CBSE Class 12 Chemistry Electrochemistry Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Suggest two materials other than hydrogen that can be used as fuels in fuel cells. [NCERT Exemplar]
Answer:
Methane and methanol can be used as fuels in fuel cells.

Teacher's Note:
1. The question says “other than hydrogen”, so do not write H₂; propane and ethanol are also acceptable fuels.

Question 2. To deposit 2 mol of Ca from CaCl₂ ............... electricity is required.
Answer:
4 F

Teacher's Note:
1. Ca²⁺ + 2e⁻ → Ca needs 2 F per mole, so 2 mol need 4 F.

Question 3. .................. gives a constant voltage throughout its life.
Answer:
Mercury cell

Teacher's Note:
1. Its overall reaction, Zn(Hg) + HgO(s) → ZnO(s) + Hg(l), involves no ions in solution, so the voltage (about 1.35 V) stays constant.

Question 4. Name the factors on which emf of a cell depends.
Answer:
Emf of a cell depends on following factors:
(a) Nature of reactants
(b) Concentration of solution in two half cells
(c) Temperature

Teacher's Note:
1. The Nernst equation shows the concentration and temperature dependence directly.

Question 5. What is the effect of temperature on the electrical conductance of metal?
Answer:
Temperature increases, electrical conductance decreases.

Teacher's Note:
1. In metals, increased vibration of the kernels at higher temperature hinders the flow of electrons.

Question 6. What is the relation between conductance and conductivity?
Answer:
Conductivity (κ) = Conductance (G) × cell constant \( \frac{l}{A} \)
i.e. κ = G × \( \frac{l}{A} \)

Teacher's Note:
1. Conductivity is the conductance of a 1 cm cube of solution; the cell constant converts one to the other.

Question 7. Reduction potentials of 4 metals A, B, C and D are −1.66 V, +0.34 V, +0.80 V and −0.76 V. What is the order of their reducing power and reactivity?
Answer:
A > D > B > C

Teacher's Note:
1. The lower (more negative) the reduction potential, the stronger the reducing agent.

Question 8. Why does a dry cell become dead even if it has not been used for a long time?
Answer:
NH₄Cl is acidic in nature. It corrodes zinc container.

Teacher's Note:
1. This is why the mercury cell, which uses a basic KOH–ZnO paste, has a longer shelf life.

Question 9. Why Na cannot be obtained by the electrolysis of aqueous NaCl solution?
Answer:
Due to low reduction potential, Na⁺ ions are not reduced at cathode. Instead, H⁺ are reduced and H₂ is obtained.

Teacher's Note:
1. Sodium is obtained by electrolysis of fused (molten) NaCl, where no water is present.

Question 10. Why Λ°_m for CH₃COOH cannot be determined experimentally?
Answer:
Molar conductivity of weak electrolytes keeps on increasing with dilution and does not become constant even at very large dilution.

Teacher's Note:
1. For weak electrolytes, Λ°_m is found indirectly using Kohlrausch’s law.

Question 11. Why is it necessary to use a salt bridge in a galvanic cell?
Answer:
To complete the inner circuit and to maintain electrical neutrality of the electrolytic solutions of the half cells.

Teacher's Note:
1. Without the salt bridge, charge would build up in the half cells and the current would stop at once.

Question 12. Why does mercury cell give a constant voltage throughout its life?
Answer:
This is because the overall cell reaction does not have any ionic concentration in it.

Teacher's Note:
1. Zn(Hg) + HgO(s) → ZnO(s) + Hg(l): with no ions in the equation, the Nernst term stays constant.

Question 13. Why does the conductivity of a solution decrease with dilution?
Answer:
Conductivity of a solution is dependent on the number of ions per unit volume. On dilution, the number of ions per unit volume decreases, hence the conductivity decreases.

Teacher's Note:
1. Molar conductivity behaves the opposite way: it increases on dilution.

Short Answer Questions (2 Marks)

Question 14. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Answer:
H⁺ + e⁻ → ½H₂, n = 1
E = E° − \( \frac{0.0591}{n} \) log \( \frac{1}{[H^+]} \)
E = 0 − \( \frac{0.0591}{1} \) × pH
E = −0.0591 × 10 V
E = −0.591 V

Teacher's Note:
1. Since log(1/[H⁺]) = pH, the hydrogen electrode potential is simply E = −0.0591 × pH at 298 K.

Question 15. If a current of 0.5 amp flows through a metallic wire for 2 hours, how many electrons would flow through the wire?
Answer:
q = i × t = 0.5 × 2 × 60 × 60 = 3600 C
96500 coulombs are equal to 6.022 × 10²³ electrons
So, 3600 coulombs = \( \frac{6.022 \times 10^{23}}{96500} \) × 3600 = 2.246 × 10²² electrons

Teacher's Note:
1. 1 F is the charge on one mole (6.022 × 10²³) of electrons; convert hours to seconds first.

Question 16. How much electricity is required in coulomb for the oxidation of 1 mole of FeO to Fe₂O₃?
Answer:
Fe²⁺ → Fe³⁺ + 1e⁻
So, 1F = 1 × 96500 C = 96500 C

Teacher's Note:
1. Only the change in oxidation state of iron (+2 to +3) matters, i.e. one electron per FeO.

Question 17. Define conductivity and molar conductivity for a solution of an electrolyte.
Answer:
Conductivity is defined as ease with which current flows through electrolyte. It is reciprocal of specific resistance.
Molar conductivity is conductance of all the ions produced by one mole of electrolyte when electrodes are at unit distance apart and have sufficient area of cross-section to hold electrolyte.

Teacher's Note:
1. Write the formulas too: κ = 1/ρ and Λ_m = κ × 1000/C.

Question 18. The resistance of conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the 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 S cm⁻¹ × 1500 Ω = 0.219 cm⁻¹

Teacher's Note:
1. Since S = Ω⁻¹, the units reduce to cm⁻¹, a quick check of the formula.

Question 19. Indicate the reactions which take place at cathode and anode in fuel cell.
Answer:
At cathode: O₂(g) + 2H₂O + 4e⁻ → 4OH⁻(aq)
At anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O + 4e⁻
The overall reaction is: 2H₂(g) + O₂(g) → 2H₂O(l)

Teacher's Note:
1. In the H₂–O₂ fuel cell the electrolyte is concentrated aqueous KOH, which is why OH⁻ appears in both half reactions.

Question 20. Explain Kohlrausch’s law of independent migration of ions.
Answer:
It states that at infinite dilution, molar conductivity of an electrolyte is equal to sum of contributions due to cation as well as anion.
e.g. Λ°_m(Na₂SO₄) = 2λ°(Na⁺) + λ°(SO₄²⁻)

Teacher's Note:
1. In general Λ°_m = ν₊λ°₊ + ν₋λ°₋; do not forget the number of ions per formula unit.

Question 21. The standard reduction potential for the Zn²⁺(aq)/Zn(s) half cell is −0.76 V. Write the reactions occurring at the electrodes when coupled with standard hydrogen electrode (SHE).
Answer:
At anode: Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode: 2H⁺ + 2e⁻ → H₂(g)
Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

Teacher's Note:
1. A negative E° means the electrode acts as the anode when paired with SHE.

Question 22. Calculate the electrode potential of a copper wire dipped in 0.1 M CuSO₄ solution at 25°C. The standard electrode potential of copper is 0.34 volt.
Answer:
The electrode reaction written as reduction potential is
Cu²⁺ + 2e⁻ → Cu   n = 2
E_Cu²⁺/Cu = E°_Cu²⁺/Cu − \( \frac{0.0591}{2} \) log \( \frac{1}{[Cu^{2+}]} \) = 0.34 − \( \frac{0.0591}{2} \) log \( \frac{1}{0.1} \) = 0.3104 V

Teacher's Note:
1. A concentration below 1 M lowers the reduction potential below E°.

Question 23. Two metals A and B have reduction potential values −0.76 V and +0.34 V respectively. Which of these will liberate H₂ from dil. H₂SO₄?
Answer:
Metal having higher oxidation potential will liberate H₂ from H₂SO₄. Thus, A will liberate H₂ from H₂SO₄.

Teacher's Note:
1. Only metals with negative reduction potential (above hydrogen in the series) displace H₂ from dilute acids.

Question 24. How does conc. of sulphuric acid change in lead storage battery when current is drawn from it?
Answer:
Concentration of sulphuric acid decreases.

Teacher's Note:
1. H₂SO₄ is consumed and water is formed during discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O.

Question 25. What type of a battery is lead storage cell? Write the anode and cathode reaction and overall reaction occurring in a lead storage battery during discharging and recharging of cell.
Answer:
It is a secondary cell.
Anode reaction: Pb + SO₄²⁻ → PbSO₄ + 2e⁻
Cathode reaction: PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O
Pb(s) + PbO₂(s) + 2H₂SO₄ \( \underset{\text{Recharging}}{\overset{\text{Discharging}}{\rightleftharpoons}} \) 2PbSO₄(s) + 2H₂O(l)

Teacher's Note:
1. On recharging, the reactions at both electrodes are simply reversed.

Question 26. E° values of MnO₄⁻, Ce⁴⁺ and Cl₂ are 1.507, 1.61 and 1.358 V respectively. Arrange these in order of increasing strength as oxidising agent.
Answer:
Cl₂ < MnO₄⁻ < Ce⁴⁺

Teacher's Note:
1. Higher reduction potential means a greater tendency to gain electrons, i.e. a stronger oxidising agent.

Short Answer Questions (3 Marks)

Question 27. Silver is uniformly electro-deposited on a metallic vessel of surface area of 900 cm² by passing a current of 0.5 ampere for 2 hours. Calculate the thickness of silver deposited.
[Given : the density of silver is 10.5 g cm⁻³ and atomic mass of Ag = 108 amu.] [CBSE Comptt. OD 2013]
Answer:
Calculation of mass of Ag deposited :
The electrode reaction is Ag⁺ + e⁻ → Ag
The quantity of electricity passed = Current × Time
= 0.5 (amp.) × 2 × 60 × 60 (sec)
= 3600 C.
From the electrode reaction, it is clear that 96500 C of electricity deposit Ag = 108 g
3600 C of electricity will deposit Ag = \( \frac{108}{96500} \) × 3600 = 4.03 g
Calculation of thickness:
Let the thickness of silver deposited be x cm.
Mass = Volume × Density
= Area × Thickness × Density (Volume = Area × thickness)
4.03 g = 900 (cm²) × x (cm) × 10.5 (g cm⁻³)
x = \( \frac{4.03\ cm}{900 \times 10.5} \) = 4.26 × 10⁻⁴ cm

Teacher's Note:
1. Always write the working formula followed by the substituted values, and carry units through; Volume = Area × thickness links mass to thickness.

Question 28. (i) How many coulombs are required to reduce 1 mole Cr₂O₇²⁻ to Cr³⁺ ?
(ii) The conductivity of 0.001 M acetic acid is 4 × 10⁻⁵ S/cm. Calculate the dissociation constant of acetic acid if Λ^°_m for acetic acid is 390 S cm² mol⁻¹. [CBSE Comptt. OD 2012]
Answer:
(i) Cr₂O₇²⁻ + 14 H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
One mole Cr₂O₇²⁻ requires 6 moles of electrons for reduction.
Thus, quantity of electricity required = 6 × 96,500
= 579000 Coulomb
= 5.79 × 10⁵ Coulomb
(ii) C = 0.001 M
κ = 4 × 10⁻⁵ S cm⁻¹
\( \Lambda_m = \frac{1000 \times \kappa}{C} \)
\( \Lambda_m = \frac{1000\ cm^3 L^{-1} \times 4 \times 10^{-5}\ S\,cm^{-1}}{0.001\ mol\ L^{-1}} \)
Λ_m = 40 S cm² mol⁻¹
Λ° = 390 S cm² mol⁻¹
α = degree of dissociation of CH₃COOH
\( \alpha = \frac{\Lambda_m}{\Lambda^{\circ}} = \frac{40}{390} = 0.1025 \)
α = 0.1025 ≈ 0.103
\( K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \)
CH₃COOH → CH₃COO⁻ + H⁺
C(1 − α)     Cα     Cα
\( K_a = \frac{C\alpha \times C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{(1-\alpha)} \)
\( = \frac{0.001 \times (0.103)^2}{1 - 0.103} \)
= 1.18 × 10⁻⁵

Teacher's Note:
1. Count electrons from the change in oxidation state: each Cr goes from +6 to +3, and there are two Cr atoms, so 6 F are needed.

Question 29. What is corrosion ? Explain the electro-chemical theory of rusting of iron and write the reactions involved in the rusting of iron. [CBSE Comptt. Delhi 2012]
Answer:
The process of slowly eating away of the metal due to attack of the moisture and atmospheric gases on the surface of the metal resulting into the formation of compound such as oxides, sulphides, carbonates, sulphates etc., is called corrosion.
The electrochemical phenomenon of rusting of iron can be described as :
At anode : Fe(s) undergoes oxidation to releases electrons.
Fe(s) → Fe²⁺(aq) + 2e⁻
At cathode : O₂(g) + 4H⁺ + 4e⁻ → 2H₂O(l)
Electrons released at anode move to another metal and reduce oxygen in presence of H⁺. It is available from H₂CO₃ formed from the dissolution of CO₂ from air into water. H⁺ in water may be available also through dissolution of other acidic oxides from the atmosphere.
This site behaves as cathode
Net reaction : Fe(s) + 2H⁺(aq) + \( \frac{1}{2} \) O₂(g) → Fe²⁺(aq) + H₂O(l)
Fe²⁺ again gets oxidised to form rust.
2Fe²⁺(aq) + \( \frac{1}{2} \) O₂(g) + 2H₂O(l) → Fe₂O₃(s) + 4H⁺ (Rust)

Teacher's Note:
1. Rusting needs both water and oxygen; the H⁺ from dissolved CO₂ explains why iron rusts faster in moist, polluted or salty air.

Question 30. A solution of CuSO₄ is electrolysed for 10 mins. with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Answer:
I = 1.5 Ampere
Time = 10 × 60 s = 600 s
Q = I × t = 1.5 × 600 = 900 C
Cu²⁺ + 2e⁻ → Cu(s)
2F amount of electricity deposit copper = 63.5 g
900 C amount of electricity deposit copper = \( \frac{63.5 \times 900}{2 \times 96500} \) = 0.296 g

Teacher's Note:
1. Convert minutes to seconds before using Q = I × t; 1 F = 96500 C.

Question 31. Predict the products of electrolysis in each of the following:
(a) An aqueous solution of AgNO₃ with platinum electrodes.
(b) An aqueous solution of CuCl₂ with Pt electrodes.
Answer:
(a) At anode (Oxidation): 4OH⁻ − 4e⁻ → 2H₂O + O₂
At cathode (Reduction): Ag⁺ + e⁻ → Ag(s)
(b) At anode (Oxidation): Cl⁻ − e⁻ → Cl(g); Cl + Cl → Cl₂
At cathode (Reduction): Cu²⁺ + 2e⁻ → Cu(s)

Teacher's Note:
1. NO₃⁻ is not oxidised, so water/OH⁻ gives O₂ at the anode in (a); in (b) Cl₂ forms because of the overpotential of O₂.

Question 32. Determine the values of equilibrium constant K_c and ΔG° for the following reaction:
Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s); E° = 1.05 V
Answer:
ΔG° = −nFE°_cell
n = 2, E°_cell = 1.05 V, F = 96500 C mol⁻¹
ΔG° = −2 × 1.05 × 96500 = −202.650 kJ
ΔG° = −RT ln K_c
ln K_c = −\( \frac{\Delta G^\circ}{RT} \) = \( \frac{202.650 \times 10^3}{8.314 \times 298} \)
K_c = 3.32 × 10³⁵

Teacher's Note:
1. Convert ΔG° to joules before dividing by RT (R = 8.314 J K⁻¹ mol⁻¹).

Question 33. The K_sp for AgCl at 298 K is 1.0 × 10⁻¹⁰. Calculate the electrode potential for Ag⁺/Ag electrode immersed in 1.0 M KCl solution. Given E°_Ag⁺/Ag = 0.80 V.
Answer:
AgCl(s) ⇄ Ag⁺ + Cl⁻
K_sp = [Ag⁺][Cl⁻]
[Cl⁻] = 1.0 M
[Ag⁺] = \( \frac{K_{sp}}{[Cl^-]} \) = \( \frac{1 \times 10^{-10}}{1} \) = 1 × 10⁻¹⁰ M
Now, Ag⁺ + e⁻ → Ag(s)
E = E° − \( \frac{0.059}{1} \) log \( \frac{1}{[Ag^+]} \)
= 0.80 − \( \frac{0.059}{1} \) log \( \frac{1}{10^{-10}} \)
= 0.80 − 0.059 × 10 = 0.21 V

Teacher's Note:
1. Very small [Ag⁺] (fixed by K_sp) lowers the electrode potential sharply; this is the basis of the Ag/AgCl electrode.

Question 34. Two electrolytic cells containing silver nitrate solution and copper sulphate solution are connected in series. A steady current of 2.5 amp was passed through them till 1.078 g of Ag were deposited. How long did the current flow? What weight of copper will be deposited? (Ag = 107.8 u, Cu = 63.5 u)
Answer:
w = z × i × t
t = \( \frac{w}{z \times i} \)
t = \( \frac{1.078 \times 1 \times 96500}{107.8 \times 2.5} \) = 386 seconds
Cu²⁺ + 2e⁻ → Cu
w = \( \frac{63.5}{2 \times 96500} \) × 2.5 × 386 = 0.3175 gram

Teacher's Note:
1. In series, the same charge passes through both cells, so masses deposited are in the ratio of their equivalent masses (Faraday’s second law).

Question 35. A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5.0 amp for 20 minutes. What mass of the nickel will be deposited at the cathode? (Ni = 58.7 u)
Answer:
w = z × i × t
z = \( \frac{58.7}{2 \times 96500} \)
i = 5.0 A, t = 20 × 60 = 1200 s
w = \( \frac{58.7}{2 \times 96500} \) × 5.0 × 1200
w = 1.825 gram

Teacher's Note:
1. Ni²⁺ needs 2 electrons, so the electrochemical equivalent z = M/(2F).

Question 36. The cell in which the following reaction occurs:
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) has E°_cell = 0.236 V.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Answer:
n = 2
ΔG° = −nFE°_cell = −2 × 96500 × 0.236 J = −45.55 kJ/mol
ΔG° = −2.303 RT log K_c
log K_c = \( \frac{\Delta G^\circ}{-2.303RT} \) = \( \frac{45.55 \times 10^3}{2.303 \times 8.314 \times 298} \) = 7.983
K_c = antilog (7.983) = 9.616 × 10⁷

Teacher's Note:
1. Two Fe³⁺ each gain one electron, so n = 2 for the reaction as written.

Long Answer Questions (5 Marks)

Question 37. (i) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
(ii) Calculate the potential for half-cell containing 0.10 M K₂Cr₂O₇ (aq), 0.20 M Cr³⁺(aq) and 1.0 × 10⁻⁴ M H⁺(aq). The half cell reaction is
Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)
and the standard electrode potential is given as E° = 1.33 V. [CBSE (AI) 2011]
Answer:
(i) Lead storage battery is a secondary battery (it can be recharged).
Anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
Cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Overall cell reaction: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
(ii) For half cell reaction
Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}} \)
Here, E° = 1.33 V, n = 6, [Cr³⁺] = 0.2 M, [Cr₂O₇²⁻] = 0.1 M, [H⁺] = 1 × 10⁻⁴ M
Substituting these values in the given expression, we get
\( E_{\text{cell}} = 1.33\ \text{V} - \frac{0.0591}{6}\log\frac{(0.20)^2}{(0.1)(10^{-4})^{14}} \)
\( = 1.33\ \text{V} - \frac{0.0591}{6}\log(4\times10^{55}) \)
\( = 1.33\ \text{V} - \frac{0.0591}{6}[\log4 + \log10^{55}] \)
\( = 1.33\ \text{V} - \frac{0.0591}{6}[2\log2 + 55\log10] \)
\( = 1.33\ \text{V} - \frac{0.0591}{6}[2\times0.3010 + 55] \)
= 1.33 V − 0.548 V = 0.782 V

Teacher's Note:
1. Low [H⁺] (raised to the power 14) sharply lowers the oxidising power of dichromate – that is why it is used in acidic medium.

Question 38. (i) Define the following terms:
(a) Limiting molar conductivity
(b) Fuel cell
(ii) Resistance of a conductivity cell filled with 0.1 mol L⁻¹ KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L⁻¹ KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L⁻¹ KCl solution. The conductivity of 0.1 mol L⁻¹ KCl solution is 1.29 × 10⁻² Ω⁻¹ cm⁻¹. [CBSE (AI) 2014]
Answer:
(i) (a) The limiting value of molar conductivity when concentration approaches to zero is called limiting molar conductivity.
(b) A fuel cell is a device which converts energy produced during the combustion of fuels like hydrogen, methane, methyl alcohol etc. directly into electrical energy. One such successful fuel cell is hydrogen-oxygen fuel cell.
(ii) For 0.1 mol L⁻¹ KCl solution,
Conductivity, κ = 1.29 × 10⁻² Ω⁻¹ cm⁻¹, Resistance, R = 100 Ω
Cell constant = Conductivity × resistance = 1.29 × 10⁻² Ω⁻¹ cm⁻¹ × 100 Ω = 1.29 cm⁻¹
For 0.02 mol L⁻¹ solution,
Resistance = 520 Ω, Cell constant = 1.29 cm⁻¹
Conductivity, \( \kappa = \frac{\text{Cell constant}}{\text{Resistance}} = \frac{1.29\ \text{cm}^{-1}}{520\ \Omega} \) = 0.00248 Ω⁻¹ cm⁻¹
Molar conductivity, \( \Lambda_m = \frac{\text{Conductivity}(\kappa)\times1000\ \text{cm}^3\ \text{L}^{-1}}{\text{Molarity}} = \frac{0.00248\ \Omega^{-1}\text{cm}^{-1}\times1000\ \text{cm}^3\ \text{L}^{-1}}{0.02\ \text{mol L}^{-1}} \)
= 124 Ω⁻¹ cm² mol⁻¹

Teacher's Note:
1. The cell constant does not change with the solution, so find it once from the standard KCl and reuse it.

Question 39. (i) State two advantages of H₂–O₂ fuel cell over ordinary cell. [CBSE 2020 (56/1/1)]
(ii) Silver is electrodeposited on a metallic vessel of total surface area 500 cm² by passing a current of 0.5 amp for two hours. Calculate the thickness of silver deposited.
[Given: Density of silver = 10.5 g cm⁻³, Atomic mass of silver = 108 amu, F = 96,500 C mol⁻¹]
Answer:
(i) Advantages of fuel cell:
(a) It is a pollution-free device since no harmful products are formed.
(b) Its efficiency is about 75% which is considerably higher than conventional cells.
(c) It is a continuous source of energy if the supply of gases is maintained. (Any Two)
(ii) \( m = ZIt = \frac{108}{96500}\times0.5\times2\times3600 = 4.029\ \text{g} \)
\( d = \frac{m}{V} \Rightarrow V = \frac{m}{d} \)
\( V = \frac{4.029\ \text{g}}{10.5\ \text{g cm}^{-3}} = 0.3837\ \text{cm}^3 \)
Let the thickness of silver deposited be x cm.
∴ V = A × x or \( x = \frac{V}{A} \)
\( x = \frac{0.3837}{500} \) = 7.67 × 10⁻⁴ cm

Teacher's Note:
1. Chain the steps: charge → mass (Faraday) → volume (density) → thickness (volume / area).

Question 40. (i) What is limiting molar conductivity? Why there is steep rise in the molar conductivity of weak electrolyte on dilution?
(ii) Calculate the emf of the following cell at 298 K:
Mg(s)|Mg²⁺ (0.1 M)||Cu²⁺ (1.0 × 10⁻³ M)|Cu(s)
[Given = E^°_Cell = 2.71 V] [CBSE OD 2016]
Answer:
(i) When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.
The change in Λ_m with dilution is due to the increase in the degree of dissociation and consequently the number of ions in the total volume of the solution that contains 1 mol of electrolyte, hence Λ_m increases steeply.
(ii) \( E_{Cell} = E^{0}_{Cell} - \frac{0.059}{n} \log \frac{[Mg^{2+}]}{[Cu^{2+}]} \)
\( = 2.71\,V - \frac{0.059}{2} \log \frac{0.1}{0.001} \)
\( = 2.71\,V - \frac{0.059}{2} \log 10^2 \)
= 2.651 V

Teacher's Note:
1. Products go in the numerator of Q: [Mg²⁺]/[Cu²⁺]; a Q greater than 1 makes E_cell smaller than E°_cell.

Question 41. (i) Calculate E^°_cell for the following reaction at 298K :
2Al(s) + 3Cu²⁺ (0.01M) → 2Al³⁺ (0.01M) + 3Cu(s)
Given : E_cell = 1.98 V
(ii) Using the E° values of A and B, predict which is better for coating the surface of iron [E°(Fe²⁺/Fe) = − 0.44V] to prevent corrosion and why ?
Given : E° (A²⁺/A) = − 2.37V : E° (B²⁺/B) = − 0.14V
OR
(i) The conductivity of 0.001 mol L⁻¹ solution of CH₃COOH is 3.905 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity and degree of dissociation (α).
Given λ° (H⁺) = 349.6 S cm² mol⁻¹ and λ° (CH₃COO⁻) = 40.9 S cm² mol⁻¹
(ii) Define electrochemical cell. What happens if external potential applied becomes greater than E^°_cell of electrochemical cell ? [CBSE OD 2016]
Answer:
(i) Al(s) | Al³⁺(aq) (0.01M) || Cu²⁺(aq) (0.01M)|Cu(s)
LHE [Al(s) → Al³⁺(aq) + 3e⁻] × 2 (Oxidation at anode)
RHE [Cu²⁺(aq) + 2e⁻ → Cu(s)] × 3 (Reduction at cathode)
∴ n = 6
\( E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3} \)
\( E^{\circ}_{cell} = E_{cell} + \frac{0.0591}{n} \log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3} \)
\( = 1.98 + \frac{0.0591}{6} \log \frac{(0.01)^2}{(0.01)^3} \)
\( = 1.98 + \frac{0.0591}{6} \log 10^2 \)
\( = 1.98 + \frac{0.0591}{6}\ 2 \log 10 \)
\( = 1.98 + \frac{0.0591}{6} \times 2 \) [∵ log 10 = 1]
= 1.98 + 0.0197
= 1.9997 V
(ii) A is better than B because its E° value is more negative.
OR
(i) C = 0.001 mol L⁻¹, κ = 3.905 × 10⁻⁵ S cm⁻¹
Molar conductivity
\( \lambda_m = \frac{\kappa \times 1000}{C} \)
\( = \frac{3.905 \times 10^{-5}\ S\,cm^{-1} \times 1000\ cm^3 L^{-1}}{0.001\ mol\ L^{-1}} \)
= 39.05 S cm² mol⁻¹ or Ω⁻¹ cm² mol⁻¹
λ° (H⁺) = 349.6 S cm² mol⁻¹,
λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹
CH₃COOH → CH₃COO⁻ + H⁺
Λ° (CH₃COOH) = Λ° CH₃COO⁻ + Λ°H⁺
= (40.9 + 349.6) S cm² mol⁻¹
= 390.5 S cm² mol⁻¹
\( \alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} \)
\( = \frac{39.05}{390.5} \)
= 0.1
(ii) Electrochemical cell is a device used for the production of electricity from energy released during spontaneous chemical reaction and use electrical energy to bring about the chemical change.
If the external potential applied becomes greater than E^°_cell of electrochemical cell, the reaction gets reversed. It starts acting as an electrolytic cell and vice-versa.

Teacher's Note:
1. For protecting iron, the coating metal must be more easily oxidised than iron (more negative E° than −0.44 V), so it corrodes instead of iron.