CBSE Class 12 Chemistry Electrochemistry Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. How is the pH of aqueous NaCl solution affected when it is electrolysed?
Answer:
When aqueous NaCl solution is electrolysed, H₂ is liberated at cathode, Cl₂ at anode and NaOH is formed in the solution. Hence pH of solution increases.

Teacher's Note:
1. Na⁺ and OH⁻ remain in solution, so the solution becomes basic.

Question 2. Which reference electrode is used to measure the electrode potential of other electrodes?
Answer:
SHE, whose electrode potential is taken as zero.

Teacher's Note:
1. SHE: Pt, H₂(1 bar) | H⁺(1 M); E° = 0.00 V at all temperatures.

Question 3. Out of zinc and tin, which one protects iron better even after cracks and why?
Answer:
Zinc protects better because oxidation of zinc is greater but that of tin is less than that of iron.

Teacher's Note:
1. Once a tin coating cracks, iron is oxidised faster than before because iron is more reactive than tin.

Question 4. Define corrosion. What is the chemical formula of rust?
Answer:
Corrosion is the slow eating away of the surface of the metal due to attack of atmospheric gases. Rust: Fe₂O₃.xH₂O.

Teacher's Note:
1. Rust is hydrated ferric oxide; both water and oxygen are needed for it to form.

Question 5. What is the electrolyte used in a dry cell?
Answer:
A paste of NH₄Cl (and ZnCl₂).

Teacher's Note:
1. The dry cell (Leclanche cell) is a primary cell giving about 1.5 V.

Question 6. How much amount of a substance is deposited by 1 coulomb? What is it called?
Answer:
One coulomb deposits Eq. wt./96500 gram. It is called electrochemical equivalent of the substance.

Teacher's Note:
1. Symbol Z: W = Z × Q, and Z = E/F.

Question 7. Why alternating current is used in place of direct current in measuring the electrolytic conductance?
Answer:
Direct current results in the electrolysis of the electrolytic solution.

Teacher's Note:
1. Electrolysis would change the concentration and polarise the electrodes, so the measured resistance would keep changing.

Question 8. Why in a concentrated solution, a strong electrolyte shows deviations from Debye-Huckel-Onsager equation?
Answer:
In concentrated solution of a strong electrolyte, the interionic forces of attraction are large.

Teacher's Note:
1. The Debye-Huckel-Onsager equation (Λ_m = Λ°_m − A√C) holds only for dilute solutions.

Question 9. What is the free energy change for (a) galvanic cell (b) electrolytic cell? [Punjab Board 2011]
Answer:
(a) ΔG < 0 (as redox reaction is spontaneous)
(b) ΔG > 0 (as redox reaction is non-spontaneous).

Teacher's Note:
1. Link the signs through ΔG = −nFE_cell: positive E means negative ΔG.

Question 10. Which type of cells are rechargeable?
Answer:
Those cells are rechargeable in which the products formed during discharge are deposited on the electrodes and these can be decomposed to give the original substances when electrical energy is supplied.

Teacher's Note:
1. Examples are the lead storage battery and the nickel-cadmium cell (secondary cells).

Question 11. What is meant by Faraday constant ?
Answer:
Faraday constant, F = 96500 coulombs. It is that quantity of electricity which deposits one gram equivalent of the substance or it is the charge carried by one mole of electrons.

Teacher's Note:
1. 1 F = N_A × charge on one electron = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C ≈ 96500 C mol⁻¹.

Question 12. How many faradays of electric charge is required to liberate 5600 cm³ of oxygen at STP? (Karnataka Board 2012)
Answer:
O²⁻ → \( \frac{1}{2} \)O₂ + 2e⁻
or 2O²⁻ → O₂ + 4e⁻
Thus, 1 mole of O₂, i.e., 22400 cm³ require 4 faradays of charge.
∴ 5600 cm³ of O₂ will require charge = \( \frac{4}{22400} \) × 5600 = 1 faraday

Teacher's Note:
1. Each O₂ molecule needs 4 electrons, so first find moles of O₂ (5600/22400 = 0.25) and multiply by 4.

Short Answer Questions (2 Marks)

Question 13. Draw a graph between Λ_m and √C for strong and weak electrolyte.
Answer:

For a strong electrolyte, Λ_m decreases linearly with √C; extrapolating the line to C = 0 gives Λ°_m.
For a weak electrolyte, Λ_m increases steeply on dilution (low √C) and the curve cannot be extrapolated to get Λ°_m.

Teacher's Note:
1. The straight line for strong electrolytes follows Λ_m = Λ°_m − A√C (Debye–Hückel–Onsager equation).

Question 14. The conductivity of 0.02 M solution of NaCl is 2.6 × 10⁻² S cm⁻¹. What is its molar conductivity?
Answer:
κ = 2.6 × 10⁻² S cm⁻¹
C = 0.02 M
Λ_m = \( \frac{\kappa \times 1000}{C(M)} \) = \( \frac{2.6 \times 10^{-2} \times 1000}{0.02} \)
= \( \frac{26 \times 100}{0.02 \times 100} \) = \( \frac{26 \times 10^{2}}{2} \)
= 13 × 10² S cm² mol⁻¹

Teacher's Note:
1. Check the unit: S cm⁻¹ × cm³ L⁻¹ ÷ mol L⁻¹ = S cm² mol⁻¹.

Question 15. Give products of electrolysis of an aqueous solution of AgNO₃ with silver electrode.
Answer:
At anode: Ag(s) → Ag⁺ + e⁻
At cathode: Ag⁺ + e⁻ → Ag(s)
Thus silver dissolves from the anode and is deposited on the cathode.

Teacher's Note:
1. With an active (silver) anode, the anode itself is oxidised; with Pt electrodes O₂ would be evolved instead.

Question 16. One faraday of electricity deposits one mol of Na from the molten salt but \( \frac{1}{3} \) mol of Al from an aluminium salt. Why?
Answer:
The reactions at cathode for the deposit of Na and Al are
Na⁺ + e⁻ → Na and Al³⁺ + 3e⁻ → Al
Thus, 1 faraday deposits 1 mol of Na whereas 3 faradays are required for depositing one mol of Al.

Teacher's Note:
1. Moles deposited by 1 F = 1/(charge on the ion); this is Faraday’s second law in mole form.

Question 17. Copper is conducting as such while copper sulphate is conducting only in molten state or in aqueous solution. Explain.
Answer:
Copper is conducting as such because it contains free electrons. CuSO₄ is conducting in molten state or in aqueous solution because it gives ions only in the molten state or aqueous solution.

Teacher's Note:
1. In solid CuSO₄ the ions are fixed in the crystal lattice and cannot move, so it does not conduct.

Question 18. Which will have greater molar conductivity and why?
Sol. A. 1 mol KCl dissolved in 200 cc of the solution.
Sol. B. 1 mol KCl dissolved in 500 cc of the solution.
Answer:
Solution B will have greater molar conductivity because Λ_m = κ × V. With dilution, κ decreases but V increases much more so that product increases.

Teacher's Note:
1. Solution B is more dilute; molar conductivity always increases on dilution.

Question 19. Solutions of two electrolytes A and B each having a concentration of 0.2 M have conductivities 2 × 10⁻² and 4 × 10⁻⁴ S cm⁻¹ respectively. Which will offer greater resistance to the flow of current and why?
Answer:
κ = G × \( \frac{l}{a} \) = \( \frac{1}{R} \) \( \frac{l}{a} \), i.e., κ ∝ \( \frac{1}{R} \). Hence, B will offer greater resistance.

Teacher's Note:
1. Lower conductivity means higher resistivity – for the same cell, resistance is inversely proportional to conductivity.

Question 20. A 0.1 M aqueous of Na₂SO₄ is diluted by adding water. What will happen to the values of its conductance (G), conductivity (κ), molar conductivity (Λ_m) and equivalent conductivity (Λ_eq)?
Answer:
κ decreases whereas G, Λ_m and Λ_eq increase with dilution.

Teacher's Note:
1. κ refers to 1 cm³ of solution, so it falls as ions per cm³ fall; Λ_m and Λ_eq refer to a fixed amount of electrolyte, so they rise.

Question 21. How does molar conductivity of a weak electrolyte vary with its concentration? Give reason also.
Answer:
Molar conductivity of a weak electrolyte increases with decrease in concentration (i.e., on dilution). The increase is small at higher concentrations but very steep at low concentrations (near infinite dilution), so the plot of Λ_m against √C is a curve, not a straight line.
Reason: A weak electrolyte is only slightly dissociated. On dilution, its degree of dissociation increases, so the number of ions in the solution containing 1 mole of electrolyte increases. Hence, Λ_m increases sharply on dilution.

Teacher's Note:
1. Because the curve rises steeply near zero concentration, Λ°_m of a weak electrolyte cannot be found by extrapolation; use Kohlrausch law instead.

Question 22. Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with decrease in concentration? [CBSE 2015]
Answer:
The molar conductivity of a solution at infinite dilution, i.e., when concentration approaches zero is called limiting molar conductivity.
With decrease in concentration, i.e., with dilution, number of ions per cm³ decreases due to large increase in volume of the solution. As conductivity is the conductance per cm³ of the solution, hence it decreases.

Teacher's Note:
1. Do not mix up the two: conductivity (κ) decreases on dilution while molar conductivity (Λ_m) increases.

Question 23. X and Y are two electrolytes. On dilution, molar conductivity of X increases 2.5 times while that of Y increases 25 times. Which of the two is a weak electrolyte and why? [CBSE 2020]
Answer:
Y is a weak electrolyte because the molar conductivity of a weak electrolyte increases rapidly on dilution whereas it increases slowly for strong electrolytes.

Teacher's Note:
1. For a strong electrolyte only interionic attraction decreases on dilution; for a weak one the degree of dissociation itself increases, giving a much larger rise.

Question 24. Write expressions for equivalent conductivity and molar conductivity of Al₂(SO₄)₃ at infinite dilution in terms of their ionic conductivities.
Answer:
Λ°_eq (Al₂(SO₄)₃) = λ°_Al³⁺ + λ°_SO4²⁻ (equivalent ionic conductivities)
Λ°_m (Al₂(SO₄)₃) = 2λ°_Al³⁺ + 3λ°_SO4²⁻ (molar ionic conductivities)

Teacher's Note:
1. In the molar expression the multipliers are the numbers of ions in the formula (2 Al³⁺, 3 SO₄²⁻).

Question 25. Following two reactions can occur at cathode in the electrolysis of aqueous sodium chloride
Na⁺ + e⁻ → Na (s), E°_Red = −2.71 V
2H₂O (l) + 2e⁻ → H₂ (g) + 2OH⁻ (aq), E°_Red = −0.83 V
Which reaction takes place preferentially and why?
Answer:
As the standard reduction potential of H₂O is greater than that of Na⁺ ion, reduction of water takes place preferentially, i.e., H₂ is liberated at cathode.

Teacher's Note:
1. At the cathode, the species with the higher (less negative) reduction potential is reduced first.

Short Answer Questions (3 Marks)

Teacher's Note:
1. Use the formula with (1 − α) in the denominator here; α = 0.114 is too large to ignore.

Question 27. Calculate the standard cell potentials of galvanic cells in which the following reaction take place:
2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
Also calculate ΔG° and equilibrium constant of the reaction. [E°_Cr³⁺/Cr = −0.74 V, E°_Cd²⁺/Cd = −0.40 V]
Answer:
E°_cell = E°_cathode − E°_anode
= −0.40 − (−0.74) = 0.34 V
ΔG° = −nFE°_cell = −6 × 96500 × 0.34 = −196860
= −196860 J mol⁻¹ = −196.86 kJ/mol
−ΔG° = 2.303 RT log K_c
196860 = 2.303 × 8.314 × 298 log K_c
or log K_c = 34.5014
K_c = antilog 34.5014 = 3.17 × 10³⁴

Teacher's Note:
1. 2 Cr atoms each lose 3 electrons, so n = 6. Never multiply E° by the stoichiometric coefficients.

Question 28. Calculate the potential of the following cell:
Sn⁴⁺(1.5 M) + Zn → Sn²⁺(0.5 M) + Zn²⁺(2 M)
Given: E°_{Sn⁴⁺/Sn²⁺} = 0.13 V, E°_Zn²⁺/Zn = −0.76 V.
Will the cell potential increase or decrease if the concentration of Sn⁴⁺ is increased?
Answer:
E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[Sn^{2+}][Zn^{2+}]}{[Sn^{4+}][Zn]} \)
= 0.89 − \( \frac{0.0591}{2} \) log \( \frac{0.5 \times 2}{1.5 \times 1} \)
= 0.89 − \( \frac{0.0591}{2} \) log \( \frac{1}{1.5} \)
= 0.895 V
On increasing the concentration of Sn⁴⁺, EMF of the cell will increase.

Teacher's Note:
1. E°_cell = 0.13 − (−0.76) = 0.89 V. Sn⁴⁺ is a reactant, so raising it lowers Q and raises E_cell.

Question 29. E°(Cu²⁺/Cu) and E°(Ag⁺/Ag) is +0.337 V and +0.799 V respectively. Make a cell whose EMF is +ve. If the concentration of Cu²⁺ is 0.01 M and E_cell at 25°C is zero, calculate the concentration of Ag⁺.
Answer:
Cu is more reactive than silver, so that the cell is Cu | Cu²⁺(0.01 M) || Ag⁺(C) | Ag
or cell reaction: Cu + 2Ag⁺ → Cu²⁺ + 2Ag
E°_cell = 0.799 − 0.337 = 0.462 V, n = 2
E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2} \)
0 = 0.462 − \( \frac{0.0591}{2} \) log \( \frac{(0.01) \times 1^2}{1 \times [Ag^+]^2} \)
log \( \frac{0.01}{[Ag^+]^2} \) = \( \frac{0.462 \times 2}{0.0591} \) = 15.63
or [Ag⁺] = 1.5 × 10⁻⁹ M

Teacher's Note:
1. At E_cell = 0 the cell is at equilibrium, so Q equals K; activities of pure solids Cu and Ag are 1.

Question 30. Calculate the potential of the cell at 298 K:
Cd | Cd²⁺(0.1 M) || H⁺(0.2 M) | Pt, H₂(0.5 atm)
Given E° for Cd²⁺/Cd = −0.403 V, R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹.
Answer:
The cell reaction is Cd + 2H⁺(0.2 M) → Cd²⁺(0.1 M) + H₂(0.5 atm)
E°_cell = 0 − (−0.403) = +0.403 V
E_cell = 0.403 − \( \frac{2.303RT}{nF} \) log \( \frac{[Cd^{2+}] \times p_{H_2}}{[Cd][H^+]^2} \)
= 0.403 − \( \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \) log \( \frac{0.1 \times 0.5}{(0.2)^2} \)
E_cell = 0.403 − 0.003 = 0.40 V

Teacher's Note:
1. A gas enters the reaction quotient as its partial pressure in atm; [Cd] = 1 for the pure solid.

Question 31. Conductivity of saturated solution of BaSO₄ at 315 K is 3.648 × 10⁻⁶ ohm⁻¹ cm⁻¹ and that of water is 1.25 × 10⁻⁶ ohm⁻¹ cm⁻¹. Ionic conductance of Ba²⁺ and SO₄²⁻ are 110 and 136.6 ohm⁻¹ cm² mol⁻¹ respectively. Calculate the solubility of BaSO₄ in g/L.
Answer:
Λ°_m(BaSO₄) = λ°(Ba²⁺) + λ°(SO₄²⁻) = 110 + 136.6 = 246.6 ohm⁻¹ cm² mol⁻¹
κ_BaSO4 = κ_BaSO4(solution) − κ_water = 3.648 × 10⁻⁶ − 1.25 × 10⁻⁶
= 2.398 × 10⁻⁶ S cm⁻¹
For a sparingly soluble salt, Λ^c_m ≈ Λ°_m = \( \frac{\kappa \times 1000}{Solubility} \)
Solubility = \( \frac{2.398 \times 10^{-6} \times 1000}{246.6} \) = 9.72 × 10⁻⁶ mol/L
Solubility = 9.72 × 10⁻⁶ × 233 = 2.26 × 10⁻³ g/L

Teacher's Note:
1. Subtract the conductivity of water first, and multiply mol/L by the molar mass of BaSO₄ (233 g/mol) to get g/L.

Question 32. An aqueous solution of copper sulphate is electrolysed using platinum electrodes in one case and copper electrodes in another case. Will the products of electrolysis be same or different? Give reason.
Answer:
In case of platinum electrodes, the products are Cu on cathode and O₂ gas on anode. This is because out of Cu²⁺ and H⁺ ions (produced from H₂O), Cu²⁺ ions have lower discharge potential.
Out of SO₄²⁻ and OH⁻ ions, OH⁻ ions have lower discharge potential (OH⁻ → OH + e⁻, 4OH → 2H₂O + O₂).
In case of copper electrodes, product at cathode is the same, viz., Cu but at anode, out of the three possible reactions, viz.,
OH⁻ → OH + e⁻, SO₄²⁻ → SO₄ + 2e⁻, Cu → Cu²⁺ + 2e⁻,
Cu is more easily oxidized than OH⁻ and SO₄²⁻ ions. Hence, anode of Cu dissolves into the solution to form Cu²⁺ ions.

Teacher's Note:
1. Products at the cathode depend on the cations; products at the anode depend on both the anions and the anode material.

Question 33. State the products of electrolysis obtained on the cathode and the anode in the following cases:
(i) A dilute solution of H₂SO₄ with platinum electrodes
(ii) An aqueous solution of AgNO₃ with silver electrodes.
Answer:
(i) H₂SO₄ (aq) → 2H⁺ (aq) + SO₄²⁻ (aq); H₂O ⇄ H⁺ + OH⁻
At cathode: H⁺ + e⁻ → H, H + H → H₂ (g)
At anode: OH⁻ → OH + e⁻, 4OH → 2H₂O (l) + O₂ (g)
Thus, H₂ is liberated at the cathode and O₂ at the anode.
(ii) AgNO₃ (s) + aq → Ag⁺ (aq) + NO₃⁻ (aq); H₂O ⇄ H⁺ + OH⁻
At cathode: Ag⁺ ions have lower discharge potential than H⁺ ions. Hence, Ag⁺ ions will be deposited on the cathode (Ag⁺ + e⁻ → Ag).
At anode: Ag anode will be attacked by NO₃⁻ ions. Hence, Ag anode will dissolve to form Ag⁺ ions in the solution (Ag → Ag⁺ + e⁻).

Teacher's Note:
1. In (ii) the net effect is transfer of silver from anode to cathode; the concentration of AgNO₃ stays the same.

Question 34. Formulate the galvanic cell in which the following reaction takes place:
Zn (s) + 2Ag⁺ (aq) → Zn²⁺ (aq) + 2Ag (s)
State (i) Which one of the electrodes is negatively charged?
(ii) The reaction taking place at each of its electrode.
(iii) The carriers of current within this cell. [CBSE 2008]
Answer:
The cell reaction is: Zn (s) + 2Ag⁺ (aq) → Zn²⁺ (aq) + 2Ag (s)
The cell is represented as: Zn (s) | Zn²⁺ (aq) || Ag⁺ (aq) | Ag (s)
(i) Anode, i.e., zinc electrode is negatively charged.
(ii) At anode: Zn (s) → Zn²⁺ (aq) + 2e⁻; At cathode: Ag⁺ (aq) + e⁻ → Ag (s)
(iii) Within the cell, the current is carried by the cations and the anions through the salt bridge. In the external circuit, electrons flow from zinc to silver and hence conventional current flows from silver to zinc.

Teacher's Note:
1. In a galvanic cell the anode is negative; in an electrolytic cell the anode is positive – a frequent source of lost marks.

Question 35. An electrochemical cell is made of aluminium and tin electrodes with their standard reduction potentials −1.66 V and 0.14 V respectively. Select the anode and the cathode, represent the cell and write the cell reaction. Find the e.m.f. of the cell.
Answer:
For E_cell to be positive, the given standard reduction potentials show that oxidation will take place on Al–electrode. Hence, Al will be anode and Sn will be cathode. The cell will be represented as:
Al (s) | Al³⁺ (aq) || Sn²⁺ (aq) | Sn (s)
Cell reaction: 2Al (s) + 3Sn²⁺ (aq) → 2Al³⁺ (aq) + 3Sn (s)
E°_cell = E°_{Sn²⁺, Sn} − E°_{Al³⁺, Al} = 0.14 − (−1.66) = 1.80 V

Teacher's Note:
1. Balance electrons (6) before writing the cell reaction, but do not multiply the E° values.

Long Answer Questions (5 Marks)

Question 36. A voltaic cell is set up at 25°C with the half cells Ag⁺ (0.001 M) Ag and Cu²⁺ (0.10 M) Cu. What should be its cell potential ? [E° = 0.46 V, log 10⁵ = 5] [CBSE Foreign 2012]
Answer:
Cu + 2Ag⁺ → Cu²⁺ + 2Ag
Half cell reactions :
Cathode (reduction) :
2Ag⁺ (0.001 M) + 2e⁻ → 2Ag(s)
Anode (oxidation) :
Cu(s) → Cu²⁺ (0.10 M) + 2e⁻
∴ n = 2
E^°_cell = 0.46 V
\( E_{cell} = E^{o}_{cell} - \frac{0.059}{n} \log \frac{[Cu^{2+}]}{[Ag^+]^2} \)
\( = 0.46 - \frac{0.059}{2} \log \frac{(0.1)}{(0.001)^2} \)
\( = 0.46\,V - \frac{0.059}{2}\,V \log 10^5 \)
\( = 0.46\,V - \frac{0.059}{2}\,V \times 5 \)
= 0.46 V − 0.1475 V
= 0.3125 V

Teacher's Note:
1. Write the half-cell reactions first to get n = 2, then square [Ag⁺]; missing the square gives log 10² and a wrong answer.

Question 37. Consider the given below figure and answer the following questions :
(a) Cell ‘A’ has E_cell = 2V and Cell ‘B’ has E_cell = 1.1V which of the two cells ‘A’ or ‘B’ will act as electrolytic cell. Which electrode reactions will occur in this cell?
(b) If cell ‘A’ has E_cell = 0.5 V and Cell ‘B’ has E_cell = 1.1 V then what will be the reactions at anode and cathode? [NCERT Exemplar]

Answer:
(a) Cell ‘B’ will act as electrolytic cell as it has lower EMF
Therefore, the electrode reactions will be :
Zn²⁺ + 2e⁻ → Zn at cathode
Cu → Cu²⁺ + 2e⁻ at anode.
(b) Now cell ‘B’ acts as galvanic cell as it has higher EMF and will push electrons into cell ‘A’.
The electrode reaction will be :
At anode : Zn → Zn²⁺ + 2e⁻
At cathode : Cu²⁺ + 2e⁻ → Cu

Teacher's Note:
1. When two cells oppose each other, the one with the higher emf drives the current and the other is forced to work as an electrolytic cell, with its reactions reversed.

Question 38. (i) What are the two classifications of batteries ? What is the difference between them ?
(ii) The resistance of 0.01 M NaCl solution at 25°C is 200 Ω. The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution.
OR
(i) What are fuel cells ? Give an example of a fuel cell.
(ii) Calculate the equilibrium constant (log K_c) and Δ_rG° for the following reaction at 298 K.
Cu(s) + 2Ag⁺(aq) ⇄ Cu²⁺(aq) + 2Ag(s)
Given E°_cell = 0.46 V and 1F = 96500 C mol⁻¹ [CBSE Comptt. OD 2016]
Answer:
(i) Classification of batteries : (a) Primary batteries (b) Secondary batteries
Primary batteries are non-chargeable batteries whereas secondary batteries are rechargeable.
(ii) \( \Lambda_m = \frac{\kappa \times 1000}{C} \) S cm² mol⁻¹
\( \kappa = \frac{1}{R}\left(\frac{l}{A}\right) \)
\( = \frac{1}{200} \times 1\ S\,cm^{-1} \)
\( = \frac{1}{200}\ S\,cm^{-1} \)
\( \Lambda_m = \frac{1 \times 1000}{200 \times 0.01} \) S cm² mol⁻¹
= 500 S cm² mol⁻¹
OR
(i) Fuel cells are the cells which converts energy of combustion of fuel directly into electricity. Example H₂ – O₂ fuel cell.
(ii) \( \log K_c = \frac{nE^{\circ}_{cell}}{0.059} \)
\( = \frac{2 \times 0.46\,V}{0.059} = \frac{0.92}{0.059} \)
log K_c = 15.59
Δ_rG° = − n FE°_cell
= −2 × 96500 C mol⁻¹ × 0.46V
= −88,780 J mol⁻¹
= −88.78 kJ mol⁻¹

Teacher's Note:
1. While solving the numerical, write the working formula followed by the data and proper units; κ = G*/R, so with G* = 1 cm⁻¹, κ is simply 1/R.

Question 39. (i) Define the following terms :
(a) Molar conductivity (Λ_m)
(b) Secondary batteries
(c) Fuel cell
(ii) State the following laws :
(a) Faraday first law of electrolysis
(b) Kohlrausch’s law of independent migration of ions
OR
(i) Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation.
(ii) For the cell reaction
Ni(s) | Ni²⁺(aq) || Ag⁺(aq) | Ag(s)
Calculate the equilibrium constant at 25 °C. How much maximum work would be obtained by operation of this cell ?
E°_Ni²⁺/Ni = 0.25 V and E°_Ag⁺/Ag = 0.80 V. [CBSE Comptt. Delhi 2015]
Answer:
(i) (a) Molar conductivity of a solution at a given concentration is the conducting power of all the ions produced by 1 mol of an electrolyte.
(b) Secondary battery can be recharged by passing current through it in opposite direction so that it can be used again.
(c) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.
(ii) (a) Faraday’s first law of electrolysis states that the amount of chemical reaction which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
(b) According to Kohlrausch law of independent migration of ions limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
OR
(i) Degree of dissociation is the ratio of molar conductivity at a specific concentration to the molar conductivity at infinite solution.
Degree of dissociation (α) = \( \frac{\Lambda^c_m}{\Lambda^0_m} \)
(ii) E°cell = E°_Ag⁺/Ag − E°_Ni²⁺/Ni
= 0.80V − 0.25V
= 0.55V
\( \log K_c = \left( \frac{nE^{\circ}_{cell}}{0.059} \right) \)
\( = \frac{2 \times 0.55}{0.059} \)
log K_c = 18.644
K_c = Antilog 18.644, K_c = 4.4 × 10¹⁸
ΔG° = − nFE°cell
= −2 × 96500 C mol⁻¹ × 0.55V
= −106,150 J mol⁻¹
Max. work = + 106150 J mol⁻¹
or 106.150 kJ mol⁻¹

Teacher's Note:
1. Write the laws exactly as stated; do not re-frame them in your own words. Maximum work obtainable from a cell equals −ΔG°.

Question 40. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K :
Sn (s) | Sn²⁺ (0.004 M) || H⁺ (0.020 M) | H₂ (g) (1 bar) | Pt (s)   (Given : E° Sn²⁺/ Sn = −0.14V)
(b) Give reasons :
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH₃COOH decreases on dilution.
OR
(a) For the reaction
2AgCl (s) + H₂ (g) (1 atm) → 2Ag(s) + 2H⁺ (0.1 M) + 2Cl⁻(0.1 M),
ΔG° = − 43600 J at 25°C.
Calculate the e.m.f. of the cell.
[log 10⁻ⁿ = −n]
(b) Define fuel cell and write its two advantages. [CBSE Delhi/OD 2018]
Answer:
(a) Sn + 2H⁺ → Sn²⁺ + H₂
\( E = E^{\circ} - \frac{0.059}{2} \log \frac{[Sn^{2+}]}{[H^+]^2} \)
\( = [0 - (-0.14)] - 0.0295 \log \frac{(0.004)}{(0.02)^2} \)
= 0.14 − 0.0295 log 10 = 0.11 V / 0.1105 V
(b) (i) Due to overpotential/ Overvoltage of O₂
(ii) The number of ions per unit volume decreases.
OR
(a) ΔG° = − nFE°
−43600 = −2 × 96500 × E°
E° = 0.226 V
E = E° − 0.059/2 log ([H⁺]²[Cl⁻]²/[H₂])
= 0.226 − 0.059/2 log[(0.1)² × (0.1)²] / 1
= 0.226 − 0.059/2 log 10⁻⁴
= 0.226 + 0.118 = 0.344 V
(b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol, etc.) directly into electrical energy are called fuel cells.
Advantages : High efficiency, non polluting (or any other suitable advantage)

Teacher's Note:
1. Write the balanced cell reaction first; it fixes n = 2 and shows that [H⁺] must be squared in the Nernst equation.

Question 41. (i) When a bright silver object is placed in the solution of gold chloride, it acquires a golden tinge but nothing happens when it is placed in a solution of copper chloride. Explain the behaviour of silver.
(ii) Consider the figure given above and answer the following questions :
(a) What is the direction of flow of electrons?
(b) Which is anode and which is cathode?
(c) What will happen if the salt bridge is removed?
(d) How will concentration of Zn²⁺ and Ag⁺ ions be affected when the cell functions?
(e) How will concentration of these ions be affected when the cell becomes dead? [CBSE Comptt. Delhi 2017]

Answer:
(i) E° value of silver is lower than that of gold, hence silver displaces gold which gets deposited on the silver object.
E° value of copper is lower than that of silver, hence silver cannot displace copper from its solution.
(ii) (a) Electrons flow from Zn to Ag plate.
(b) Zn as anode and Ag acts as cathode
(c) Cell will stop functioning
(d) Concentration of Zn²⁺ ions will increase and that of Ag⁺ ions will decrease.
(e) No change

Teacher's Note:
1. A metal displaces another from solution only if its own E° is lower (it is the better reducing agent); a dead cell is at equilibrium, so concentrations stop changing.