn^{2+}}+5O_2+8H_2O$. Evolves $O_2$. …

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12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.$\mathrm{H_2O_2}$ reduces $\mathrm{KMnO_4}$ in acidic medium to 'x' and in basic medium to 'y'. What are x and y? (A) $x = \mathrm{MnO_2}$, $y = \mathrm{Mn^{2+}}$ (B) $x = \mathrm{Mn^{2+}}$, $y = \mathrm{MnO_2}$ (C) $x = \mathrm{MnO_4^{2-}}$, $y = \mathrm{Mn^{2+}}$ (D) $x = \mathrm{MnO_2}$, $y = \mathrm{MnO_4^{2-}}$

    ›Reveal solutionSolution

    H$_2$O$_2$'s reducing action on permanganate depends on medium: acidic gives Mn$^{2+}$, basic/neutral gives MnO$_2$ — so $x=\mathrm{Mn^{2+}}$, $y=\mathrm{MnO_2}$.

    Concept and Intuition

    KMnO$_4$ is a versatile oxidizer whose reduced product depends heavily on the pH of the medium, because the reduction half-reactions available to Mn(VII) change with H$^+$/OH$^-$ concentration. H$_2$O$_2$ here plays the reducing agent, itself being oxidized to O$_2$. In strongly acidic conditions, permanganate is reduced all the way to the pale pink Mn$^{2+}$ ion (a 5-electron reduction). In neutral or basic conditions, the reduction stops at the brown MnO$_2$ precipitate (a 3-electron reduction), since the deeper reduction to Mn$^{2+}$ requires the H$^+$ ions that acidic medium supplies.

    Step-by-Step Solution

    1. Acidic medium: $2\mathrm{MnO_4^-} + 6\mathrm{H^+} + 5\mathrm{H_2O_2} \to 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} + 5\mathrm{O_2}$ — so $x = \mathrm{Mn^{2+}}$. …

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