Oxidation Reduction Electronic

8 MCQs9-step worked example

Source: NCERT Redox ReactionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The electronic concept of oxidation and reduction redefines these classical ideas in terms of electron transfer — and getting the direction wrong is a quiet mark-loser.

The core idea (NCERT Class 11 Chemistry, Chapter 8 Part 2, page 2): Oxidation is the loss of electrons by a species. Reduction is the gain of electrons. The two always occur together — you cannot have one without the other. This is why the combined process is called a redox reaction.

A species that loses electrons is oxidised and acts as a reducing agent (it causes reduction in something else). A species that gains electrons is reduced and acts as an oxidising agent. The terminology is deliberately inverse: the oxidising agent itself gets reduced.

Where aspirants lose marks: confusing the agent label with the process. When a question asks "identify the reducing agent," many students pick the species that gets reduced — the exact opposite. The rule is mechanical: the reducing agent is the one that donates electrons (gets oxidised). The oxidising agent is the one that accepts electrons (gets reduced).

Consider the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu. Zinc loses 2 electrons (oxidised → reducing agent). Cu²⁺ gains 2 electrons (reduced → oxidising agent). No ambiguity once you track the electron flow.

NEET relevance: Questions on electronic concepts of oxidation and reduction typically appear as recall or conceptual-application items — identify the oxidising/reducing agent, determine which species is oxidised/reduced, or recognise electron transfer in a given reaction. The skill tested is definitional precision, not calculation.

Watch-out: Half-reaction notation makes the electron direction explicit. Always write the half-reactions if the full equation feels ambiguous. Oxidation half: species → species⁺ⁿ + ne⁻. Reduction half: species⁺ⁿ + ne⁻ → species.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

In the electronic concept of redox reactions, oxidation is defined as:

MCQ 2Easy RecallPractice

A species that donates electrons in a redox reaction is called:

MCQ 3Easy RecallPractice

In the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, which species is reduced?

MCQ 4Direct ApplicationPractice

In the reaction 2Na + Cl₂ → 2NaCl, identify the oxidising agent.

MCQ 5Direct ApplicationPractice

In the half-reaction Mg → Mg²⁺ + 2e⁻, magnesium is:

MCQ 6Direct ApplicationPractice

In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, the reducing agent is:

MCQ 7Concept TrapPractice

A student claims: "In every redox reaction, the oxidising agent is oxidised." This statement is:

MCQ 8Concept TrapPractice

Consider the reaction: 2H₂ + O₂ → 2H₂O. Which statement correctly identifies both the electronic change AND the agent role?

Worked Example

1
Given
In the reaction: MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂ Identify which species is oxidised and which is reduced. Name the oxidising agent and the reducing agent.
2
Required
(a) The species oxidised and the species reduced. (b) The oxidising agent and the reducing agent.
3
Concept
Electronic concept of redox: loss of electrons = oxidation; gain of electrons = reduction. The oxidised species is the reducing agent; the reduced species is the oxidising agent (NCERT Class 11 Chemistry, Chapter 8 Part 2, page 2).
4
Formula
No numerical formula needed. The method is: assign oxidation states → identify which element's oxidation state increases (oxidised) and which decreases (reduced).
5
Substitution (Oxidation-state tracking)
- **Mn in MnO₂:** Mn is +4 (since each O is −2, and 2(−2) + Mn = 0 → Mn = +4). - **Mn in MnCl₂:** Mn is +2 (since each Cl is −1, and 2(−1) + Mn = 0 → Mn = +2). - Mn goes from +4 → +2: **gain of 2 electrons → reduced.** - **Cl in HCl:** Cl is −1. - **Cl in Cl₂:** Cl is 0. - Cl goes from −1 → 0: **loss of 1 electron per Cl atom → oxidised.**
6
Calculation
Not a numerical problem. The electron-transfer analysis is complete above. Note: Oxidation states like +4, +2, −1, 0 are exact integers (counting the formal charge assignment); they do not involve significant-figure considerations.
7
Final answer
- **Oxidised species:** Cl⁻ (in HCl), going from −1 to 0. - **Reduced species:** Mn⁴⁺ (in MnO₂), going from +4 to +2. - **Reducing agent:** HCl (contains the Cl⁻ that donates electrons). - **Oxidising agent:** MnO₂ (contains the Mn⁴⁺ that accepts electrons).
8
Common trap
The agent-label inversion: students identify MnO₂ as the reducing agent because Mn "changes more." But the reducing agent is the species that is oxidised (loses electrons) — that is HCl here, not MnO₂. Always ask: "Who lost the electrons?" That species is the reducing agent.
9
Similar NEET-style question
In the reaction: 2FeCl₃ + H₂S → 2FeCl₂ + S + 2HCl, identify the oxidising and reducing agents. (Answer: Fe³⁺ in FeCl₃ is reduced → oxidising agent. S²⁻ in H₂S is oxidised → reducing agent.) ---

Before solving, remember these

Definition

Oxidation, reduction, redox

Oxidation = loss of electrons / gain of O or loss of H. Reduction = gain of electrons / loss of O or gain of H. Redox = both happen simultaneously.

-- NCERT, p. 2

Key Fact

Types of redox reactions

Combination (2 substances → 1), decomposition (1 → ≥2), displacement (active element displaces less active from compound), disproportionation (same element both oxidised and reduced).

-- NCERT, p. 10

Formulas

Cell EMF

E_{cell}^{\circ} = E_{cath}^{\circ} - E_{anode}^{\circ}

Both as reduction potentials. E°_cell > 0 → spontaneous.

Symbol	Quantity	SI Unit
E°_cell	standard cell EMF	V
E°_red	reduction potential	V

Valid when

Standard conditions (1 M, 1 bar, 298 K)
Both half-reactions as reductions

Faraday's law of electrolysis

m = \frac{M I t}{n F}

Mass deposited at electrode. M = molar mass; I = current; t = time; n = electrons per ion.

Symbol	Quantity	SI Unit
m	mass deposited	g
M	molar mass	g/mol
I	current	A
t	time	s
n	electrons per ion	-
F	Faraday	C/mol

Valid when

Steady current
Single product

ΔG from EMF

Δ G = - n F E

Connection between thermodynamics and electrochemistry. F = 96485 C/mol.

Symbol	Quantity	SI Unit
ΔG	Gibbs energy change	J
n	electrons transferred	-
F	Faraday 96485	C/mol
E	cell EMF	V

Valid when

Single redox process

Molar conductivity

Λ_{m} = \frac{1000 κ}{C}

Molar conductivity from specific conductance. Increases with dilution as more ions are free.

Symbol	Quantity	SI Unit
Λ_m	molar conductivity	S cm^2/mol
κ	specific conductance	S/cm
C	molarity	mol/L

Valid when

Aqueous solution
Single electrolyte

Nernst equation

E = E^{\circ} - \frac{0.0591}{n} lo g Q

Cell potential at non-standard conditions. n = electrons transferred. At equilibrium E=0, Q=K.

Symbol	Quantity	SI Unit
E	cell potential	V
E°	standard	V
n	electrons	-
Q	reaction quotient	-

Valid when

298 K (else use RT/F)
Single redox process

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Nernst n electrons cumulative error

Category: Negative Marking

Multi-step Nernst problem: identify electrons, write Q correctly, plug into 0.0591/n. Each sub-step has factor errors.

When it triggers

Cell EMF problem at non-standard conditions.

How to avoid

Step-by-step: (1) write balanced redox; (2) count n electrons; (3) compute Q from concentrations; (4) plug into Nernst. Verify by checking limits: at standard conditions Q=1, log Q=0, E=E°.

Trap

KMnO4 oxidation product depends on medium

Category: Inorganic Exception

Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).

When it triggers

Question gives KMnO4 oxidation in unspecified or specific medium.

How to avoid

Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.

Mistake

Uses n=1 in Faraday's law for ions like Cu²⁺ (actual n=2) or Al³⁺ (n=3).

Root cause: concept gap

Correction

n = number of electrons per ion to deposit. Cu²⁺ + 2e⁻ → Cu: n=2. Al³⁺ + 3e⁻: n=3. m = MIt/(nF).

Mistake

Uses incorrect electron count n in Nernst equation.

Root cause: formula misuse

Correction

n = electrons transferred per balanced redox equation. For Cu²⁺ + 2e⁻ → Cu: n=2. For Mn(VII) → Mn(II): n=5.

Past Year Questions

17 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

If the rate constant of a reaction is 0.03 s–1, how much time does it take for 7.2 mol L–1 concentration of the reactant to get reduced to 0.9 mol L–1? (Given: log 2 = 0.301)

121.0 s

269.3 s

323.1 s

4210 s

NTA Answer: Option 2(final)

NEET 2025

Consider the following compounds : KO , H O and H SO 2 2 2 2 4 The oxidation state of the underlined elements in them are, respectively,

1+4, –4, and +6

2+1, –1, and +6

3+2, –2, and +6

4+1, –2, and +4

NTA Answer: Option 2(final)

NEET 2025

Match List-I with List-II List-I List-II (Example) (Type of Solution) A. Humidity I. Solid in solid B. Alloys II. Liquid in gas C. Amalgams III. Solid in gas D. Smoke IV. Liquid in solid Choose the correct answer from the options given below:

1A-III, B-II, C-I, D-IV

2A-II, B-IV, C-I,D-III

3A-II, B-I, C-IV, D-III

4A-III, B-I, C-IV, D-II

NTA Answer: Option 3(final)

NEET 2024Revised key

Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

1Both Statement I and Statement II are correct.

2Both Statement I and Statement II are incorrect.

3Statement I is correct but Statement II is incorrect.

4Statement I is incorrect but Statement II is correct.

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Match List I with List II. List I List II (Conversion) (Number of Faraday required) A. 1 mol of H O to O I. 3F 2 2 B. 1 mol of MnO− 4 to Mn2+ II. 2F C. 1.5 mol of Ca from molten CaCl III. 1F 2 D. 1 mol of FeO to Fe O IV. 5F 2 3 Choose the correct answer from the options given below:

1A-II, B-IV, C-I, D-III

2A-III, B-IV, C-I, D-II

3A-II, B-III, C-I, D-IV

4A-III, B-IV, C-II, D-I

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol–1, 1 F = 96487 C)

13.15 g

20.315 g

331.5 g

40.0315 g

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

The products A and B obtained in the following reactions, respectively, are 3ROH + PCl → RCl + A 3 ROH + PCl → RCl + HCl + B 5

1POCl and H PO 3 3 3

2POCl and H PO 3 3 4

3H PO and POCl 3 4 3

4H PO and POCl 3 3 3

NTA Answer: Option 4(revised_final)

NEET 2023

The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm–1 cm–1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of cell constant is

13.28 cm–1

21.26 cm–1

33.34 cm–1

41.34 cm–1

NTA Answer: Option 2(final)

NEET 2023

The number of  bonds,  bonds and lone pair of electrons in pyridine, respectively are:

112, 3, 0

211, 3, 1

312, 2, 1

411, 2, 0

NTA Answer: Option 2(final)

NEET 2022

At 298 K, the standard electrode potentials of Cu2+ / Cu, Zn2+ / Zn, Fe2+ / Fe and Ag+ / Ag are 0.34 V, –0.76 V, –0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?

12CuSO (aq) + 2Ag(s) → 2Cu(s) + Ag SO (aq) 4 2 4

2CuSO (aq) + Zn(s) → ZnSO (aq) + Cu(s) 4 4

3CuSO (aq) + Fe(s) → FeSO (aq) + Cu(s) 4 4

4FeSO (aq) + Zn(s) → ZnSO (aq) + Fe(s) 4 4

NTA Answer: Option 1(final)

NEET 2022

Which of the following statement is not correct about diborane?

1Both the Boron atoms are sp2 hybridised.

2There are two 3-centre-2-electron bonds.

3The four terminal B-H bonds are two centre two electron bonds.

4The four terminal Hydrogen atoms and the two Boron atoms lie in one plane.

NTA Answer: Option 1(final)

NEET 2022

Given below are half cell reactions: - 13 - M n O −4 + 8 H + + 5 e − → M n 2 + + 4 H 2 O E ºM n 2 + /M n O −4 = − 1 .5 1 0 V 1 2 O 2 + 2 H + + 2 e − → H 2 O Eº =+1.223V O2/H2O Will the permanganate ion, M n O –4 liberate O from water in the presence of an acid? 2

1No, because E cell =−2.733V

2Yes, because E c e ll = + 0 .2 8 7 V

3No, because E c e ll = – 0 .2 8 7 V

4Yes, because E c e ll = + 2 .7 3 3 V

NTA Answer: Option 2(final)

NEET 2022

If radius of second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of third Bohr orbit of Li2+ ion?

1158.7 Å

2158.7 pm

315.87 pm

41.587 pm

NTA Answer: Option 2(final)

NEET 2021

Which one among the following is the correct option for right relationship between C and C for one mole P V of ideal gas?

1C = RC V P

2C + C = R P V

3C – C = R P V

4C = RC P V

NTA Answer: Option 3(final)

NEET 2021

The molar conductance of NaCl, HCl and CH COONa 3 at infinite dilution are 126.45, 426.16 and cm2 mol–1 91.0 S respectively. The molar conductance of CH COOH at infinite dilution is. 3 Choose the right option for your answer. cm2 mol–1

1540.48 S cm2 mol–1

2201.28 S cm2 mol–1

3390.71 S cm2 mol–1

4698.28 S

NTA Answer: Option 3(final)

NEET 2021

Which one of the following polymers is prepared by addition polymerisation?

1Dacron

2Teflon

3Nylon-66

4Novolac

NTA Answer: Option 2(final)

NEET 2021

The molar conductivity of 0.007 M acetic acid is 20 cm2 mol–1. S What is the dissociation constant of acetic acid? Choose the correct option. Λ° =350Scm2 mol−1  H+   Λ° mol−1 =50Scm2  CHCOO−  3 2.50×10−5 molL−1

11.75×10−4 molL−1

22.50×10−4 molL−1

31.75×10−5 molL−1

4Answer

NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Compute Λ_m from κ and C. Or compute Λ°_m of weak electrolyte from strong electrolyte values via Kohlrausch.

Direct ApplicationMedium

Common distractors

inverts c multiplier

Uses Λ = κC instead of κ/C

forgets 1000 factor

Drops 1000 in unit conversion

Apply Nernst equation to compute non-standard cell EMF given Q. n = electrons transferred.

Multi StepMedium

Common distractors

wrong n electrons

Uses incorrect electron count from half-reactions

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