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 , 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:

Oxalic acid titration equation: Statement I is true because the reaction between oxalic acid and KMnO4 is slow at room temperature and requires initial heating to around 60. No heating is required for FAS titration. Statement II is true because Mn2+ acts as an autocatalyst. Heating FAS would cause atmospheric oxygen to prematurely oxidize Fe2+ to Fe3+, leading to experimental errors. Sees: Oxalic acid vs FAS titration with permanganate. Shortcut: Oxalic acid requires heat + autocatalysis. FAS titration must be kept cold to avoid aerial oxidation. Class 11 Chemistry: Redox Reactions

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 , but no heating is required in Ferr

Oxalic acid titration equation: Statement I is true because the reaction between oxalic acid and KMnO4 is slow at room temperature and requires initial heating to around 60. No heating is required for FAS titration. Statement II is true because Mn2+ acts as an autocatalyst. Heating FAS would cause atmospheric oxygen to prematurely oxidize Fe2+ to Fe3+, leading to experimental errors. Sees: Oxalic acid vs FAS titration with permanganate. Shortcut: Oxalic acid requires heat + autocatalysis. FAS ti

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 , 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:

Oxalic acid titration equation: Statement I is true because the reaction between oxalic acid and KMnO4 is slow at room temperature and requires initial heating to around 60. No heating is required for FAS titration. Statement II is true because Mn2+ acts as an autocatalyst. Heating FAS would cause atmospheric oxygen to prematurely oxidize Fe2+ to Fe3+, leading to experimental errors. Sees: Oxalic acid vs FAS titration with permanganate. Shortcut: Oxalic acid requires heat + autocatalysis. FAS titration must be kept cold to avoid aerial oxidation. Class 11 Chemistry: Redox Reactions

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 , but no heating is required in Ferr

Oxalic acid titration equation: Statement I is true because the reaction between oxalic acid and KMnO4 is slow at room temperature and requires initial heating to around 60. No heating is required for FAS titration. Statement II is true because Mn2+ acts as an autocatalyst. Heating FAS would cause atmospheric oxygen to prematurely oxidize Fe2+ to Fe3+, leading to experimental errors. Sees: Oxalic acid vs FAS titration with permanganate. Shortcut: Oxalic acid requires heat + autocatalysis. FAS ti

Given below are two statements : Statement I: In the oxalic acid vs KMnO4 (in the presence of dil H2SO4 ) titration the…

Year	2026	2025	2024	Total
Questions	4	4	6	14

Q54 jee_main_2026_21_jan_evening Stoichiometry and Redox Titrations

Aqueous

HCl

reacts with

MnO₂(s)

to form

MnCl₂(aq)

,

Cl₂(g)

and

H₂O(l)

. What is the weight (in g) of

Cl₂

liberated when 8.7 g of

MnO₂(s)

is reacted with excess aqueous

HCl

solution? (Given Molar mass in

g mol⁻¹: Mn = 55, Cl = 35.5, O = 16, H = 1

)

A. $(1) 7.1$ $(1) \ 7.1$
B. $(2) 71$ $(2) \ 71$
C. $(3) 21.3$ $(3) \ 21.3$
D. $(4) 14.2$ $(4) \ 14.2$

Solution

Related Formula

Moles = Given MassMolar Mass

Core Logic

Balanced reaction:

MnO₂ + 4HCl arrow MnCl₂ + Cl₂ + 2H₂O

Molar mass of $MnO₂ = 55 + 32 = 87 g/mol$ . Moles of $MnO₂ = (8.7)/(87) = 0.1 mol$ .

From stoichiometry, 1 mole of $MnO₂$ $\text{MnO}_2$ produces 1 mole of $Cl₂$ $\text{Cl}_2$. Therefore, moles of $Cl₂$ $\text{Cl}_2$ produced = $0.1 mol$ $0.1 \text{ mol}$.

Step 1: Final Calculation

Weight of $Cl₂ = 0.1 \times 71 = 7.1 g$ .

Pattern Recognition

Sees: stoichiometry calculation with manganese dioxide and hydrochloric acid. Trap: Forgetting stoichiometric coefficients for chlorine gas.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q73 jee_main_2026_23_january_evening Titration

200 cc

of

x \times 10⁻³ M

potassium dichromate is required to oxidise

750 cc

of

0.6 M

Mohr's salt solution in acidic medium. Here x =

Numerical Answer. Answer: 375 to 375

Solution

Related Formula

Equivalents of Oxidizing Agent = Equivalents of Reducing Agent

N₁ V₁ = N₂ V₂

Normality (N) = Molarity (M) \times n-factor

Core Logic

In acidic medium, potassium dichromate ( $K₂Cr₂O₇$ $K_2Cr_2O_7$) reduces from $Cr⁺⁶$ $Cr^{+6}$ to $Cr⁺³$ $Cr^{+3}$. Change in oxidation state = 3 per Cr atom. For $K₂Cr₂O₇$ $K_2Cr_2O_7$, $n-factor = 2 \times 3 = 6$ $n\text{-factor} = 2 \times 3 = 6$. Normality of dichromate = $6 \times (x \times 10⁻³) N$ $6 \times (x \times 10^{-3}) \text{ N}$.

Mohr's salt is $(NH₄)₂Fe(SO₄)₂ \cdot 6H₂O$ $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$. The active reducing species is $Fe²⁺$ $Fe^{2+}$, which oxidizes to $Fe³⁺$ $Fe^{3+}$. Change in oxidation state = 1. For Mohr's salt, $n-factor = 1$ $n\text{-factor} = 1$. Normality of Mohr's salt = $1 \times 0.6 = 0.6 N$ $1 \times 0.6 = 0.6 \text{ N}$.

Step 1: Applying Law of Equivalence

Milliequivalents of K₂Cr₂O₇ = Milliequivalents of Mohr's salt

(6 \times x \times 10⁻³ N) \times (200 cc) = (0.6 N) \times (750 cc)

$1.2x = 450$ $1.2x = 450$

x = (450)/(1.2) = 375

Pattern Recognition

Memorize the $n$ $n$-factors for redox titrations: $K₂Cr₂O₇$ $K_2Cr_2O_7$ in acid is always $n=6$ $n=6$. $KMnO₄$ $KMnO_4$ in acid is $n=5$ $n=5$. Mohr's salt ( $Fe²⁺$ $Fe^{2+}$) is always $n=1$ $n=1$.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: d and f Block Elements

Q75 jee_main_2026_24_january_morning Balancing Redox Equations

X and Y are the number of electrons involved, respectively during the oxidation of

I^-

to

I₂

and

S²⁻

to S by acidified

K₂Cr₂O₇

. The value of X + Y is ____.

Numerical Answer. Answer: 12 to 12

Solution

Core Logic

Dichromate ion ( $Cr₂O₇²⁻$ $Cr_2O_7^{2-}$) in acidic medium acts as a strong oxidizing agent, getting reduced to $Cr³⁺$ $Cr^{3+}$. This reduction process consumes $6$ $6$ electrons per dichromate ion: $Cr₂O₇²⁻ + 14H^+ + 6e^- arrow 2Cr³⁺ + 7H₂O$

Reaction 1: Oxidation of Iodide $6I^- arrow 3I₂ + 6e^-$ $6I^- \rightarrow 3I_2 + 6e^-$ Overall balanced reaction: $Cr₂O₇²⁻ + 14H^+ + 6I^- arrow 2Cr³⁺ + 3I₂ + 7H₂O$ Number of electrons involved (X) = 6.

Reaction 2: Oxidation of Sulfide $3S²⁻ arrow 3S + 6e^-$ $3S^{2-} \rightarrow 3S + 6e^-$ Overall balanced reaction: $Cr₂O₇²⁻ + 14H^+ + 3S²⁻ arrow 2Cr³⁺ + 3S + 7H₂O$ Number of electrons involved (Y) = 6.

Step 1: Final Summation

Sum of $X + Y = 6 + 6 = 12$ $X + Y = 6 + 6 = 12$.

Pattern Recognition

Dichromate ( $Cr⁶⁺$ $Cr^{6+}$ to $Cr³⁺$ $Cr^{3+}$, two atoms) always demands 6 electrons in a balanced equation block. The reducing agent must therefore supply exactly 6 electrons to match.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The d- and f-Block Elements

Q71 jee_main_2026_28_january_morning Iodometric Titrations

500~mL

of

1.2~M~KI

solution is mixed with

500~mL

of

0.2~M~KMnO₄

solution in basic medium. The liberated iodine was titrated with standard

0.1~M~Na₂S₂O₃

solution in the presence of starch indicator till the blue color disappeared. The volume (in

L

) of

Na₂S₂O₃

consumed is _____.

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Equivalents of KMnO₄ = Equivalents of I₂ = Equivalents of Na₂S₂O₃

Core Logic

In a basic medium, permanganate ( $MnO₄^-$ $MnO_4^-$) is reduced to $MnO₂$ $MnO_2$, with an n-factor of 3. Iodide ( $I^-$ $I^-$) is oxidized to $I₂$ $I_2$.\nThe liberated $I₂$ $I_2$ is titrated with thiosulfate ( $S₂O₃²⁻$ $S_2O_3^{2-}$), which is oxidized to tetrathionate ( $S₄O₆²⁻$ $S_4O_6^{2-}$) with an n-factor of 1 per thiosulfate molecule.

Step 1: Determine the Limiting Reagent

Equivalents of $KMnO₄ = M \times V(L) \times n-factor = 0.2 \times 0.5 \times 3 = 0.3~eq$ .\nEquivalents of $KI = M \times V(L) \times n-factor = 1.2 \times 0.5 \times 1 = 0.6~eq$ .\n $KMnO₄$ $KMnO_4$ is the limiting reagent. It will completely react to liberate exactly $0.3$ $0.3$ equivalents of $I₂$ $I_2$.

Step 2: Titration with Thiosulfate

Equivalents of $Na₂S₂O₃$ $Na_2S_2O_3$ consumed = Equivalents of $I₂$ $I_2$ liberated = $0.3~eq$ $0.3\mathrm{~eq}$.\n $0.3 = M \times V(L) \times n-factor$ \n $0.3 = 0.1 \times V \times 1$ $$0.3 = 0.1 \times V \times 1$$\n $V = (0.3)/(0.1) = 3~L$ $$V = \frac{0.3}{0.1} = 3\mathrm{~L}$$

Pattern Recognition

In iodometric titrations, equivalents transfer sequentially 1:1. The n-factor of $KMnO₄$ $KMnO_4$ shifts radically based on medium ( $5$ $5$ acidic, $3$ $3$ neutral/weakly basic, $1$ $1$ strongly basic). Here, basic medium means $n=3$ $n=3$.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q34 jee_main_2025_29_jan_evening Stoichiometry and Equivalents

0.1 M

solution of KI reacts with excess of

H₂SO₄

and

KIO₃

solution according to the equation:

5I⁻ + IO₃⁻ + 6H⁺ arrow 3I₂ + 3H₂O

Identify the correct statements: (A)

200 mL

of KI solution reacts with

0.004 mol

of

KIO₃

(B)

200 mL

of KI solution reacts with

0.006 mol

of

H₂SO₄

(C)

0.5 L

of KI solution produced

0.005 mol

of

I₂

(D) Equivalent weight of

KIO₃

is equal to

Molecular weight5

Choose the correct answer from the options given below:

A. (A) and (D) only
B. (B) and (C) only
C. (A) and (B) only
D. (C) and (D) only

Solution

Core Logic

Let's find the moles of $I⁻$ $I^{-}$ in $200 mL$ $200\text{ mL}$ of $0.1 M$ $0.1\text{ M}$ KI:

Moles = 0.1 \times 0.2 = 0.02 mol

According to the stoichiometry: * $5 moles I⁻ arrow 1 mole IO₃⁻$ Therefore, $0.02 mol I⁻$ $0.02\text{ mol } I^{-}$ reacts with $(0.02)/(5) = 0.004 mol$ $\frac{0.02}{5} = 0.004\text{ mol}$ of $KIO₃$ $KIO_{3}$. Statement (A) is correct.

For Statement (D), Iodine goes from $+5$ $+5$ state in $IO₃⁻$ $IO_{3}^{-}$ to $0$ $0$ state in $I₂$ $I_{2}$. The change in oxidation state per iodine atom is $5$ $5$.

Equivalent weight = Molecular weight5

Statement (D) is correct.

Step 1: Check Statement B and C

For $200 mL$ $200\text{ mL}$ KI ( $0.02 mol$ $0.02\text{ mol}$):

5 moles I⁻ requires 3 moles H₂SO₄ implies 0.02 \times (3)/(5) = 0.012 mol

Hence, Statement (B) is false. For $0.5 L$ $0.5\text{ L}$ KI ( $0.05 mol$ $0.05\text{ mol}$):

0.05 \times (3)/(5) = 0.03 mol I₂

Hence, Statement (C) is false.

Pattern Recognition

Always focus on checking the n-factor calculation directly from oxidation states for quick elimination of equivalence statements.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Redox Reactions Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Law of Equivalence

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Summation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine the Limiting Reagent

Step 2: Titration with Thiosulfate

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Check Statement B and C

Pattern Recognition

Chapter Mix

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