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This MCQ module is based on: Conductance Electrolysis
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Conductance Electrolysis
🎓 Class 12
Chemistry
CBSE
Theory
Ch 2 – Electrochemistry
⏱ ~14 min
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Conductance Electrolysis
2.4 Conductance of Electrolytic Solutions
Metals conduct electricity through mobile electrons. Electrolytes, by contrast, conduct through mobile ions moving in opposite directions when a potential is applied. Measuring how well an ionic solution carries current tells us a great deal about the ions present, their concentration, and the extent to which a weak electrolyte has dissociated.
2.4.1 Resistance, Conductance, Conductivity
For a conductor of length ℓ and cross-sectional area A, the resistance obeys the familiar relation
\(R \;=\; \rho\,\dfrac{\ell}{A}\)
where ρ (rho) is the resistivity. Unit of R: ohm (Ω). Conductance G is simply the reciprocal: \(G = 1/R\), measured in siemens (S) where 1 S = 1 Ω⁻¹.
The reciprocal of resistivity is the conductivity κ (kappa):
\(\kappa \;=\; \dfrac{1}{\rho} \;=\; G\cdot\dfrac{\ell}{A}\) (SI unit: S m⁻¹; often quoted as S cm⁻¹ too)
The quantity ℓ/A is called the cell constant (G*) of the conductivity cell and is determined once using a standard KCl solution of known κ.
2.4.2 Molar Conductivity Λm
Conductivity depends on the number of ions per unit volume. To compare electrolytes fairly, we normalise it by concentration:
\(\Lambda_{m} \;=\; \dfrac{\kappa}{c}\)
with c in mol m⁻³, giving Λm in S m² mol⁻¹. A convenient lab form is
\(\Lambda_{m}\,(\text{S cm}^{2}\text{mol}^{-1}) \;=\; \dfrac{\kappa\,(\text{S cm}^{-1}) \times 1000}{c\,(\text{mol L}^{-1})}\)
2.4.3 How Λm Changes with Dilution
Upon dilution, the number of ions per mole of electrolyte either stays roughly the same (strong electrolyte) or grows (weak electrolyte, because more of it ionises). Either way, Λm rises as c decreases, but the detailed shape of the curve is diagnostic.
Strong electrolytes — Debye–Hückel–Onsager equation
For NaCl, KCl, HCl etc., Λm falls only slightly from its infinite-dilution value Λ°m as concentration increases. The relationship is almost linear in √c:
\(\Lambda_{m} \;=\; \Lambda^{\circ}_{m} - A\sqrt{c}\)
where A is a constant for a given solvent and temperature. Extrapolating a plot of Λm vs √c to c = 0 gives Λ°m, the limiting molar conductivity.
Weak electrolytes — dramatic rise on dilution
For CH₃COOH, NH₄OH and other weak electrolytes, the low-concentration solution has many more ions (because dilution pushes the dissociation equilibrium forward by Le Chatelier). Λm therefore rises sharply as c → 0 and the √c plot curves upward — the limiting value cannot be reached by simple extrapolation.
2.4.4 Kohlrausch's Law of Independent Migration of Ions
Kohlrausch's Law: At infinite dilution, each ion contributes independently to the total molar conductivity:
\(\Lambda^{\circ}_{m} \;=\; \nu_{+}\lambda^{\circ}_{+} + \nu_{-}\lambda^{\circ}_{-}\)
where ν+, ν− are the number of cations / anions per formula unit and λ° are their limiting ionic conductivities.This lets us calculate Λ°m of a weak electrolyte (whose curve cannot be extrapolated) from strong-electrolyte data. For example:
\(\Lambda^{\circ}_{m}(\text{CH}_{3}\text{COOH}) = \Lambda^{\circ}_{m}(\text{HCl}) + \Lambda^{\circ}_{m}(\text{CH}_{3}\text{COONa}) - \Lambda^{\circ}_{m}(\text{NaCl})\)
Degree of dissociation & dissociation constant
For a weak electrolyte at concentration c with degree of dissociation α:
\(\alpha \;=\; \dfrac{\Lambda_{m}}{\Lambda^{\circ}_{m}} \qquad\text{and}\qquad K_{a} \;=\; \dfrac{c\alpha^{2}}{1-\alpha}\)
Worked Examples — Conductance
Example 2.8 — Conductivity from resistance
A conductivity cell with cell constant 0.367 cm⁻¹ filled with 0.001 M KCl solution shows a resistance of 1500 Ω at 298 K. Find κ and Λm.
\(\kappa = G\cdot\dfrac{\ell}{A} = \dfrac{1}{R}\cdot G^{*} = \dfrac{0.367}{1500} = 2.447\times10^{-4}\,\text{S cm}^{-1}\)
\(\Lambda_{m} = \dfrac{1000\,\kappa}{c} = \dfrac{1000 \times 2.447\times10^{-4}}{0.001} = 244.7\,\text{S cm}^{2}\text{mol}^{-1}\)
Answer: κ = 2.45 × 10⁻⁴ S cm⁻¹; Λm ≈ 245 S cm² mol⁻¹.
Example 2.9 — Λ°m of acetic acid by Kohlrausch's law
Given Λ°m(HCl) = 426, Λ°m(CH₃COONa) = 91, Λ°m(NaCl) = 126 S cm² mol⁻¹, compute Λ°m(CH₃COOH).
\(\Lambda^{\circ}_{m}(\text{CH}_{3}\text{COOH}) = 426 + 91 - 126 = 391\,\text{S cm}^{2}\text{mol}^{-1}\)
Answer: 391 S cm² mol⁻¹ — close to the experimentally accepted value.
Example 2.10 — Degree of dissociation of acetic acid
The molar conductivity of 0.10 M acetic acid is 5.2 S cm² mol⁻¹. Using Λ°m = 390.5 S cm² mol⁻¹, find α and Ka.
\(\alpha = \dfrac{\Lambda_{m}}{\Lambda^{\circ}_{m}} = \dfrac{5.2}{390.5} = 0.0133\)
\(K_{a} = \dfrac{c\alpha^{2}}{1-\alpha} \approx c\alpha^{2} = (0.10)(0.0133)^{2} = 1.77\times10^{-5}\)
Answer: α ≈ 1.33 %, Ka ≈ 1.8 × 10⁻⁵ — the standard textbook value for CH₃COOH.
2.5 Electrolytic Cells & Electrolysis
Electrolysis is the chemical change driven by passing an electric current through a molten or dissolved electrolyte. An external battery pushes electrons into the cathode (reduction site) and pulls them out of the anode (oxidation site). This is the reverse of what a galvanic cell does.
For instance, passing current through molten NaCl:
\(\text{Cathode: Na}^{+} + e^{-} \to \text{Na}(\ell); \quad \text{Anode: 2Cl}^{-} \to \text{Cl}_{2}(g) + 2e^{-}\)
produces metallic sodium and chlorine gas — the industrial Downs process.
What actually gets deposited?
In an aqueous solution, several species can be reduced at the cathode or oxidised at the anode. The one that wins depends on:
- Nature of the electrolyte — molten NaCl gives Na at the cathode; aqueous NaCl gives H₂ at the cathode (water is easier to reduce than Na⁺) and Cl₂ at the anode (overvoltage effect).
- Concentration — dilute aqueous NaCl gives O₂ at the anode; concentrated brine gives Cl₂ (this is the chlor-alkali process).
- Nature of the electrode — an inert electrode (Pt, graphite) just transfers electrons; a reactive copper anode in CuSO₄ dissolves preferentially, allowing Cu to be electrorefined.
- Overvoltage — the extra voltage needed to sustain gas evolution. It is why Cl₂ (not O₂) is liberated from concentrated brine despite thermodynamic preference for O₂.
2.5.1 Faraday's Laws of Electrolysis (1833)
First Law: The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed.
\(m = Z\cdot Q = Z\cdot I\cdot t\)
where Z is the electrochemical equivalent (mass deposited per coulomb).Second Law: When the same quantity of electricity passes through different electrolytes in series, the masses deposited are proportional to their equivalent weights (E = M/n).
\(\dfrac{m_{1}}{m_{2}} = \dfrac{E_{1}}{E_{2}} = \dfrac{M_{1}/n_{1}}{M_{2}/n_{2}}\)
The Faraday constant F is the charge on one mole of electrons:
\(F \;=\; N_{A}\cdot e \;=\; (6.022\times10^{23})(1.602\times10^{-19}\,\text{C}) \;\approx\; 96\,500\,\text{C mol}^{-1}\)
So to deposit one mole of a substance carrying charge n, we must pass n × 96,500 C. For example, reducing 1 mol Cu²⁺ to Cu needs 2F = 193,000 C; depositing 1 mol Al³⁺ needs 3F = 289,500 C.
Applications
- Electroplating: a thin coating of Ag, Au, Ni, Cr on jewellery, car parts — the article is the cathode in a solution of the plating metal's salt.
- Electrometallurgy: extraction of Na, K, Ca, Mg by electrolysing their molten chlorides; Al from molten Al₂O₃ dissolved in cryolite (Hall–Héroult process).
- Electrorefining: impure Cu anode dissolves in CuSO₄ and pure Cu deposits on the cathode; noble impurities (Ag, Au) settle as "anode mud" — often the revenue offset that pays for the process.
- Chlor-alkali industry: concentrated brine → Cl₂(g) + NaOH(aq) + H₂(g).
Worked Examples — Electrolysis
Example 2.11 — Mass of Cu deposited
Calculate the mass of Cu deposited when 0.5 A flows through CuSO₄ solution for 30 minutes. (MCu = 63.5 g mol⁻¹)
Charge Q = I × t = 0.5 × 30 × 60 = 900 C. For Cu²⁺ + 2e⁻ → Cu, n = 2 so 2 × 96500 = 193,000 C deposits 63.5 g.
\(m = \dfrac{63.5}{193000}\times 900 = 0.296\,\text{g}\)
Answer: m ≈ 0.296 g.
Example 2.12 — Time to deposit 1 g of Ag
How long must a current of 2 A pass through a AgNO₃ solution to deposit 1 g of silver? (MAg = 108, n = 1)
Charge for 1 g Ag = (1/108) × 96500 = 893.5 C. Then t = Q/I = 893.5/2 = 446.8 s ≈ 7.45 min.
Answer: about 7 min 27 s.
Example 2.13 — Electrolysis of molten Al₂O₃ (Hall–Héroult)
How much electric charge is needed to produce 5.4 kg of Al from molten Al₂O₃?
n(Al) = 5400/27 = 200 mol. Each Al³⁺ needs 3 electrons, so mol e⁻ = 600.
\(Q = 600 \times 96500 = 5.79\times10^{7}\,\text{C}\)
Answer: Q ≈ 5.8 × 10⁷ C. This enormous charge demand explains why aluminium smelters are always built beside cheap hydroelectric power.
Example 2.14 — Series cells (Faraday's 2nd Law)
The same current passes through AgNO₃ and CuSO₄ solutions in series and deposits 1.08 g of Ag. How much Cu is simultaneously deposited?
Equivalent weights: E(Ag) = 108/1 = 108; E(Cu) = 63.5/2 = 31.75.
\(\dfrac{m_{Cu}}{m_{Ag}} = \dfrac{E_{Cu}}{E_{Ag}}\;\Rightarrow\;m_{Cu} = 1.08 \times \dfrac{31.75}{108} = 0.3175\,\text{g}\)
Answer: 0.32 g Cu.
Example 2.15 — Current from mass deposited
A current passed through a CuSO₄ cell for 1 hour deposits 1.5 g of Cu. Find the current.
Q = (1.5/63.5) × 2 × 96500 = 4559 C. I = Q/t = 4559/3600 = 1.27 A.
Answer: I ≈ 1.27 A.
Activity 2.2 — Electroplate a Key with Copper
L3 Apply
Predict: If you make an iron key the cathode and a copper strip the anode, both dipped in CuSO₄ solution, what will happen after 5 minutes of current flow? Will the mass of the copper strip change?
- Fill a beaker with 100 mL of 0.5 M CuSO₄ solution.
- Clean an iron key with steel-wool and connect it to the negative terminal of a 4.5 V dry cell.
- Connect a clean copper strip to the positive terminal and dip both into the solution.
- Leave the cell running for 5 min; remove, rinse and dry. Weigh before and after.
Expected: The key acquires a thin, uniform copper coating; the copper strip loses mass almost equal to the mass gained by the key. At the anode Cu → Cu²⁺ + 2e⁻ replenishes the solution; at the cathode Cu²⁺ + 2e⁻ → Cu deposits on the key. This is a miniature version of industrial electroplating used for car bumpers, jewellery and electronic connectors.
Interactive: Electrolysis Calculator L3 Apply
Enter the current, time and metal details. The tool computes the charge and mass deposited using m = (M·I·t)/(n·F), with F = 96,500 C mol⁻¹.
Current I (A):
Time t (min):
M (g/mol):
n:
Enter values and press Compute.
Competency-Based Questions
A student measures the resistance of 0.001 M KCl solution as 1500 Ω in a conductivity cell of cell constant 0.367 cm⁻¹ (κ_KCl expected ≈ 0.000147 S cm⁻¹ at 298 K; the student's value is used as a check). Separately, she measures Λm of 0.001 M HCl as 421.3 S cm² mol⁻¹ and finds Λ°m(HCl) = 425.9, Λ°m(CH₃COONa) = 91.0, Λ°m(NaCl) = 126.4 S cm² mol⁻¹.
Q1. L1 Remember The SI unit of specific conductance (κ) is:
Answer: B. κ = 1/ρ, so its unit is the reciprocal of Ω m, i.e., S m⁻¹.
Q2. L3 Apply Use Kohlrausch's law to compute Λ°m(CH₃COOH). (2 marks)
Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COONa) − Λ°m(NaCl) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹.
Q3. L3 Apply State Faraday's first law and find the mass of silver deposited by 0.965 A current in 1 hour through AgNO₃. (3 marks)
First law: m ∝ Q. Q = 0.965 × 3600 = 3474 C. For 1 mol Ag (108 g) needs 96500 C → m = (108/96500) × 3474 = 3.89 g Ag.
Q4. L3 Apply Why does Λm of a weak electrolyte rise sharply on dilution whereas for a strong electrolyte it changes only slightly? (3 marks)
A strong electrolyte is already fully dissociated; dilution only reduces inter-ionic attraction slightly, giving a small rise. A weak electrolyte has only a few % of ions; dilution drives the dissociation equilibrium forward, producing many more ions and a huge rise in Λm.
Q5. L4 Analyse In aqueous NaCl electrolysis with inert electrodes, why is Cl₂ evolved at the anode instead of O₂, even though E°(O₂/H₂O) is lower than E°(Cl₂/Cl⁻)? (3 marks)
Oxygen evolution has a large overvoltage on common electrodes — the actual voltage needed is far above the thermodynamic E°. Combined with high Cl⁻ concentration in brine, this makes Cl₂ kinetically favoured. This is the basis of the industrial chlor-alkali process.
Assertion-Reason Questions
Assertion (A): The molar conductivity of a weak electrolyte cannot be obtained by graphical extrapolation of its Λm vs √c plot to c = 0.
Reason (R): At very low concentrations, Λm of a weak electrolyte rises sharply due to increasing dissociation, so the curve does not become linear.
Answer: A. This is precisely why Kohlrausch's law (ion additivity) is needed for weak electrolytes.
Assertion (A): In an electrolytic cell, the cathode is the negative terminal.
Reason (R): The external source pushes electrons into the cathode to reduce cations.
Answer: A. The polarity is opposite to that of a galvanic cell, where the cathode is positive.
Assertion (A): Depositing 1 mol of Al needs 3 times as much charge as depositing 1 mol of Ag.
Reason (R): Al³⁺ requires three electrons to become Al, while Ag⁺ needs only one to become Ag.
Answer: A. Charge per mole = n × F, so Al needs 3F = 289,500 C whereas Ag needs just 1F = 96,500 C.
Frequently Asked Questions - Conductance Electrolysis
What is the main concept covered in Conductance Electrolysis?
In NCERT Class 12 Chemistry Chapter 2 (Electrochemistry), "Conductance Electrolysis" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Conductance Electrolysis useful in real-life or applied chemistry?
Real-life applications of "Conductance Electrolysis" from NCERT Class 12 Chemistry Chapter 2 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Conductance Electrolysis?
Key reactions in "Conductance Electrolysis" (NCERT Class 12 Chemistry Chapter 2 Electrochemistry) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 2?
NCERT Class 12 Chemistry Chapter 2 (Electrochemistry) is structured so each part builds chemical understanding sequentially. "Conductance Electrolysis" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Conductance Electrolysis?
CBSE board questions from "Conductance Electrolysis" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Conductance Electrolysis" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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