Fuel Cells

8 MCQs9-step worked example

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

Lesson

A fuel cell converts chemical energy of a fuel directly into electrical energy without combustion. The key difference from a conventional galvanic cell: reactants are continuously supplied from an external source rather than stored inside the cell.

The hydrogen–oxygen fuel cell is the NCERT-standard example (NCERT Class 12 Chemistry Chapter 2, page 24). At the anode, hydrogen is oxidised; at the cathode, oxygen is reduced. The overall reaction is:

2H₂(g) + O₂(g) → 2H₂O(l)

The electrodes are porous carbon or platinum, and the electrolyte is typically concentrated KOH or a proton-exchange membrane.

Half-reactions (alkaline medium):

Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻
Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻

Why NEET cares about fuel cells:

Efficiency advantage. Fuel cells bypass the Carnot-cycle limitation of heat engines. They convert chemical energy to electrical energy in a single step, achieving efficiencies of 60–70% — higher than thermal power plants (~40%).
Clean product. The only product of a hydrogen–oxygen fuel cell is water. No CO₂, no SO₂, no particulates.
Continuous operation. Unlike a dry cell or lead-acid battery that runs down, a fuel cell operates as long as fuel and oxidant are supplied. It is not "recharged" — fresh reactants are fed in.

Common confusion: Students conflate fuel cells with rechargeable batteries. A fuel cell is NOT rechargeable — it does not store energy internally. A lead-acid battery stores reactants inside and can be recharged by reversing the reaction; a fuel cell requires an external fuel supply.

Watch-out: Questions may test whether a fuel cell is a primary cell, secondary cell, or neither. It is classified as a primary cell (non-rechargeable) in NCERT treatment, but distinguished from ordinary primary cells by continuous fuel supply.

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 a hydrogen–oxygen fuel cell, the overall cell reaction produces:

MCQ 2Easy RecallPractice

A fuel cell differs from an ordinary galvanic cell primarily because:

MCQ 3Easy RecallPractice

In a hydrogen–oxygen fuel cell using alkaline (KOH) electrolyte, the cathode reaction is:

MCQ 4Direct ApplicationPractice

A fuel cell achieves higher efficiency than a thermal power plant because:

MCQ 5Direct ApplicationPractice

Which statement correctly classifies a hydrogen–oxygen fuel cell?

MCQ 6Direct ApplicationPractice

The electrode material commonly used in a hydrogen–oxygen fuel cell is:

MCQ 7Concept TrapPractice

A student claims: "A fuel cell is like a rechargeable battery because you can keep using it indefinitely." Which correction is accurate?

MCQ 8Concept TrapPractice

Why is water the only product in a hydrogen–oxygen fuel cell, while burning hydrogen in air also produces water but with significant energy lost as heat?

Worked Example

1
Given
A hydrogen–oxygen fuel cell with alkaline (KOH) electrolyte operates at standard conditions. The student is asked to write the half-reactions and the overall reaction, then explain why this cell cannot be classified as a secondary cell.
2
Required
(a) Anode and cathode half-reactions in alkaline medium. (b) Overall cell reaction. (c) Classification with justification.
3
Concept
In a fuel cell, hydrogen is oxidised at the anode and oxygen is reduced at the cathode. The electrolyte medium determines the form of the half-reactions (OH⁻ ions appear in alkaline medium). Classification depends on whether the reaction is reversible by applying external current.
4
Formula
No numerical formula is needed. The half-reactions are: - Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻ - Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
5
Substitution
Adding the two half-reactions: 2H₂(g) + 4OH⁻ + O₂(g) + 2H₂O → 4H₂O + 4e⁻ + 4OH⁻ + (−4e⁻)
6
Calculation
Cancel species appearing on both sides: 4OH⁻ cancels, 4e⁻ cancels, and 2H₂O cancels from the 4H₂O product. Overall: 2H₂(g) + O₂(g) → 2H₂O(l)
7
Final answer
(a) Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻; Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (b) Overall: 2H₂(g) + O₂(g) → 2H₂O(l) (c) Primary cell — the reaction is not reversed by applying external current. Continuous operation comes from external fuel supply, not electrochemical recharging.
8
Common trap
Confusing "refilling fuel" with "recharging." Recharging means reversing the cell reaction electrochemically (as in a lead-acid battery). A fuel cell never reverses its reaction — it is mechanically resupplied.
9
Similar NEET-style question
"A hydrogen–oxygen fuel cell uses an acid electrolyte (H₃PO₄). Write the half-reactions at each electrode and explain how they differ from the alkaline-medium half-reactions." (Answer: Anode: H₂ → 2H⁺ + 2e⁻; Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O. The ion carrier changes — H⁺ migrates through the electrolyte in acid medium, OH⁻ in alkaline — but the overall reaction remains 2H₂ + O₂ → 2H₂O.) ---

Before solving, remember these

Key Fact

Common batteries

Dry cell (Leclanché): Zn anode, MnO₂/C cathode, ~1.5 V. Lead accumulator: Pb anode, PbO₂ cathode, H₂SO₄ electrolyte, ~2 V/cell. Fuel cell: H₂/O₂ → H₂O + electrical energy.

-- NCERT Class 12 Chemistry, Ch. 2, p. 24

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

Sources

NCERT refs: Class 12 Chemistry Chapter 2, p.24

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