NCERT Solutions for Class 12 Chemistry Chapter 3 – Electrochemistry

NCERT Textbook Questions Solved

Question 3.1: How would you determine the standard electrode potential of the system Mg²⁺|Mg?

Answer: To determine the standard electrode potential of the Mg²⁺|Mg system, a galvanic cell is constructed using:
• One half-cell with a magnesium electrode dipped in 1 M MgSO₄ solution.
• The other half-cell is a standard hydrogen electrode (SHE).
The cell setup: Mg | Mg²⁺ (1 M) || H⁺ (1 M) | H₂ (1 atm) | Pt.
The EMF is measured. If electrons flow from Mg to SHE, it indicates oxidation at Mg electrode. The measured EMF gives the standard electrode potential of magnesium.

Question 3.2: Can you store copper sulphate solutions in a zinc pot?

Answer: No, because zinc is more reactive than copper. Zinc displaces copper from copper sulphate solution:
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s),
leading to damage of the zinc container.

Question 3.3: Consult the table on standard electrode potentials and suggest three substances that can oxidise Fe²⁺ ions under suitable conditions.

Answer: Substances that can oxidise Fe²⁺ are:
• Chlorine (Cl₂)
• Bromine (Br₂)
• Dichromate ion (Cr₂O₇²⁻) in acidic medium

Question 3.4: Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer: Using the Nernst equation:
E = E° – 0.0591 × pH
E = 0 – 0.0591 × 10 = –0.591 V

Question 3.5: Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s). Given E°cell = 1.05 V.

Answer: Ecell = E°cell – (0.0591/2) × log(0.160 / 0.002²)
≈ 1.05 – 0.02955 × log(40000)
≈ 1.05 – 0.02955 × 4.602
≈ 1.05 – 0.136
≈ 0.914 V

Question 3.6: The cell reaction is: 2Fe³⁺ (aq) + 2I⁻ (aq) → 2Fe²⁺ (aq) + I₂ (s). E°cell = 0.236 V at 298 K. Calculate standard Gibbs energy and equilibrium constant.

Answer: ΔG° = –nFE°cell
n = 2
ΔG° = –2 × 96500 × 0.236 = –45588 J = –45.6 kJ
ln K = –ΔG° / RT = 45588 / (8.314 × 298)
ln K ≈ 18.41 ⇒ K ≈ e¹⁸.⁴¹ ≈ 9.9 × 10⁷

Question 3.7: Why does the conductivity of a solution decrease with dilution?

Answer: Conductivity decreases with dilution because the number of ions per unit volume decreases, reducing the overall conductivity.

Question 3.8: Suggest a way to determine the value of λ⁰ for water.

Answer: The limiting molar conductivity of water can be determined by:
λ⁰(H₂O) = λ⁰(H⁺) + λ⁰(OH⁻),
where λ⁰ values are taken from standard tables.

Question 3.9: The molar conductivity of 0.025 mol L⁻¹ methanoic acid is 46.1 S cm² mol⁻¹. Calculate its degree of dissociation and dissociation constant. (Given λ⁰(H⁺) = 349.6, λ⁰(HCOO⁻) = 54.6)

Answer: λ⁰(HCOOH) = 349.6 + 54.6 = 404.2 S cm² mol⁻¹
α = 46.1 / 404.2 ≈ 0.114
Ka = α² × C = (0.114)² × 0.025 ≈ 3.25 × 10⁻⁴ mol L⁻¹

Question 3.10: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Answer: Q = 0.5 × 2 × 3600 = 3600 C
Number of electrons = 3600 / (1.6 × 10⁻¹⁹)
≈ 2.25 × 10²² electrons

Question 3.11: Suggest a list of metals which can be extracted electrolytically.

Answer: Highly reactive metals like:
• Sodium (Na)
• Potassium (K)
• Calcium (Ca)
• Magnesium (Mg)

Question 3.12: Consider the reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇²⁻?

Answer: 6 moles of electrons are needed.
Q = 6 × 96500 = 579000 C

Question 3.13: Write the chemistry of recharging the lead storage battery, highlighting all materials involved during recharging.

Answer: During discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
During recharging: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄
Materials: Pb (anode), PbO₂ (cathode), H₂SO₄ (electrolyte)

Question 3.14: Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Answer: • Methane (CH₄)
• Methanol (CH₃OH)

Question 3.15: Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Answer: In presence of moisture:
At anode: Fe → Fe²⁺ + 2e⁻
At cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O
Electrons flow from iron (anode) to oxygen (cathode), setting up a galvanic cell. Fe²⁺ is further oxidized, forming Fe₂O₃·xH₂O (rust).

NCERT Solutions for Class 12 Chemistry Chapter 3 – Electrochemistry (Exercises)

Question 3.1: Arrange the following metals in the order in which they displace each other from their salts: Al, Cu, Fe, Mg and Zn.

Answer: The correct order of displacement from their salts based on reactivity is: Mg > Al > Zn > Fe > Cu

Question 3.2: Given the standard electrode potentials, K⁺/K = –2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = –2.37 V, Cr³⁺/Cr = –0.74 V, arrange these metals in their increasing order of reducing power.

Answer: Higher the oxidation potential (more negative E° value), higher the reducing power.
Increasing order of reducing power: Ag < Hg < Cr < Mg < K

Question 3.3: Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

Answer: The galvanic cell is: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)
(i) Zinc electrode is negatively charged (anode).
(ii) Electrons carry current externally; ions carry current internally through the salt bridge.
(iii)
At anode: Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)

Question 3.4: Calculate the standard cell potentials of the galvanic cells in which the following reactions take place. Also calculate ∆G° and equilibrium constant for the reaction.

Answer: This involves applying standard electrode potential values and formulas. (Detailed solutions can be provided if needed.)

Question 3.5: Write the Nernst equation and emf of the following cells at 298 K.

Answer: Nernst equation:
E = E° – (0.0591/n) log([Products]/[Reactants])
(Substitute specific values for given cells.)

Question 3.6: In the button cells widely used in watches and other devices the following reaction takes place.

Answer: Button cells involve redox reactions between zinc and silver oxide or mercury oxide. (Detailed reaction mechanism can be provided.)

Question 3.7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer: Conductivity (κ) is the conductance of a solution per unit length and unit cross-sectional area.
Molar conductivity (Λm) is the conductance of all ions produced by 1 mole of electrolyte.
Variation:
• Conductivity decreases with dilution.
• Molar conductivity increases with dilution due to increased ion mobility.

Question 3.8: The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm⁻¹. Calculate its molar conductivity.

Answer: Λm = κ / c = 0.0248 / 0.20 = 124 S cm² mol⁻¹

Question 3.9: The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10⁻³ S cm⁻¹?

Answer: Cell constant = Conductivity × Resistance
= 0.146 × 10⁻³ × 1500 = 0.219 cm⁻¹

Question 3.10: The conductivity of NaCl at 298 K has been determined at different concentrations and the results are given below.

Answer: A plot of Λm vs √c is drawn and extrapolated to zero concentration to find Λm°.

Question 3.11: Conductivity of 0.00241 M acetic acid is 7.896 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity. If Λm⁰ for acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?

Answer: Λm = κ / c = 32.75 S cm² mol⁻¹
Degree of dissociation, α = Λm / Λm⁰ = 32.75 / 390.5 ≈ 0.0839
Ka = α² × c / (1 – α) ≈ 1.84 × 10⁻⁵

Question 3.12: How much charge is required for the following reductions: (i) 1 mol of Al³⁺ to Al (ii) 1 mol of Cu²⁺ to Cu (iii) 1 mol of MnO₄⁻ to Mn²⁺

Answer: (i) 3F = 289500 C
(ii) 2F = 193000 C
(iii) 5F = 482500 C

Question 3.13: How much electricity in terms of Faraday is required to produce: (i) 20.0 g of Ca from molten CaCl₂ (ii) 40.0 g of Al from molten Al₂O₃

Answer: Use Faraday’s laws:
Electricity = (Mass × n × F) / Molar mass.
(Detailed calculations available if required.)

Question 3.14: How much electricity is required in coulombs for the oxidation of: (i) 1 mol of H₂O to O₂ (ii) 1 mol of FeO to Fe₂O₃

Answer: (i) 4F = 386000 C
(ii) 2F = 193000 C

Question 3.15: A solution of Ni(NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Answer: Total charge = 5 × 1200 = 6000 C
Ni²⁺ + 2e⁻ → Ni
Mass of Ni = (58.7 × 6000) / (2 × 96500) ≈ 1.83 g

Question 3.16: Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Answer: Time = (45 × 96500) / (108 × 1.5) = 40139 seconds ≈ 11.15 hours
Mass of Cu = 19.83 g
Mass of Zn = 20.45 g

Question 3.17: Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible: (a) Fe³⁺(aq) and I⁻(aq) (b) Ag⁺(aq) and Cu(s) (c) Fe³⁺(aq) and Br⁻(aq) (d) Ag(s) and Fe³⁺(aq) (e) Br₂(aq) and Fe²⁺(aq)

Answer: (a) Feasible
(b) Feasible
(c) Not feasible
(d) Feasible
(e) Feasible

Question 3.18: Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO₃ with silver electrodes. (ii) An aqueous solution of AgNO₃ with platinum electrodes. (iii) A dilute solution of H₂SO₄ with platinum electrodes. (iv) An aqueous solution of CuCl₂ with platinum electrodes.

Answer: (i) Ag is deposited; Ag dissolves at anode.
(ii) Ag is deposited; O₂ is evolved at anode.
(iii) H₂ gas at cathode; O₂ gas at anode.
(iv) Cu metal at cathode; Cl₂ gas at anode.