Electrolytic Metallic Conduction

8 MCQs1 revision card9-step worked example

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

Lesson

Electrolytic conduction and metallic conduction both involve charge transport, but the carrier and mechanism differ fundamentally — and NEET exploits exactly this distinction.

Metallic conduction uses free electrons as charge carriers. Electrons flow through the metal lattice without any chemical change to the conductor. Increasing temperature increases lattice vibrations, scattering electrons more frequently, so resistance increases with temperature. No matter transfer occurs.

Electrolytic conduction uses ions as charge carriers. When an electrolyte (molten or dissolved in water) conducts, cations migrate toward the cathode and anions toward the anode. This migration constitutes current, and chemical change occurs at the electrodes — matter is deposited or liberated. Unlike metals, increasing temperature for an electrolyte typically decreases resistance (increases conductivity) because more ions dissociate and ionic mobility rises.

Faraday's laws of electrolysis quantify the chemical change. The first law states that the mass deposited at an electrode is directly proportional to the charge passed. The second law states that for the same charge, the mass deposited is proportional to the equivalent weight (molar mass divided by the number of electrons per ion). Combined:

m = (M × I × t) / (n × F)

where M is molar mass, I is current, t is time, n is the number of electrons required per ion (not per mole of compound — this is where the common mistake lives), and F = 96485 C/mol (NCERT Class 12 Chemistry Chapter 2, page 20).

The high-frequency mistake: using n = 1 for every ion. For Cu²⁺ deposition, n = 2 (Cu²⁺ + 2e⁻ → Cu). For Al³⁺, n = 3. Using n = 1 inflates the calculated mass by a factor of 2 or 3, landing you on a distractor.

Watch-out: when a problem states "electrolysis of CuSO₄ solution," identify the ion being deposited (Cu²⁺) and count its charge — that gives you n directly.

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 metallic conduction, the charge carriers are:

MCQ 2Easy RecallPractice

Which of the following statements about electrolytic conduction is correct?

MCQ 3Easy RecallPractice

In Faraday's law of electrolysis, m = MIt/(nF), the quantity 'n' represents:

MCQ 4Direct ApplicationPractice

During electrolysis of aqueous CuSO₄ using inert electrodes, 0.50 A of current is passed for 965 s. The mass of copper deposited at the cathode is: (Cu = 63.5 g/mol, F = 96500 C/mol)

MCQ 5Direct ApplicationPractice

How long (in seconds) must a current of 2.0 A be passed through molten AlCl₃ to deposit 0.270 g of aluminium at the cathode? (Al = 27.0 g/mol, F = 96500 C/mol)

MCQ 6Direct ApplicationPractice

The same quantity of electricity is passed through solutions of AgNO₃ and CuSO₄ in series. If 1.08 g of Ag is deposited (Ag = 108 g/mol), the mass of Cu deposited is: (Cu = 63.5 g/mol)

MCQ 7Concept TrapPractice

A student claims: "Passing the same current for the same time through molten NaCl and molten MgCl₂ will deposit equal masses of Na and Mg because both are Group I/II metals." What is wrong with this reasoning?

MCQ 8CalculationPractice

During electrolysis of aqueous CuSO₄, a current of 1.50 A is passed for 32 min 10 s. Calculate the mass of Cu deposited and the volume of O₂ liberated at STP at the anode. (Cu = 63.5 g/mol, F = 96500 C/mol, molar volume at STP = 22400 mL/mol)

Quick recall before you leave

Worked Example

1
Given
- I = 0.500 A - t = 3860 s - M(Ag) = 108 g/mol - F = 96500 C/mol - Electrode reaction: Ag⁺ + e⁻ → Ag, so n = 1
2
Required
Mass of silver deposited, m.
3
Concept
Faraday's first law: the mass deposited is proportional to the total charge passed. For a specific ion, m = MIt/(nF).
4
Formula
m = MIt/(nF)
5
Substitution
m = (108 × 0.500 × 3860) / (1 × 96500)
6
Calculation
Numerator: 108 × 0.500 = 54.0; 54.0 × 3860 = 208440 Denominator: 1 × 96500 = 96500 m = 208440 / 96500 = 2.160 g **Note on exact values:** n = 1 is an exact integer (electron count per Ag⁺ ion) and F = 96500 C/mol is given as a defined constant for this problem. Neither limits significant figures.
7
Final answer
m = 2.16 g (3 significant figures, limited by I = 0.500 A and the given molar mass).
8
Common trap
If you mistakenly used n = 2 (confusing Ag⁺ with Cu²⁺), you would get m = 1.08 g — exactly half the correct answer. Always check the charge on the specific ion being deposited.
9
Similar NEET-style question
"A current of 1.00 A is passed through molten CaCl₂ for 9650 s. Calculate the mass of calcium deposited at the cathode. (Ca = 40.0 g/mol, F = 96500 C/mol)" [Answer: Ca²⁺ + 2e⁻ → Ca, n = 2. m = (40.0 × 1.00 × 9650)/(2 × 96500) = 2.00 g] ---

Before solving, remember these

Law

Faraday's laws of electrolysis

1st law: mass deposited (m) ∝ quantity of charge passed (Q = It). 2nd law: m₁/m₂ = E₁/E₂ where E is equivalent mass. Combined: m = (M/nF)·Q = (M/nF)·I·t.

-- NCERT Class 12 Chemistry, Ch. 2, p. 20

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.20

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Electrolytic Metallic Conduction

Lesson

Practice MCQs

Quick recall before you leave

Faraday's Law of Electrolysis

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Sources

Free NEET study resources