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NCERT Exercises and Solutions: Electrochemistry

🎓 Class 12 Chemistry CBSE Theory Ch 2 – Electrochemistry ⏱ ~8 min
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NCERT Exercises and Solutions: Electrochemistry

Chapter 2 Summary — The Key Ideas at a Glance

Electrochemistry studies the two-way transfer between chemical and electrical energy. A galvanic cell harnesses a spontaneous redox reaction to deliver electrical energy; an electrolytic cell uses electrical energy to drive a non-spontaneous reaction.
  • Cell notation: Anode | Anode solution ‖ Cathode solution | Cathode. EMF = E°cathode − E°anode (both reduction potentials).
  • Standard Hydrogen Electrode: 2H⁺(1 M) + 2e⁻ → H₂(1 bar), E° = 0 — reference for all electrode potentials.
  • Nernst equation: Ecell = E°cell − (0.0591/n) log Q at 298 K.
  • Equilibrium: log Kc = n E°cell/0.0591. Gibbs energy: ΔG° = −nFE°cell = −RT ln Kc.
  • Conductivity κ = G·(ℓ/A); Molar conductivity Λm = 1000κ/c.
  • Strong electrolytes follow Λm = Λ°m − A√c (Debye–Hückel–Onsager). Weak electrolytes rise sharply on dilution.
  • Kohlrausch's law: Λ°m = ν₊λ°₊ + ν₋λ°₋. Degree of dissociation α = Λm/Λ°m, Ka = cα²/(1−α).
  • Faraday's laws: m = (M·I·t)/(n·F); F = 96,500 C mol⁻¹. Equivalent-weight ratio = mass-ratio for series cells.
  • Primary batteries: dry cell (1.5 V), mercury cell (1.35 V). Secondary: lead storage (2 V/cell), NiCd, Li-ion.
  • Fuel cell: 2H₂ + O₂ → 2H₂O, E° = 1.23 V, ~70% efficient, only water as exhaust.
  • Corrosion = electrochemical oxidation of metal. Prevention: painting, galvanising, alloying, sacrificial anode.

Keyword Grid

Galvanic cellChemical → electrical (spontaneous, ΔG < 0).
Electrolytic cellElectrical → chemical (non-spontaneous).
AnodeSite of oxidation; (−) in galvanic, (+) in electrolytic.
CathodeSite of reduction; (+) in galvanic, (−) in electrolytic.
Salt bridgeU-tube of KCl/KNO₃ that maintains electroneutrality.
SHEStandard hydrogen electrode; E° = 0.00 V reference.
Nernst equationE = E° − (0.0591/n) log Q.
ΔG° = −nFE°Links thermodynamics to EMF.
Specific conductance κG·(ℓ/A); unit S m⁻¹.
Molar conductivity Λmκ/c; unit S m² mol⁻¹.
Λ°mLimiting molar conductivity (c → 0).
Kohlrausch's lawΛ°m = sum of ionic contributions.
Faraday (F)96,500 C mol⁻¹ of electrons.
Electrochemical equivalent ZMass deposited per coulomb.
OvervoltageExtra voltage needed to evolve gases.
GalvanisationZn coating protects iron cathodically.
Sacrificial anodeMg/Zn bolted to Fe; oxidises in its place.
Fuel cellContinuous supply; 2H₂ + O₂ → 2H₂O.

NCERT-Style Exercises (Solved)

Q 2.1 — Formulating a cell from electrode potentials

Write the cell notation and compute E°cell for the redox reaction Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s), given E°(Mg²⁺/Mg) = −2.37 V and E°(Cu²⁺/Cu) = +0.34 V.

Mg has the more negative E° → it is oxidised (anode). Cu²⁺ is reduced (cathode).

\(\text{Mg}(s)\;|\;\text{Mg}^{2+}(aq)\;||\;\text{Cu}^{2+}(aq)\;|\;\text{Cu}(s)\)
\(E^{\circ}_{cell} = 0.34 - (-2.37) = +2.71\,\text{V}\)
Q 2.2 — Identifying anode and cathode

For the cell Zn | Zn²⁺(1 M) ‖ Ag⁺(1 M) | Ag, identify the anode, cathode, positive terminal, and compute E°cell. Given E°(Zn²⁺/Zn) = −0.76 V, E°(Ag⁺/Ag) = +0.80 V.

Left electrode Zn → anode (oxidation, negative terminal). Right electrode Ag → cathode (reduction, positive terminal).

\(E^{\circ}_{cell} = 0.80 - (-0.76) = +1.56\,\text{V}\)

Net reaction: Zn + 2Ag⁺ → Zn²⁺ + 2Ag.

Q 2.3 — Spontaneity from E° values

Will Fe(s) displace Cu²⁺ from an aqueous CuSO₄ solution? Given E°(Fe²⁺/Fe) = −0.44 V, E°(Cu²⁺/Cu) = +0.34 V.

For Fe + Cu²⁺ → Fe²⁺ + Cu:

\(E^{\circ}_{cell} = 0.34 - (-0.44) = +0.78\,\text{V} \;>\;0\)

Positive E°cell → spontaneous. Yes, Fe readily displaces Cu²⁺ (the basis of the classic blue-to-colourless test).

Q 2.4 — Cell for a given reaction

Represent the cell for the reaction 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s) in cell notation. How many electrons are transferred per unit reaction?

\(\text{Al}(s)\;|\;\text{Al}^{3+}(aq)\;||\;\text{Cu}^{2+}(aq)\;|\;\text{Cu}(s)\)

Each Al loses 3 e⁻; the balanced equation shows 2 Al → 6 e⁻; 3 Cu²⁺ accept 6 e⁻. So n = 6.

Q 2.5 — Arranging metals by reducing power

Using the following E° values, arrange K, Ca, Mg, Al, Zn, Fe, Cu, Ag in order of increasing reducing power: E° (V) = −2.93 (K⁺/K), −2.87 (Ca²⁺/Ca), −2.37 (Mg²⁺/Mg), −1.66 (Al³⁺/Al), −0.76 (Zn²⁺/Zn), −0.44 (Fe²⁺/Fe), +0.34 (Cu²⁺/Cu), +0.80 (Ag⁺/Ag).

Lower (more positive) E° = weaker reducer; more negative E° = stronger reducer.

Increasing reducing power: Ag < Cu < Fe < Zn < Al < Mg < Ca < K.

Q 2.6 — Nernst equation application

Compute Ecell at 298 K for Zn | Zn²⁺(0.1 M) ‖ Cu²⁺(0.01 M) | Cu.

n = 2; Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10; E°cell = 1.10 V.

\(E_{cell} = 1.10 - \dfrac{0.0591}{2}\log(10) = 1.10 - 0.02955 = 1.0705\,\text{V}\)

Answer: ≈ 1.07 V.

Q 2.7 — Gibbs energy of cell reaction

Calculate ΔG° and Kc for the reaction Fe + Cu²⁺ → Fe²⁺ + Cu (E°cell = 0.78 V, n = 2) at 298 K.

\(\Delta G^{\circ} = -nFE^{\circ} = -(2)(96500)(0.78) = -150\,540\,\text{J} = -150.5\,\text{kJ}\)
\(\log K_{c} = \dfrac{n\,E^{\circ}}{0.0591} = \dfrac{2 \times 0.78}{0.0591} = 26.4\;\Rightarrow\;K_{c} \approx 2.5 \times 10^{26}\)
Q 2.8 — Concentration cell

Calculate the EMF of the cell Cu | Cu²⁺(0.001 M) ‖ Cu²⁺(0.1 M) | Cu at 298 K.

E°cell = 0 (same metal). n = 2, Q = 0.001/0.1 = 0.01.

\(E_{cell} = 0 - \dfrac{0.0591}{2}\log(0.01) = -\dfrac{0.0591}{2}(-2) = 0.0591\,\text{V}\)

Answer: 59.1 mV.

Q 2.9 — Conductivity and molar conductivity

The resistance of a conductivity cell filled with 0.02 M KCl solution is 82.4 Ω. The cell constant is 0.367 cm⁻¹. Calculate the conductivity and molar conductivity of the solution.

\(\kappa = \dfrac{G^{*}}{R} = \dfrac{0.367}{82.4} = 4.454\times10^{-3}\,\text{S cm}^{-1}\)
\(\Lambda_{m} = \dfrac{1000\,\kappa}{c} = \dfrac{1000 \times 4.454\times10^{-3}}{0.02} = 222.7\,\text{S cm}^{2}\text{mol}^{-1}\)
Q 2.10 — Kohlrausch's law

Given Λ°m(HCl) = 426, Λ°m(NaCl) = 126, Λ°m(CH₃COONa) = 91 S cm² mol⁻¹, find Λ°m(CH₃COOH).

\(\Lambda^{\circ}_{m}(\text{CH}_{3}\text{COOH}) = \Lambda^{\circ}_{m}(\text{HCl}) + \Lambda^{\circ}_{m}(\text{CH}_{3}\text{COONa}) - \Lambda^{\circ}_{m}(\text{NaCl})\)
\(= 426 + 91 - 126 = 391\,\text{S cm}^{2}\text{mol}^{-1}\)
Q 2.11 — Degree of dissociation

The molar conductivity of a 0.025 M methanoic acid (HCOOH) solution is 46.1 S cm² mol⁻¹. Calculate its degree of dissociation and dissociation constant. Given λ°(H⁺) = 349.6 and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.

\(\Lambda^{\circ}_{m} = 349.6 + 54.6 = 404.2\,\text{S cm}^{2}\text{mol}^{-1}\)
\(\alpha = \dfrac{\Lambda_{m}}{\Lambda^{\circ}_{m}} = \dfrac{46.1}{404.2} = 0.1141\)
\(K_{a} = \dfrac{c\alpha^{2}}{1-\alpha} = \dfrac{(0.025)(0.1141)^{2}}{1-0.1141} = 3.67\times10^{-4}\)

Answer: α ≈ 11.4 %, Ka ≈ 3.67 × 10⁻⁴.

Q 2.12 — Why does Λm rise with dilution?

Explain why Λm of a weak electrolyte rises steeply on dilution while Λm of a strong electrolyte rises only slightly.

A strong electrolyte is already ~100% ionised at finite concentration. The slight rise in Λm on dilution comes only from a reduction in interionic attractions (Debye–Hückel–Onsager effect). For a weak electrolyte, dilution shifts the dissociation equilibrium forward (Le Chatelier), generating many more ions; the rise in Λm is therefore dramatic.

Q 2.13 — Faraday's first law

How much charge is required for the reduction of 1 mol of Cu²⁺ to Cu? How long will it take when a current of 1.5 A is passed?

Cu²⁺ + 2e⁻ → Cu. Charge for 1 mol Cu = 2F = 2 × 96500 = 1.93 × 10⁵ C.

\(t = \dfrac{Q}{I} = \dfrac{1.93\times10^{5}}{1.5} = 1.287\times10^{5}\,\text{s} \approx 35.7\,\text{hours}\)
Q 2.14 — Mass of Ag deposited

A current of 0.5 A is passed through a solution of AgNO₃ for 30 minutes. Calculate the mass of silver deposited. (MAg = 108 g mol⁻¹, n = 1).

Q = It = 0.5 × 30 × 60 = 900 C.

\(m = \dfrac{M\cdot Q}{nF} = \dfrac{108\times 900}{1\times 96500} = 1.007\,\text{g}\)

Answer: ≈ 1.01 g Ag.

Q 2.15 — Series cells (Faraday's 2nd law)

The same quantity of electricity that deposits 1.08 g of Ag also deposits how much Al from Al³⁺? (MAg = 108, n = 1; MAl = 27, n = 3).

Equivalent weights: E(Ag) = 108; E(Al) = 27/3 = 9.

\(\dfrac{m_{Al}}{m_{Ag}} = \dfrac{E_{Al}}{E_{Ag}} \;\Rightarrow\; m_{Al} = 1.08 \times \dfrac{9}{108} = 0.090\,\text{g}\)

Answer: 0.09 g Al.

Q 2.16 — EMF of lead storage battery

The standard EMF of a single lead storage cell is 2.04 V. Compute ΔG° (in kJ) for the overall discharge reaction involving n = 2.

\(\Delta G^{\circ} = -nFE^{\circ} = -(2)(96500)(2.04) = -393\,720\,\text{J} \approx -393.7\,\text{kJ}\)
Q 2.17 — Hydrogen–oxygen fuel cell

Write the anode, cathode and overall reactions for the H₂–O₂ fuel cell (KOH electrolyte) and state two advantages over a conventional thermal power plant.

\(\text{Anode: 2H}_{2}(g) + 4\text{OH}^{-} \to 4\text{H}_{2}\text{O}(\ell) + 4e^{-}\)
\(\text{Cathode: O}_{2}(g) + 2\text{H}_{2}\text{O}(\ell) + 4e^{-} \to 4\text{OH}^{-}\)
\(\text{Overall: 2H}_{2}(g) + \text{O}_{2}(g) \to 2\text{H}_{2}\text{O}(\ell)\)

Advantages: (i) Higher conversion efficiency (~70%) vs. ~40% for thermal plants — no Carnot limit because there is no heat-to-work step. (ii) No pollutants (NOx, SOx, CO, particulates) — only water is produced.

Q 2.18 — Corrosion and its prevention

Explain the electrochemical mechanism of rusting of iron and list any three methods of prevention.

A drop of water (slightly acidic because of dissolved CO₂) on an iron surface behaves as a miniature galvanic cell. At the anodic region, Fe → Fe²⁺ + 2e⁻. The electrons travel through the metal to the cathodic region at the edge of the drop, where O₂ + 4H⁺ + 4e⁻ → 2H₂O. The Fe²⁺ is then oxidised further by O₂ to hydrated Fe₂O₃·xH₂O (rust).

Prevention methods:

  • Barrier coating — paint, grease or polymer film.
  • Galvanisation — zinc coating that oxidises preferentially (more negative E° than Fe).
  • Sacrificial anode — a block of Mg or Zn bolted to the iron structure.
  • (Also: alloying into stainless steel, electroplating with Cr or Sn.)
Q 2.19 — Cell reaction from EMF

A cell has E°cell = 1.23 V and n = 4. Compute log Kc and Kc at 298 K.

\(\log K_{c} = \dfrac{nE^{\circ}}{0.0591} = \dfrac{4 \times 1.23}{0.0591} = 83.25\)

Answer: Kc ≈ 10⁸³·²⁵ ≈ 1.8 × 10⁸³ — essentially complete reaction, as expected for H₂–O₂ combustion.

Q 2.20 — Electrolysis of molten NaCl

What mass of Na and what volume of Cl₂ (at STP, 22.4 L mol⁻¹) will be produced by passing a current of 10 A through molten NaCl for 1 hour?

Q = It = 10 × 3600 = 3.6 × 10⁴ C. Mol e⁻ = Q/F = 36000/96500 = 0.373 mol.

Na⁺ + e⁻ → Na, so mol Na = 0.373; mass Na = 0.373 × 23 = 8.58 g.

2Cl⁻ → Cl₂ + 2e⁻, so mol Cl₂ = 0.373/2 = 0.187; volume (STP) = 0.187 × 22.4 = 4.19 L.

Frequently Asked Questions - NCERT Exercises and Solutions: Electrochemistry

What are the key NCERT exercise types in Chapter 2 Electrochemistry?
NCERT Class 12 Chemistry Chapter 2 Electrochemistry exercises cover definitions, structure-property relationships, reaction mechanisms, numerical problems, and predict-the-product questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should focus on reaction mechanisms, IUPAC nomenclature, and stoichiometric reasoning to score full marks.
How should students approach reaction mechanism questions in Electrochemistry?
For reaction mechanism questions in NCERT Class 12 Chemistry Chapter 2 Electrochemistry: (1) identify the substrate, reagent, solvent, and conditions, (2) draw curly-arrow electron movement at each step, (3) label intermediates (carbocation, carbanion, free radical, transition state), (4) state stereochemistry where relevant. The MyAiSchool solutions show full curly-arrow mechanisms for every multi-step reaction.
What are the most-asked CBSE board questions from Chapter 2?
From NCERT Class 12 Chemistry Chapter 2 (Electrochemistry), the most-asked CBSE board questions test conceptual understanding, structure-property logic, distinguishing tests, IUPAC naming, and short numerical problems. 5-mark questions usually combine structural reasoning + mechanism + application. The MyAiSchool exercise set tags each question by mark weight and Bloom level for prioritized prep.
How do I balance chemical equations in NCERT exercises?
For balancing equations in NCERT Class 12 Chemistry Chapter 2: (1) write skeletal equation with correct formulas, (2) balance atoms other than H, O first, (3) balance O, then H (using H2O for organic reactions; or in acidic/basic medium for redox), (4) balance charge using e- in redox, (5) cross-check atom count and charge on both sides. The MyAiSchool solutions include balanced redox half-reactions where applicable.
What are common mistakes students make in Chapter 2 exercises?
Common mistakes in NCERT Class 12 Chemistry Chapter 2 (Electrochemistry) include: (1) incorrect IUPAC names (wrong locants or suffix), (2) skipping reaction conditions (catalyst, temperature, solvent), (3) wrong mechanism arrows, (4) sign errors in numerical (Ecell, Kc, ΔG), (5) forgetting stereochemistry (retention/inversion/racemisation). The MyAiSchool solutions flag these traps for each question.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Chemistry Chapter 2 Electrochemistry solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with chemical reasoning, curly-arrow mechanisms, IUPAC naming verifications, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes.
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