Q.85 Chemistry · Section A ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 85 Solution

Topic: Electrochemistry · Official NTA Booklet Code T1

Question

Match List I (Conversion) with List II (Number of Faraday required). List I: A. 1 mol of H₂O to O₂; B. 1 mol of MnO₄⁻ to Mn²⁺; C. 1.5 mol of Ca from molten CaCl₂; D. 1 mol of FeO to Fe₂O₃. List II: I. 3F; II. 2F; III. 1F; IV. 5F. Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (3) : A-II, B-IV, C-I, D-III

📚 Electrochemistry Easy

📋 Given Parameters & Problem Setup

Conversion A 1 mol of H2O to O2

Conversion B 1 mol of MnO4⁻ to Mn²⁺

Conversion C 1.5 mol of Ca from molten CaCl2

Conversion D 1 mol of FeO to Fe2O3

⚡

The 30-Second Exam Hall Shortcut: ⚡ Faraday Charge Multiplication Hack

⏱️ Save ~40 seconds

Faradays required = (moles) × (electron transfer n)! H2O -> 1/2 O2 (2e⁻ -> 2F, II); MnO4⁻ -> Mn²⁺ (5e⁻ -> 5F, IV); 1.5 mol Ca²⁺ (1.5 × 2e⁻ -> 3F, I); 1 mol Fe²⁺ -> Fe³⁺ (1e⁻ -> 1F, III)!

1 Formula: Q = n_electrons × F for 1 mole of reaction.

2 Conversion A: H2O -> 1/2 O2 + 2 H⁺ + 2e⁻. Exactly 2 moles of electrons -> 2F (A-II).

3 Conversion B: Mn(+7)O4⁻ + 5e⁻ -> Mn²⁺. Oxidation state changes from +7 to +2 (5e⁻ transfer) -> 5F (B-IV).

4 Conversion C: Ca²⁺ + 2e⁻ -> Ca. 1 mole Ca requires 2F. For 1.5 moles Ca: 1.5 × 2 = 3F (C-I).

5 Conversion D: FeO (Fe²⁺) -> 1/2 Fe2O3 (Fe³⁺) + e⁻. Fe(+2) to Fe(+3) transfers 1e⁻ per mole of FeO -> 1F (D-III).

6 Match: A-II, B-IV, C-I, D-III -> Option (3) in 10 seconds!

🛡️

Why You Can 100% Trust This Shortcut: NCERT Chemistry · Electrochemistry (Faraday’s laws of electrolysis and charge required for oxidation/reduction).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Faraday’s Law Relating Electrons and Charge

One Faraday (1 F ≈ 96487 C) is the electrical charge carried by 1 mole of electrons. The number of Faradays required to oxidize or reduce a substance is given by:

Q (\text{in Faradays}) = n_{\text{substance}} \times |\Delta(\text{O.N.})| = n_{\text{substance}} \times z

Step 2

Calculation for Conversion A: 1 mol of H2O to O2

In water, oxygen has oxidation state -2. In O2, oxygen has oxidation state 0. Writing the half-reaction for 1 mole of H2O:

\text{H}_2\text{O} \longrightarrow \frac{1}{2}\text{O}_2 + 2\text{H}^\oplus + 2e^\ominus \implies 2\;\text{moles of } e^\ominus = 2\;\text{F} \quad (\text{II})

Step 3

Calculation for Conversion B: 1 mol of MnO4⁻ to Mn²⁺

In permanganate ion MnO4⁻, manganese is in the +7 oxidation state. In acidic reduction to Mn²⁺:

\overset{+7}{\text{Mn}}\text{O}_4^\ominus + 8\text{H}^\oplus + 5e^\ominus \longrightarrow \overset{+2}{\text{Mn}}^{2+} + 4\text{H}_2\text{O} \implies 5\;\text{moles of } e^\ominus = 5\;\text{F} \quad (\text{IV})

Step 4

Calculation for Conversion C: 1.5 mol of Ca from CaCl2

Calcium in molten CaCl2 is Ca²⁺. Deposition of 1 mole of Ca requires 2 moles of electrons:

\text{Ca}^{2+} + 2e^\ominus \longrightarrow \text{Ca}; \quad Q = 1.5\;\text{mol} \times 2 = 3\;\text{F} \quad (\text{I})

Step 5

Calculation for Conversion D: 1 mol of FeO to Fe2O3

In FeO, iron is in the +2 oxidation state. In Fe2O3, iron is in the +3 oxidation state. For 1 mole of FeO:

\text{FeO} \longrightarrow \frac{1}{2}\text{Fe}_2\text{O}_3 + e^\ominus \implies 1\;\text{mole of } e^\ominus = 1\;\text{F} \quad (\text{III})

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Mismatches MnO4⁻ (requires 5F, not 1F).

⚠️

Student Fallacy (Why candidates pick this trap): Miscalculating permanganate electron transfer.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Assigns 1F to H2O oxidation instead of 2F.

⚠️

Student Fallacy (Why candidates pick this trap): Thinking 1 mol H2O yields 1 electron.

3 (3)

✓ 100% Correct

Verdict & Scientific Reason:

Exact stoichiometric match across all four electrochemical conversions.

💡

Why this is airtight: None.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Mismatches both H2O and FeO.

⚠️

Student Fallacy (Why candidates pick this trap): Arbitrary guessing.

⚠️

NTA Trap Alert: Confusing 1 mol of O2 with 1 mol of H2O

Producing 1 mol of O2 requires 4 F (from 2 mol H2O). But the question specifies 1 mol of H2O to O2 (which produces 1/2 mol O2), requiring exactly 2 F!

🛡️ How to avoid the -1 mark: Check the starting material: 1 mol of H2O has 1 oxygen atom (-2 -> 0), which releases 2 electrons = 2 F.

📖

Core NEET Syllabus Standard

Chemistry Core Physical Chemistry · Electrochemistry Standard Benchmark

Core Scientific Principle Tested:

"The quantity of electricity required for oxidation or reduction is given by Q = n × z × F, where z is the number of electrons involved in the half-reaction."

Syllabus Focus: Faraday’s Laws of Electrolysis: Quantitative Aspects of Electrolysis

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

H2O gives two, MnO4 gives five; Ca times one-and-a-half gives three, Fe gives one to thrive!

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NEET 2024 Chemistry Question 85 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Faraday Charge Multiplication Hack

Step-by-Step Rigorous Workflow & Mathematical Proof

Faraday’s Law Relating Electrons and Charge

Calculation for Conversion A: 1 mol of H2O to O2

Calculation for Conversion B: 1 mol of MnO4⁻ to Mn²⁺

Calculation for Conversion C: 1.5 mol of Ca from CaCl2

Calculation for Conversion D: 1 mol of FeO to Fe2O3

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing 1 mol of O2 with 1 mol of H2O

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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