Q.Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Showing the 12 most recent of 15 on this concept.

1. KCET 2025Set D-41 markMCQ
   Q.Which one of the following reactions has $\Delta H = \Delta U$ ? (A) $\mathrm{CaCO_3\ (s) \xrightarrow{\Delta} CaO\ (s) + CO_2\ (g)}$ (B) $\mathrm{C_6H_6\ (l) + \dfrac{15}{2} O_2\

   (g) \longrightarrow 6CO_2\

   (g) + 3H_2O\ (l)}$(C)$\mathrm{2HI_{(g)} \rightleftharpoons H_2\

   (g) + I_2\ (g)}$(D)$\mathrm{N_2\

   (g) + 3H_2\

   (g) \rightleftharpoons 2NH_3\ (g)}$

   ›Reveal solutionSolution

   $\Delta H - \Delta U = \Delta n_g RT$, so the two are equal only for the reaction with zero change in the number of moles of gas.

   Step 1 — The relation between $\Delta H$ and $\Delta U$.

   By definition $H = U + PV$. For a reaction at constant temperature and pressure involving ideal gases, $PV = n_g RT$, so

   $\Delta H = \Delta U + \Delta (PV) = \Delta U + \Delta n_g\,RT$

   where

   $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

   Only gases count — solids and liquids have negligible molar volume, so they contribute essentially nothing to $\Delta(PV)$.

   Step 2 — The condition asked for.

   Since $R$ and $T$ are never zero,

   $\Delta H = \Delta U \iff \Delta n_g = 0$

   So we simply count gas moles on each side of each reaction.

   Step 3 — Test every option.

   (A) $\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$

   Gaseous reactants: 0. Gaseous products: 1 ($\mathrm{CO_2}$).

   $\Delta n_g = 1 - 0 = +1 \ne 0$

   (B) $\mathrm{C_6H_6(l) + \tfrac{15}{2}O_2(g) \to 6CO_2(g) + 3H_2O(l)}$

   Gaseous reactants: $\tfrac{15}{2} = 7.5$ (benzene is a liquid). Gaseous products: 6 (water is liquid).

   $\Delta n_g = 6 - 7.5 = -1.5 \ne 0$

   (C) $\mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2(g)}$

   Gaseous reactants: 2. Gaseous products: $1 + 1 = 2$.

   $\Delta n_g = 2 - 2 = \mathbf{0}\ \checkmark$

   (D) $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$

   Gaseous reactants: $1 + 3 = 4$. Gaseous products: 2. …

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2. COMEDK 2025Set 2025-M1 markMCQ
   Q.Choose the correct statement. (A) Calcium and Magnesium metals are manufactured by electrolysis of aqueous solutions of their salts (B) The oxo-anion $\mathrm{ClO}_3^{-}$does not undergo disproportionation reaction (C) Nitride ion $\mathrm{N}^{-3}$ cannot act as an oxidising agent (D) In alkaline medium the reduction of $\mathrm{MnO}_4^{-}$ion involves gain of 5 electrons

   ›Reveal solutionSolution

   The question tests fundamental inorganic chemistry concepts: electrolysis of active metals, oxo-anion stability, oxidation states, and redox behaviour. The correct statement is (C), because the nitride ion ($\mathrm{N}^{3-}$) is already in its lowest oxidation state and cannot be reduced further, so it cannot act as an oxidising agent.

   ---

   Concept and Intuition

   Each option touches a different principle. We need to check each one carefully:

   • Option (A) — Electrolysis of aqueous solutions of active metals like Ca and Mg usually produces hydrogen at the cathode instead of the metal, because water is more easily reduced. So they are manufactured by electrolysis of molten salts, not aqueous.
   • Option (B) — Disproportionation requires an element in an intermediate oxidation state. Chlorine in $\mathrm{ClO}_3^{-}$ is +5; it can both increase and decrease its oxidation state under suitable conditions, so disproportionation is possible.
   • Option (C) — An oxidising agent gains electrons (is reduced). The nitride ion $\mathrm{N}^{3-}$ has nitrogen in its lowest possible oxidation state (−3). It cannot accept more electrons, so it cannot act as an oxidising agent. It can only act as a reducing agent.
   • Option (D) — In alkaline medium, $\mathrm{MnO}_4^{-}$ is reduced to $\mathrm{MnO}_2$ (not $\mathrm{Mn}^{2+}$), which involves a gain of 3 electrons, not 5.

   ---

   Step-by-step reasoning

   1. Option (A):

      Calcium and magnesium are highly electropositive metals. In aqueous solution, water is reduced more easily than $\mathrm{Ca}^{2+}$ or $\mathrm{Mg}^{2+}$ (standard reduction potentials: $\mathrm{Ca}^{2+}/\mathrm{Ca} = -2.87\,\mathrm{V}$, $\mathrm{Mg}^{2+}/\mathrm{Mg} = -2.37\,\mathrm{V}$, while $2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow \mathrm{H}_2 + 2\mathrm{OH}^-$ is about $-0.83\,\mathrm{V}$). So electrolysis of aqueous solutions gives $\mathrm{H}_2$ at the cathode, not the metal. They are produced by electrolysis of molten salts. Hence (A) is false.

   2. Option (B):

      In $\mathrm{ClO}_3^{-}$, chlorine has oxidation state +5. Chlorine can exist in states from −1 to +7. Since +5 is intermediate, $\mathrm{ClO}_3^{-}$ can disproportionate, e.g., in hot concentrated alkali:

   $4\mathrm{ClO}_3^{-} \rightarrow 3\mathrm{ClO}_4^{-} + \mathrm{Cl}^{-}$

   Here chlorine goes from +5 to +7 (oxidation) and to −1 (reduction). So (B) is false.

   3. Option (C): …

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3. KCET 2024Set B-21 markMCQ
   Q.In the reaction between moist SO$_2$ and acidified permanganate solution : (A) SO$_2$ is oxidised to SO$_4^{2-}$, MnO$_4^-$ is reduced to Mn$^{2+}$ (B) SO$_2$ is reduced to S, MnO$_4^-$ is oxidised to MnO$_4$ (C) SO$_2$ is oxidised to SO$_3^{2-}$, MnO$_4^-$ is reduced to MnO$_2$ (D) SO$_2$ is reduced to H$_2$S, MnO$_4^-$ is oxidised to MnO$_4$

   ›Reveal solutionSolution

   In any redox pair the oxidising agent is reduced: acidified permanganate goes to $Mn^{2+}$, which forces $SO_2$ to be oxidised to $SO_4^{2-}$.

   Step 1 — Identify who oxidises whom

   $KMnO_4$ in acid medium is one of the strongest common oxidising agents. Being the oxidising agent, it is itself reduced. $SO_2$ (sulphur in the intermediate oxidation state $+4$) can go up to $+6$, so it acts as the reducing agent and is oxidised.

   Step 2 — Track the oxidation numbers

   Manganese in $MnO_4^-$: let $x$ be Mn's oxidation number.

   $x + 4(-2) = -1 \;\Rightarrow\; x = +7$

   In acidic medium permanganate is reduced by 5 electrons all the way to $Mn^{2+}$ ($+7 \to +2$). (Only in neutral/alkaline medium does it stop at $MnO_2$, Mn $= +4$ — the trap in option (C).)

   $MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$

   Sulphur in $SO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$. In $SO_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$. So sulphur loses 2 electrons:

   $SO_2 + 2H_2O \longrightarrow SO_4^{2-} + 4H^+ + 2e^-$

   (The "moist" in the question matters — the water is a reactant supplying the extra oxygens.)

   Step 3 — Combine (balance electrons: $\times 2$ and $\times 5$)

   $2MnO_4^- + 5SO_2 + 2H_2O \longrightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$ …

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4. COMEDK 2024Set 2024-A1 markMCQ
   Q.Identify the oxidation reaction in which acidified $\mathrm{KMnO}_4$ is required (A) Conversion of iodide ions to iodine (B) Oxidation of iodide ions to iodate ions (C) Oxidation of manganous salt to manganese dioxide (D) Conversion of thiosulphate ions to sulphate ions

   ›Reveal solutionSolution

   The reaction that specifically needs an acidic (acidified) medium is the oxidation of $I^-$ to $I_2$; the other conversions occur in neutral or faintly alkaline medium.

   Behaviour of $KMnO_4$ depends on the medium:

   • Acidic medium ($MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$): oxidises iodide to iodine — $10I^- + 2MnO_4^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$. …

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5. KCET 2023Set D-21 markMCQ
   Q.For the formation of which compound in Ellingham diagram $\Delta G^\circ$ becomes more and more negative with increase in temperature? (A) CO (B) FeO (C) ZnO (D) Cu$_2$O

   ›Reveal solutionSolution

   The slope of an Ellingham line is $-\Delta S^\circ$; only the C → CO formation has a positive $\Delta S^\circ$ (gas moles increase), so only its $\Delta G^\circ$ becomes more negative as $T$ rises.

   1. The Ellingham diagram in one equation

   An Ellingham diagram plots $\Delta G^{\circ}$ of oxide formation against $T$. From the Gibbs–Helmholtz relation:

   $\Delta G^{\circ} = \Delta H^{\circ} - T\,\Delta S^{\circ}$

   Treating $\Delta H^\circ$ and $\Delta S^\circ$ as roughly constant, this is a straight line whose

   $\text{slope} = -\Delta S^{\circ}$

   So the sign of $\Delta S^{\circ}$ decides everything:

   • $\Delta S^{\circ} < 0 \Rightarrow$ positive slope ($\Delta G^\circ$ becomes less negative with $T$)
   • $\Delta S^{\circ} > 0 \Rightarrow$ negative slope ($\Delta G^\circ$ becomes more negative with $T$) ← what we want

   2. Entropy change of each formation reaction

   Metal oxides — (B) FeO, (C) ZnO, (D) $\mathrm{Cu_2O}$:

   $2\mathrm{Fe(s)} + \mathrm{O_2(g)} \to 2\mathrm{FeO(s)}$

   $2\mathrm{Zn(s)} + \mathrm{O_2(g)} \to 2\mathrm{ZnO(s)}$

   $4\mathrm{Cu(s)} + \mathrm{O_2(g)} \to 2\mathrm{Cu_2O(s)}$

   In each case 1 mole of gaseous $\mathrm{O_2}$ is consumed and no gas is produced — gaseous randomness is destroyed, so $\Delta S^{\circ}$ is negative. Their lines slope upward: $\Delta G^\circ$ becomes less negative as $T$ increases.

   (The ZnO line does bend even more steeply upward above the boiling point of Zn, but it never turns downward.)

   Carbon monoxide — (A) CO:

   $2\mathrm{C(s)} + \mathrm{O_2(g)} \to 2\mathrm{CO(g)}$ …

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6. COMEDK 2023Set 2023-E1 markMCQ
   Q.In neutral medium $\mathrm{KMnO}_4$ oxidises $\mathrm{MnSO}_4$ to _________ (A) $\mathrm{Mn}_2 \mathrm{O}_3$ (B) $\mathrm{Mn}_3 \mathrm{O}_4$ (C) $\mathrm{MnO}_2$ (D) $\mathrm{K}_2 \mathrm{MnO}_4$

   ›Reveal solutionSolution

   (In acidic medium MnO4- goes to Mn2+; in strongly alkaline medium to the green manganate MnO4^2-.)

   Concept: in NEUTRAL (or faintly alkaline) medium, permanganate is reduced from Mn(VII) to Mn(IV), i.e. to MnO2 (a brown precipitate). This is the classic reaction with manganese(II) salts:

   2 KMnO4 + 3 MnSO4 + 2 H2O -> 5 MnO2 + K2SO4 + 2 H2SO4 …

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7. COMEDK 2023Set 2023-M1 markMCQ
   Q.In dilute alkaline solution $\mathrm{MnO}_4^{-}$ changes to (A) $\mathrm{MnO}_2$ (B) $\mathrm{MnO}_4^{2-}$ (C) $\mathrm{MnO}$ (D) $\mathrm{Mn}_2 \mathrm{O}_3$

   ›Reveal solutionSolution

   [!TLDR]

   Dilute (faintly) alkaline permanganate is reduced to brown MnO$_2$, so the answer is (A).

   Concept

   The fate of MnO$_4^-$ depends on the medium (CBSE/NCERT Class 12 d- and f-Block Elements). In acidic medium it goes to Mn$^{2+}$ (+2); in neutral or faintly/dilute alkaline medium it goes to MnO$_2$ (+4); only in strongly alkaline medium does it stop at the manganate ion MnO$_4^{2-}$ (+6).

   Solution

   In dilute alkaline solution the relevant half-reaction is a three-electron reduction:

   $\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$ …

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8. KCET 2022Set B-31 markMCQ
   Q.In which of the following compounds, an element exhibits two different oxidation states? (A) $N_2H_4$ (B) $N_3H$ (C) $NH_2CONH_2$ (D) $NH_4NO_3$

   ›Reveal solutionSolution

   Assign oxidation numbers to nitrogen in each compound; only $NH_4NO_3$ contains nitrogen in two chemically distinct sites, giving it two different oxidation states in the same formula unit.

   Concept. An element can show two oxidation states within one compound only if it occupies two chemically different positions (e.g. a cation and an anion). We assign oxidation numbers using $H=+1$, $O=-2$, and the requirement that the charges sum to the ion/molecule charge.

   Step 1 — (A) Hydrazine, $N_2H_4$.

   $2x + 4(+1) = 0 \Rightarrow x = -2$

   Both nitrogens are equivalent ($H_2N-NH_2$): a single oxidation state $-2$.

   Step 2 — (B) Hydrazoic acid, $HN_3$ (written $N_3H$).

   $3x + (+1) = 0 \Rightarrow x = -\tfrac{1}{3}$

   This is an average (fractional) value; conventionally this compound is quoted as having the single average state $-1/3$, not two clean, different oxidation states, and CBSE/KCET treat the intended answer as the salt below.

   Step 3 — (C) Urea, $NH_2CONH_2$.

   Both $-NH_2$ groups are identical, each nitrogen is $-3$. One oxidation state only.

   Step 4 — (D) Ammonium nitrate, $NH_4NO_3$. …

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9. KCET 2021Set B-21 markMCQ
   Q.A colourless, neutral, paramagnetic oxide of Nitrogen ‘P’ on oxidation gives reddish brown gas Q. Q on cooling gives colourless gas R. R on reaction with P gives blue solid S. Identify P, Q, R, S, respectively (A) N$_2$O NO NO$_2$ N$_2$O$_5$ (B) N$_2$O NO$_2$ N$_2$O$_4$ N$_2$O$_3$ (C) NO NO$_2$ N$_2$O$_4$ N$_2$O$_3$ (D) NO NO N$_2$O$_4$ N$_2$O$_5$

   ›Reveal solutionSolution

   Paramagnetic + colourless + neutral pins P = NO; then NO → NO₂ (brown) → N₂O₄ (colourless dimer) → N₂O₃ (blue solid) with NO.

   1. Identify P from the three clues.

   The key word is paramagnetic — it requires an unpaired electron.

   • $\mathrm{N_2O}$ has an even number of electrons and is diamagnetic. So P cannot be $\mathrm{N_2O}$ — this alone eliminates options (A) and (B).
   • NO has an odd total electron count (7 + 8 = 15), leaving one unpaired electron in a $\pi^*$ orbital $\Rightarrow$ paramagnetic. It is also colourless (as a gas) and neutral (neither acidic nor basic).

   $\boxed{P = \mathrm{NO}}$

   2. Q — oxidation of NO.

   Nitric oxide is oxidised by air/oxygen at once (this is the brown fume you see when NO meets air):

   $2\mathrm{NO} + \mathrm{O_2} \longrightarrow 2\mathrm{NO_2} \quad (\text{reddish-brown gas})$

   $\boxed{Q = \mathrm{NO_2}}$

   This eliminates option (D), which repeats NO as Q.

   3. R — cooling $\mathrm{NO_2}$.

   $\mathrm{NO_2}$ is itself an odd-electron (brown, paramagnetic) molecule. On cooling it dimerises — the two odd electrons pair up in an N–N bond — and the colour disappears:

   $2\mathrm{NO_2} \;\rightleftharpoons\; \mathrm{N_2O_4} \quad (\text{colourless, diamagnetic})$

   $\boxed{R = \mathrm{N_2O_4}}$

   4. S — R with P gives the blue solid. …

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10. KCET 2021Set B-21 markMCQ
    Q.Which of the following is not true for oxidation? (A) addition of oxygen (B) addition of electronegative element (C) removal of hydrogen (D) removal of electronegative element

    ›Reveal solutionSolution

    Recall the four classical definitions of oxidation; the odd one out — removal of an electronegative element — is actually a reduction.

    Step 1 — The classical definitions

    Before the electron-transfer definition, oxidation was defined operationally. An element/compound is oxidised when it undergoes any of:

    #	Oxidation is …	Example
    1	addition of oxygen	$2Mg + O_2 \to 2MgO$
    2	addition of an electronegative element	$Fe + S \to FeS$ ; $2Fe + 3Cl_2 \to 2FeCl_3$
    3	removal of hydrogen	$H_2S + Cl_2 \to 2HCl + S$
    4	removal of an electropositive element	$2KI + H_2O_2 \to 2KOH + I_2$

    The modern unifying statement behind all four: oxidation is loss of electrons / an increase in oxidation number. Each of the four moves above raises the oxidation number of the species in question.

    Step 2 — Screen the options against this list

    • (A) addition of oxygen — definition 1. Oxygen is electronegative; adding it pulls electron density away, raising the oxidation number. TRUE for oxidation.
    • (B) addition of electronegative element — definition 2. Same logic as oxygen. TRUE for oxidation.
    • (C) removal of hydrogen — definition 3. Hydrogen is the electropositive partner; losing it raises the oxidation number. TRUE for oxidation. …

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11. KCET 2020Set A-11 markMCQ
    Q.If an aqueous solution of NaF is electrolyzed between inert electrodes, the product obtained at anode is (A) $O_2$ (B) $F_2$ (C) $H_2$ (D) Na

    ›Reveal solutionSolution

    During electrolysis of an aqueous NaF solution, water is oxidised more easily than fluoride ions at the anode, so oxygen gas ($O_2$) is produced instead of fluorine.

    The key idea here is selective discharge of ions — not every ion present in solution actually gets discharged at the electrodes. In aqueous electrolysis, water itself can compete with the dissolved ions. The anode is where oxidation happens (loss of electrons), so we compare the ease of oxidation of the species present: fluoride ions ($F^-$) and water molecules ($H_2O$).

    Fluoride ions are notoriously difficult to oxidise. In fact, fluorine gas is such a strong oxidising agent that its production from aqueous solution is practically impossible — water would be oxidised first. The standard electrode potentials tell the story clearly.

    Standard oxidation potentials (at 25°C, 1 M, 1 atm):

    $\begin{aligned} 2F^- &\rightarrow F_2 + 2e^- \quad E^\circ = -2.87\ \text{V} \\ 2H_2O &\rightarrow O_2 + 4H^+ + 4e^- \quad E^\circ = -1.23\ \text{V} \end{aligned}$

    A less negative (or more positive) oxidation potential means the species is more easily oxidised. Here, water has a much less negative value ($-1.23$ V vs $-2.87$ V), so water is oxidised preferentially.

    Let’s walk through the reasoning step by step.

    1. Identify all species present in solution. NaF dissociates completely in water:

    $\text{NaF} \rightarrow \text{Na}^+ + \text{F}^-$

    So the solution contains $\text{Na}^+$, $\text{F}^-$, and water molecules ($H_2O$). At the anode (positive electrode), oxidation occurs — we look for species that can lose electrons.

    2. List possible oxidation reactions at the anode.
       • Oxidation of fluoride ions:

    $2F^- \rightarrow F_2 + 2e^-$

    • Oxidation of water:

    $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

    3. Compare the ease of oxidation using standard potentials. The more positive (or less negative) the oxidation potential, the more readily the species gives up electrons. Water’s oxidation potential ($-1.23$ V) is significantly higher (less negative) than that of fluoride ions ($-2.87$ V). This means water is much easier to oxidise. …

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12. KCET 2020Set A-11 markMCQ
    Q.Sulphide ore on roasting gives a gas X. X reacts with $Cl_2$ in the presence of activated charcoal to give Y. Y is : (A) $SOCl_2$ (B) $SO_2Cl_2$ (C) $S_2Cl_2$ (D) $SCl_6$

    ›Reveal solutionSolution

    Roasting a sulphide gives $\mathrm{SO_2}$; $\mathrm{SO_2+Cl_2}$ over activated charcoal is the standard preparation of sulphuryl chloride $\mathrm{SO_2Cl_2}$.

    Step 1 — Identify X (roasting).

    Roasting = heating an ore strongly in the presence of excess air, used for sulphide ores. Sulphur is oxidised to $\mathrm{SO_2}$:

    $2\,\mathrm{ZnS}+3\,\mathrm{O_2}\ \xrightarrow{\ \Delta\ }\ 2\,\mathrm{ZnO}+2\,\mathrm{SO_2}\uparrow$

    (The same happens with $\mathrm{PbS}$, $\mathrm{Cu_2S}$, etc.) The evolved gas is therefore

    $X=\mathrm{SO_2}$

    Step 2 — Identify Y ($\mathrm{SO_2}+\mathrm{Cl_2}$).

    The classic preparation of sulphuryl chloride is the direct union of sulphur dioxide and chlorine in the presence of a catalyst — activated charcoal (or camphor):

    $\mathrm{SO_2}+\mathrm{Cl_2}\ \xrightarrow[\text{activated charcoal}]{}\ \mathrm{SO_2Cl_2}$

    Mechanistically, sulphur in $\mathrm{SO_2}$ is in the $+4$ state and is oxidised to $+6$ in $\mathrm{SO_2Cl_2}$; chlorine is reduced from $0$ to $-1$. The charcoal simply provides a surface — it is a catalyst, not a reactant.

    Step 3 — Rule out the other options. …

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