0.1 M solution of KI reacts with excess of H2SO4 and KIO3 solution according to the equation: Identify the correct…

Year	2026	2025	2024	Total
Questions	4	4	6	14

Q27 jee_main_2025_28_jan_morning Titration and Autocatalysis

Given below are two statements : Statement I: In the oxalic acid vs

KMnO₄

(in the presence of dil

H₂SO₄

) titration the solution needs to be heated initially to

60°C

, but no heating is required in Ferrous ammonium sulphate (FAS) vs

KMnO₄

titration (in the presence of dil

H₂SO₄

) Statement II : In oxalic acid vs

KMnO₄

titration, the initial formation of

MnSO₄

takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs

KMnO₄

, heating oxidizes

Fe²⁺

into

Fe³⁺

by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below:

A. $Statement I is false but Statement II is true.$
B. $Both Statement I and Statement II are true.$
C. $Statement I is true but Statement II is false$
D. $Both Statement I and Statement II are false.$

Solution

Related Formula

Oxalic acid titration equation:

2MnO₄^- + 5(COO)₂²⁻ + 16H^+ arrow 10CO₂ + 2Mn²⁺ + 8H₂O

Core Logic

Statement I is true because the reaction between oxalic acid and $KMnO₄$ $\mathrm{KMnO}_4$ is slow at room temperature and requires initial heating to around $60°C$ $60^{\circ}\mathrm{C}$. No heating is required for FAS titration.

Statement II is true because $Mn²⁺$ $\mathrm{Mn}^{2+}$ acts as an autocatalyst. Heating FAS would cause atmospheric oxygen to prematurely oxidize $Fe²⁺$ $\mathrm{Fe}^{2+}$ to $Fe³⁺$ $\mathrm{Fe}^{3+}$, leading to experimental errors.

Pattern Recognition

Sees: Oxalic acid vs FAS titration with permanganate. Shortcut: Oxalic acid requires heat + autocatalysis. FAS titration must be kept cold to avoid aerial oxidation.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q28 jee_main_2025_28_jan_morning Types of Redox Reactions

Match the List-I with List-II

List-I (Redox Reaction)		List-II (Type of Redox Reaction)
A	$CH₄(g) + 2O₂(g) Δ CO₂(g) + 2H₂O(l)$	(I)	Disproportionation reaction
B	$2NaH(s) Δ 2Na(s) + H₂(g)$	(II)	Combination reaction
C	$V₂O₅(s) + 5Ca(s) arrow 2V(s) + 5CaO(s)$	(III)	Decomposition reaction
D	$2H₂O₂(aq) Δ 2H₂O(l) + O₂(g)$	(IV)	Displacement reaction

Choose the correct answer from the options given below:

A. $A-II, B-III, C-IV, D-I$ $\text{A-II, B-III, C-IV, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$
D. $A-IV, B-I, C-II, D-III$ $\text{A-IV, B-I, C-II, D-III}$

Solution

Core Logic

Let us evaluate each redox pair:

A: $CH₄ + 2O₂ arrow CO₂ + 2H₂O$ involves elements combining with oxygen (combustion/combination), corresponding to (II).
B: $2NaH arrow 2Na + H₂$ shows a single compound breaking down into its elements, matching (III) Decomposition.
C: $V₂O₅ + 5Ca arrow 2V + 5CaO$ represents calcium displacing vanadium from its oxide, matching (IV) Displacement.
D: $2H₂O₂ arrow 2H₂O + O₂$ represents oxygen in $-1$ $-1$ state going simultaneously to $-2$ $-2$ and $0$ $0$ states, which is (I) Disproportionation.

Pattern Recognition

Sees: Classic classification table of redox classes. Shortcut: Identify $H₂O₂$ $\mathrm{H}_2\mathrm{O}_2$ decomposition as the benchmark example of disproportionation (D-I) to immediately eliminate wrong choices.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q26 jee_main_2025_24_jan_evening Standard Reduction Potentials

Based on the data given below:

E°_Cr₂O₇²⁻/Cr³⁺ = 1.33 V E°_Cl₂/Cl⁻ = 1.36 V

E°_MnO₄⁻/Mn²⁺ = 1.51 V E°Cr³⁺/Cr = -0.74 V

the strongest reducing agent is :

A. $Mn²⁺$ $\mathrm{Mn}^{2+}$
B. $Cr$ $\mathrm{Cr}$
C. $MnO₄⁻$ $\mathrm{MnO}_{4}^{-}$
D. $Cl⁻$ $\mathrm{Cl}^{-}$

Solution

Related Formula

$Reducing Power \propto 1Standard Reduction Potential (E°red)$

Core Logic

A stronger reducing agent undergoes oxidation more easily, which corresponds to the lowest standard reduction potential value among the given options.

Comparing the given values:

$E°_MnO₄⁻/Mn²⁺ = +1.51 V$
$E°_Cl₂/Cl⁻ = +1.36 V$
$E°_Cr₂O₇²⁻/Cr³⁺ = +1.33 V$
$E°Cr³⁺/Cr = -0.74 V$

Since $Cr³⁺/Cr$ $\mathrm{Cr}^{3+}/\mathrm{Cr}$ has the lowest standard reduction potential ( $-0.74 V$ $-0.74\mathrm{ V}$), elemental metallic $Cr$ $\mathrm{Cr}$ is the most easily oxidized and is therefore the strongest reducing agent.

Pattern Recognition

To find the strongest reducing agent, simply look for the lowest or most negative reduction potential. Ensure you pick the species on the right side of the reduction half-reaction (the reduced form, which will act as the reducer).

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: Electrochemistry

Q64 jee_main_2024_01_february_morning Types of Redox Reactions

Which of the following reactions are disproportionation reactions? (A)

Cu^+ arrow Cu²⁺ + Cu

(B)

3MnO₄²⁻ + 4H^+ arrow 2MnO₄^- + MnO₂ + 2H₂O

(C)

2KMnO₄ arrow K₂MnO₄ + MnO₂ + O₂

(D)

2MnO₄^- + 3Mn²⁺ + 2H₂O arrow 5MnO₂ + 4H^+

Choose the correct answer from the options given below:

A. $(A), (B)$ $\text{(A), (B)}$
B. $(B), (C), (D)$ $\text{(B), (C), (D)}$
C. $(A), (B), (C)$ $\text{(A), (B), (C)}$
D. $(A), (D)$ $\text{(A), (D)}$

Solution

Core Logic

When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, the species is said to undergo a disproportionation reaction.

Step 1: Check Reaction (A)

Reaction (A): $Cu^+ arrow Cu²⁺ + Cu$ $Cu^+ \rightarrow Cu^{2+} + Cu$ Oxidation state of $Cu$ $Cu$ goes from +1 to +2 (oxidation) and +1 to 0 (reduction). This is disproportionation.

Step 2: Check Reaction (B)

Reaction (B): $3MnO₄²⁻ + 4H^+ arrow 2MnO₄^- + MnO₂ + 2H₂O$ Oxidation state of $Mn$ $Mn$ in $MnO₄²⁻$ $MnO_4^{2-}$ is +6. In $MnO₄^-$ $MnO_4^-$ it is +7, and in $MnO₂$ $MnO_2$ it is +4. The +6 state oxidizes to +7 and reduces to +4. This is disproportionation.

Step 3: Check Reactions (C) and (D)

Reaction (C): $2KMnO₄ arrow K₂MnO₄ + MnO₂ + O₂$ $Mn$ $Mn$ in $KMnO₄$ $KMnO_4$ is +7, forming +6 ( $K₂MnO₄$ $K_2MnO_4$) and +4 ( $MnO₂$ $MnO_2$). Here, Oxygen is oxidized from -2 to 0. This is an intramolecular redox reaction, NOT disproportionation.

Reaction (D): $2MnO₄^- + 3Mn²⁺ + 2H₂O arrow 5MnO₂ + 4H^+$ $Mn$ $Mn$ +7 and $Mn$ $Mn$ +2 both convert to $Mn$ $Mn$ +4. This is a comproportionation reaction, the reverse of disproportionation.

Pattern Recognition

Disproportionation requires the SAME element in a SINGLE oxidation state in the reactants to split into a HIGHER and LOWER oxidation state in the products.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q87 jee_main_2024_01_february_morning Oxidation Number

Lowest Oxidation number of an atom in a compound

A₂B

is -2. The number of an electron in its valence shell is

Numerical Answer. Answer: 6 to 6

Solution

Core Logic

In the binary compound $A₂B$ $A_2B$, the atom 'B' exhibits a -2 oxidation state, which is its lowest possible oxidation state (meaning it has achieved a stable noble gas configuration by accepting 2 electrons).

A₂B arrow 2A^+ + B²⁻

If the element B gains 2 electrons to complete its octet (8 electrons), it must originally have $8 - 2 = 6$ $8 - 2 = 6$ electrons in its valence shell in the neutral atomic state.

Step 1: Determine Valence Electrons

$B²⁻$ $B^{2-}$ has a complete octet. Neutral state B = $8 - 2 = 6$ $8 - 2 = 6$ valence electrons. (This corresponds to Group 16 elements like Oxygen or Sulfur).

Pattern Recognition

Lowest negative oxidation state = (Group Number - 18) for p-block. If lowest ox state is -2, then Group number is $18 - 2 = 16$ $18 - 2 = 16$. Group 16 elements have 6 valence electrons.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Solution & Explanation

Core Logic

Step 1: Check Statement B and C

Pattern Recognition

Chapter Mix

Reference Study Guides

More Redox Reactions Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Check Reaction (A)

Step 2: Check Reaction (B)

Step 3: Check Reactions (C) and (D)

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Determine Valence Electrons

Pattern Recognition

Chapter Mix

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