Based on the data given below: the strongest reducing agent is :

Q: Based on the data given below: the strongest reducing agent is :

Reducing Power propto (1/Standard Reduction Potential ) (E_red) 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_4-/Mn2+ = +1.51 V * E_2/Cl- = +1.36 V * E_2O72-/Cr3+ = +1.33 V * E^3+/Cr = -0.74 V Since Cr3+/Cr has the lowest standard reduction potential (-0.74 V), elemental metallic Cr is the most easily oxidized and is therefore the strongest reducing agent. To f

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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