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. 52Electrolytic Metallic Conduction
8 MCQs3 revision cards9-step worked example
Source: NCERT Redox Reactions and ElectrochemistryPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 26 Sep 2026
Electrolytic Metallic Conduction, explained for NEET
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 52).
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.
Can you answer these Electrolytic Metallic Conduction 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:
Show answer and why every option is right or wrong
Answer: D. Metals conduct electricity through the movement of free (delocalised) electrons in the metallic lattice. No ions are involved (NCERT Class 12 Chemistry Chapter 2, section on conductance in electrolytic solutions).
Why A is wrong: A is wrong because cations are charge carriers in electrolytic conduction, not metallic conduction.
Why B is wrong: B is wrong because anions are charge carriers in electrolytic conduction, not metallic conduction.
Why C is wrong: C is wrong because both cations and anions serve as carriers in electrolytic conduction; metallic conduction involves only electrons.
MCQ 2Easy RecallPractice
Which of the following statements about electrolytic conduction is correct?
Show answer and why every option is right or wrong
Answer: A. For electrolytic conductors, higher temperature increases ionic dissociation and mobility, raising conductivity and thus decreasing resistance (NCERT Class 12 Chemistry Chapter 2).
Why B is wrong: B is wrong because electrolytic conduction causes chemical change at electrodes — ions are oxidised or reduced, depositing or liberating matter.
Why C is wrong: C is wrong because electrolytic conduction uses ions (cations and anions) as charge carriers, not free electrons.
Why D is wrong: D is wrong because resistance increases with temperature for metallic conductors, not electrolytic conductors.
MCQ 3Easy RecallPractice
In Faraday's law of electrolysis, m = MIt/(nF), the quantity 'n' represents:
Show answer and why every option is right or wrong
Answer: B. In the formula m = MIt/(nF), n is the number of electrons transferred per ion of the species being deposited — for Cu²⁺, n = 2; for Al³⁺, n = 3 (NCERT Class 12 Chemistry Chapter 2, page 52).
Why A is wrong: A is wrong because n is not the number of moles of electrolyte; it specifically refers to the electron count per ion in the electrode reaction.
Why C is wrong: C is wrong because Avogadro's number (6.022 × 10²³) is a constant unrelated to the variable n in this formula.
Why D is wrong: D is wrong because the total moles of electrons passed is It/F; the variable n is the per-ion electron requirement, which divides the total to give mass deposited.
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)
Show answer and why every option is right or wrong
Answer: C. Cu²⁺ + 2e⁻ → Cu, so n = 2. m = MIt/(nF) = (63.5 × 0.50 × 965)/(2 × 96500) = 30638.75/193000 = 0.1588 g (NCERT Class 12 Chemistry Chapter 2, page 52).
Why A is wrong: A is wrong because it uses n = 1 and also doubles the time or current somewhere, giving a value 4× too large. Always check: Cu²⁺ needs n = 2.
Why B is wrong: B is wrong because it uses n = 1 instead of n = 2 for Cu²⁺. This is the common mistake of ignoring the charge on the ion — Cu²⁺ requires 2 electrons per ion (mistake: using n = 1 doubles the mass).
Why D is wrong: D is wrong because it uses n = 4, which does not correspond to any copper ion. Cu²⁺ → Cu requires exactly 2 electrons.
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)
Show answer and why every option is right or wrong
Answer: C. Al³⁺ + 3e⁻ → Al, so n = 3. Rearranging m = MIt/(nF): t = mnF/(MI) = (0.270 × 3 × 96500)/(27.0 × 2.0) = 78165/54.0 = 1447.5 s (NCERT Class 12 Chemistry Chapter 2, page 52).
Why A is wrong: A is wrong because it uses n = 2 instead of n = 3, as if aluminium were divalent: 0.270 × 2 × 96500/(27.0 × 2.0) = 965.0 s.
Why B is wrong: B is wrong because it uses n = 9 (squaring 3 or misapplying the formula), which triples the required time. Al³⁺ requires exactly 3 electrons.
Why D is wrong: D is wrong: 2895.0 s is twice the correct time, which would need 6 electrons per Al atom; Al³⁺ + 3e⁻ → Al needs 3.
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)
Show answer and why every option is right or wrong
Answer: B. Same charge Q passed through both. For Ag⁺ (n=1): Q = m×n×F/M = (1.08 × 1 × 96500)/108 = 965 C. For Cu²⁺ (n=2): m = MQ/(nF) = (63.5 × 965)/(2 × 96500) = 61277.5/193000 = 0.3175 g. This is Faraday's second law in action (NCERT Class 12 Chemistry Chapter 2, page 51).
Why A is wrong: A is wrong because it uses n = 1 for Cu²⁺ instead of n = 2. Cu²⁺ requires 2 electrons per ion; using n = 1 doubles the calculated mass.
Why C is wrong: C is wrong because it treats copper as monovalent AND doubles the charge. Cu²⁺ is divalent (n = 2); the mass should be about one-third of the Ag mass deposited, not more than it.
Why D is wrong: D is wrong because it uses n = 4 for copper, which has no chemical basis. Cu²⁺ → Cu is a 2-electron reduction.
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?
Show answer and why every option is right or wrong
Answer: D. Faraday's law m = MIt/(nF) depends on both M and n. Na⁺ (n=1) and Mg²⁺ (n=2) differ in n, and their molar masses differ (23 vs 24.3). Equal charge gives m(Na) = 23Q/F and m(Mg) = 24.3Q/(2F) ≈ 12.15Q/F — roughly half (NCERT Class 12 Chemistry Chapter 2, page 52).
Why A is wrong: A is wrong because it claims n is the same for both. Na⁺ requires 1 electron and Mg²⁺ requires 2 — the difference in n is the central reason equal charge deposits unequal masses.
Why B is wrong: B is wrong because the question is about the fundamental law, not practical current efficiency. Even at 100% efficiency, unequal masses deposit because M and n differ.
Why C is wrong: C is wrong because Faraday's laws apply to all electrolysis — aqueous and molten. There is no restriction to aqueous solutions.
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)
Show answer and why every option is right or wrong
Answer: A. A is correct. t = 32 × 60 + 10 = 1930 s, so Q = It = 1.50 × 1930 = 2895 C. At the cathode, Cu²⁺ + 2e⁻ → Cu, so n = 2 and m(Cu) = MQ/(nF) = 63.5 × 2895/(2 × 96500) = 183832.5/193000 = 0.952 g. At the anode, 2H₂O → O₂ + 4H⁺ + 4e⁻, so n = 4 and moles of O₂ = Q/(nF) = 2895/(4 × 96500) = 0.0075 mol. Volume = 0.0075 × 22400 = 168.0 mL. Carry the final multiplication out fully: 0.0075 × 22400 is 168, and 75 × 224 = 16800 is the quickest way to see it (NCERT Class 12 Chemistry Chapter 2, page 52).
Why B is wrong: B is wrong because it uses n = 1 for Cu²⁺, doubling the copper mass, and n = 2 for O₂, doubling the gas volume. Cu²⁺ needs 2 electrons and O₂ liberation needs 4 electrons per molecule.
Why C is wrong: C is wrong because the copper mass is right (n = 2) but the oxygen uses n = 2 instead of n = 4, which doubles the volume to 336 mL. The anode half-reaction 2H₂O → O₂ + 4H⁺ + 4e⁻ requires four electrons per O₂ molecule.
Why D is wrong: D is wrong because it uses n = 4 for Cu²⁺, halving the copper mass. Cu²⁺ → Cu requires exactly 2 electrons, not 4. The oxygen volume here is right.
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Electrolytic Metallic Conduction: quick recall before you leave
How do you solve a Electrolytic Metallic Conduction question? A 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]
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What to remember before solving Electrolytic Metallic Conduction questions
Which Electrolytic Metallic Conduction formulas do you need for NEET?
Faraday's law of electrolysis
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
Where do students lose marks on Electrolytic Metallic Conduction?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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).
More in Redox Reactions and Electrochemistry: 3 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
Electrolytic Metallic Conduction questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys.
NEET 2023
13.28 cm⁻¹
21.26 cm⁻¹
33.34 cm⁻¹
41.34 cm⁻¹
NTA Answer: Option 2(final)
All 21 past-paper questions from Redox Reactions and Electrochemistry →
Sources
NCERT refs: Class 12 Chemistry Chapter 2, p.52
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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