CBSE Class 12 Chemistry Electrochemistry Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Write the chemical formula of rust.
Answer:
Fe₂O₃.xH₂O

Teacher's Note:
1. Rust is hydrated iron(III) oxide – the water content x is variable.

Question 2. What is galvanization?
Answer:
The process of coating zinc over iron is called galvanization.

Teacher's Note:
1. Zinc protects iron even after scratching, because zinc (lower E°) is oxidised in preference to iron.

Question 3. What is cathodic protection?
Answer:
A process in which a metal is connected with a more reactive metal which acts as anode and the metal to be protected acts as a cathode is known as cathodic protection.

Teacher's Note:
1. The more reactive metal is called a sacrificial anode, e.g., Mg blocks attached to iron pipes and ship hulls.

Question 4. Which type of a metal can be used in cathodic protection of iron against rusting?
Answer:
A metal which is more electropositive than iron such as Al, Zn, Mg can be used in cathodic protection of iron against rusting.

Teacher's Note:
1. Pick a metal with a lower (more negative) E° than Fe²⁺/Fe (−0.44 V).

Question 5. Using the E° values of X and Y, predict which is better for coating the surface of iron to prevent rust and why?
Given: E°_(Fe²⁺/Fe) = −0.44 V, E°_(X²⁺/X) = −2.36 V, E°_(Y²⁺/Y) = −0.14 V [CBSE 2019 (56/4/3)]
Answer:
X, as its standard reduction potential is less than Y. Therefore, it will undergo oxidation more easily than Y.

Teacher's Note:
1. Y (−0.14 V) is above iron, so if the coating breaks iron would corrode faster; X (−2.36 V) is below iron and acts as a sacrificial metal.

Teacher's Note:
1. When the coating metal has a higher E° than iron, iron becomes the anode at a scratch and corrodes faster than uncoated iron.

Teacher's Note:
1. Derive units instead of memorising: κ = G*/R gives S cm⁻¹, and Λ_m = κ/C gives S cm² mol⁻¹.

Question 8. In an aqueous solution how does specific conductivity of electrolytes change with addition of water? [NCERT Exemplar]
Answer:
Conductivity decreases because number of ions per unit volume decreases.

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

Question 9. Under what condition is E_Cell = 0 or Δ_rG = 0? [NCERT Exemplar]
Answer:
At equilibrium, that is, when the cell is completely discharged.

Teacher's Note:
1. Since Δ_rG = −nFE_cell, both become zero together, at equilibrium (Q = K).

Question 10. Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell? [NCERT Exemplar]

Answer:
‘A’ will have negative polarity; ‘B’ will have positive polarity.

Teacher's Note:
1. An electrode joined to the zinc anode (−) of the galvanic cell becomes negative, and the one joined to the copper cathode (+) becomes positive.

Teacher's Note:
1. Link each cell to its one key feature: mercury cell – constant voltage (no ions in the overall reaction); fuel cell – about 70% efficiency.

Question 12. What are secondary cells? [NCERT Exemplar]
Answer:
Those cell which are rechargeable, i.e., in which products can convert back into reactants.

Teacher's Note:
1. Give an example to make the answer complete, e.g. the lead storage battery or the nickel-cadmium cell.

Short Answer Questions (2 Marks)

Question 13. The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. [CBSE Delhi 2011]
Answer:
According to electrochemical theory of rusting, the impure iron surface behaves like small electrochemical cell in the presence of water containing dissolved oxygen or carbon dioxide. In this cell pure iron acts as anode and impure iron surface acts as cathode. Moisture having dissolved CO₂ or O₂ acts as an electrolyte. The reactions are given below.
At anode: Fe → Fe²⁺ + 2e⁻;   E°_Fe²⁺/Fe = −0.44 V
At cathode: 2H⁺ + \( \frac{1}{2} \)O₂ + 2e⁻ → H₂O;   E°_H⁺/O2/H2O = 1.23 V
Overall reaction: Fe + 2H⁺ + \( \frac{1}{2} \)O₂ → Fe²⁺ + H₂O;   E°_cell = 1.67 V
The Fe²⁺ ions are further oxidised by atmospheric oxygen to Fe³⁺ ions, which comes out in the form of hydrated ferric oxide (rust).
2Fe²⁺ + \( \frac{1}{2} \)O₂ + 2H₂O → Fe₂O₃ + 4H⁺
Fe₂O₃ + xH₂O → Fe₂O₃.xH₂O (Rust)

Teacher's Note:
1. The H⁺ comes from H₂CO₃ (dissolved CO₂), which is why rusting is faster in acidic, moist air.

Question 14. Define the following terms :
(i) Fuel cell
(ii) Limiting molar conductivity (Λ^°_m) [CBSE OD 2014]
Answer:
(i) Galvanic cells that are designed to convert the energy of combustion of fuels (methane, methanol etc.) directly into electrical energy are called fuel cells.
(ii) When the concentration approaches zero, the molar conductivity is known as limiting molar conductivity. It is represented by Λ^°_m.

Teacher's Note:
1. Keep the phrase “directly into electrical energy” in the fuel cell definition; it is the key word examiners look for.

Question 15. (i) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution :
Ag⁺(aq) + e⁻ → Ag(s)    E° = +0.80 V
H⁺(aq) + e⁻ → \( \frac{1}{2} \)H₂(g)    E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why ?
(ii) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration ? [CBSE Delhi 2015]
Answer:
(i) Ag⁺(aq) + e⁻ → Ag(s); E° = +0.80 V
H⁺(aq) + e⁻ → \( \frac{1}{2} \)H₂(g); E° = 0.00 V
On the basis of their standard reduction potential (E°) values, cathode reaction is given by the one with higher E° values.
Thus Ag⁺(aq) + e⁻ → Ag(s) reaction will be more feasible at cathode.
(ii) Limiting molar conductivity : When the concentration approaches zero, the molar conductivity is known as limiting molar conductivity. It is represented by Λ^°_m.
The conductivity decreases with decrease in concentration due to decrease in the no. of ions that carry the current in a solution.

Teacher's Note:
1. At the cathode the species with the higher reduction potential is reduced first; at the anode the one with the lower reduction potential is oxidised first.

Question 16. (i) On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti⁴⁺ species may be used to oxidise Fe(II) to Fe(III).
Ti⁴⁺ + e⁻ → Ti³⁺    E° = +0.01 V
Fe³⁺ + e⁻ → Fe²⁺    E° = +0.77 V
(ii) Based on the data arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state. (Give a brief reason)
E°Cr³⁺/Cr²⁺ = −0.4 V
E°Mn³⁺/Mn²⁺ = +1.5 V
E°Fe³⁺/Fe²⁺ = +0.8 V [CBSE SQP 2018-2019]
Answer:
(i) Since Ti⁴⁺/Ti³⁺ has lower reduction potential than Fe³⁺/Fe²⁺, it cannot be reduced in comparison with Fe³⁺/Fe²⁺ ions.
Hence Ti⁴⁺ cannot oxidise Fe²⁺ to Fe³⁺.
(ii) As the value of reduction potential increases the stability of +2 oxidation state increases.
Therefore correct order of stability is
Cr³⁺/Cr²⁺ < Fe³⁺/Fe²⁺ < Mn³⁺/Mn²⁺

Teacher's Note:
1. A higher E°(M³⁺/M²⁺) means M³⁺ readily takes an electron, so the +2 state is the stable one: Cr²⁺ < Fe²⁺ < Mn²⁺.

Question 17. Calculate the emf of the following cell at 298 K
Cr(s)/Cr³⁺ (0.1M)//Fe²⁺ (0.01M)/Fe(s)
[Given: E^°_cell = + 0.30 V]
OR
The conductivity of 10⁻³ mol/L acetic acid at 25°C is 4.1 × 10⁻⁵ S cm⁻¹. Calculate its degree of dissociation if Λ^°_m for acetic acid at 25°C is 390.5 S cm² mol⁻¹. [CBSE SQP 2017]
Answer:
2Cr(s) + 3 Fe²⁺(aq.) → 3Fe(s) + 2Cr³⁺ (aq.)
n = 6
\( E_{Cell} = E^{o}_{Cell} - \frac{2.303RT}{nF} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} \)
\( E_{Cell} = 0.30\,V - \frac{0.059}{6}\,V \log \frac{[10^{-1}]^2}{[10^{-2}]^3} \)
E_Cell = 0.26 V
OR
\( \Lambda_m = \frac{1000\kappa}{C} \)
\( \Lambda_m = \frac{1000\ cm^3 L^{-1} \times 4.1 \times 10^{-5}\ S\,cm^{-1}}{10^{-3}\ mol\ L^{-1}} \)
= 41 S cm² mol⁻¹
\( \alpha = \frac{\Lambda^c_m}{\Lambda^0_m} \)
\( \alpha = \frac{41}{390.5} \) = 0.105

Teacher's Note:
1. Raise each concentration to the power of its coefficient in the balanced equation: [Cr³⁺]² and [Fe²⁺]³, with n = 6.

Question 18. Suggest a list of metals that are extracted electrolytically. [NCERT]
Answer:
Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

Teacher's Note:
1. These metals have large negative E° values, so no common chemical reducing agent can reduce their ions; only electrolysis of the fused salt works.

Question 19. A current of 1.50 A was passed through an electrolytic cell containing AgNO₃ solution with inert electrodes. The weight of silver deposited was 1.50 g. How long did the current flow ? (Molar mass of Ag = 108 g mol⁻¹, 1F = 96500 C mol⁻¹). [CBSE Comptt. Delhi/OD 2018]
Answer:
Quantity of charge required to deposit 108 g of silver = 96500 C
Quantity of charge required to deposit 1.50 g of silver = \( \frac{96500}{108} \) × 1.50 = 1340.28 C
Time taken = \( \frac{Q}{I} \) = \( \frac{1340.28}{1.50} \) = 893.5 s
(or by any other suitable method)

Teacher's Note:
1. Ag⁺ + e⁻ → Ag needs one faraday per mole, so 108 g of Ag needs 96500 C; then use t = Q/I.

Question 20. How can you increase the reduction potential of an electrode for the reaction:
Mⁿ⁺(aq) + ne⁻ → M(s)
Answer:
Nernst equation is:
E_Mⁿ⁺/M = E°_Mⁿ⁺/M − \( \frac{0.0591}{n} \) log \( \frac{1}{[M^{n+}]} \)
E_Mⁿ⁺/M can be increased by
(a) Increase in concentration of Mⁿ⁺ ions in solution.
(b) By increasing the temperature (when [Mⁿ⁺] > 1 M).

Teacher's Note:
1. A larger [Mⁿ⁺] makes log(1/[Mⁿ⁺]) smaller, so less is subtracted from E°.

Question 21. Calculate emf of the following cell at 298 K:
Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s)
[Given: E°_cell = 3.17 V]
Answer:
n = 2
The Nernst equation for the cell is:
E = E° − \( \frac{0.059}{2} \) log \( \frac{[Mg^{2+}]}{[Ag^+]^2} \)
= 3.17 − \( \frac{0.059}{2} \) log \( \frac{0.130}{(0.0001)^2} \)
= 3.17 − 0.21 = 2.96 V

Teacher's Note:
1. Solids (Mg, Ag) do not appear in the Nernst expression, and [Ag⁺] is squared because of the coefficient 2.

Question 22. Suggest a way to determine the Λ°_m value of water.
Answer:
Λ°_m(H₂O) = λ°(H⁺) + λ°(OH⁻)
It can be determined from the value of Λ°_m(HCl), Λ°_m(NaOH) and Λ°_m(NaCl). Then,
Λ°_m(H₂O) = Λ°_m(HCl) + Λ°_m(NaOH) − Λ°_m(NaCl)

Teacher's Note:
1. Check by adding ions: (H⁺ + Cl⁻) + (Na⁺ + OH⁻) − (Na⁺ + Cl⁻) leaves only H⁺ + OH⁻.

Question 23. How much electricity in terms of Faraday is required to produce 40 gram of Al from Al₂O₃? (Atomic mass of Al = 27 g/mol)
Answer:
Al³⁺ + 3e⁻ → Al
27 gram of Al require electricity = 3F
40 gram of Al require electricity = \( \frac{3F}{27} \) × 40 = 4.44 F

Teacher's Note:
1. Take n from the charge on the ion being reduced (Al³⁺ needs 3 electrons per atom).

Question 24. Predict the product of electrolysis of an aqueous solution of CuCl₂ with an inert electrode.
Answer:
CuCl₂(s) + aq → Cu²⁺ + 2Cl⁻
H₂O → H⁺ + OH⁻
At cathode (Reduction): Cu²⁺ will be reduced in preference to H⁺ ions.
Cu²⁺ + 2e⁻ → Cu(s)
At anode (Oxidation): Cl⁻ ions will be oxidised in preference to OH⁻ ions.
Cl⁻ → ½Cl₂ + e⁻
Thus, Cu will be deposited at cathode and Cl₂ will be liberated at anode.

Teacher's Note:
1. Cu²⁺ (E° = +0.34 V) is reduced more easily than water; Cl₂ forms at the anode because of overpotential of O₂.

Question 25. Calculate Λ°_m for CaCl₂ and MgSO₄ from the following data:
λ°(Ca²⁺) = 119.0, λ°(Mg²⁺) = 106.0, λ°(Cl⁻) = 76.3 and λ°(SO₄²⁻) = 160.0 S cm² mol⁻¹
Answer:
Λ°_m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻)
= 119 + (2 × 76.3) = 271.6 S cm² mol⁻¹
Λ°_m(MgSO₄) = λ°(Mg²⁺) + λ°(SO₄²⁻)
= 106 + 160 = 266 S cm² mol⁻¹

Teacher's Note:
1. Multiply each ionic conductivity by the number of those ions in one formula unit: two Cl⁻ in CaCl₂, but only one SO₄²⁻ in MgSO₄.

Short Answer Questions (3 Marks)

Question 26. A strip of nickel metal is placed in a 1 molar solution of Ni(NO₃)₂ and a strip of silver metal is placed in a 1 molar solution of AgNO₃. An electrochemical cell is created when the two solutions are connected by a salt bridge and the two strips are connected by wires to a voltmeter.
(i) Write the balanced equation for the overall reaction occurring in the cell and calculate the cell potential.
(ii) Calculate the cell potential, E, at 25°C for the cell if the initial concentration of Ni(NO₃)₂ is 0.100 molar and the initial concentration of AgNO₃ is 1.00 molar.
[E°_Ni²⁺/Ni = −0.25 V; E°_Ag⁺/Ag = 0.80 V, log 10⁻¹ = −1] [CBSE (F) 2012]
Answer:
(i) At anode: Ni → Ni²⁺ + 2e⁻
At cathode: [Ag⁺ + e⁻ → Ag] × 2
Cell reaction: Ni + 2Ag⁺ → Ni²⁺ + 2Ag
E°_cell = E°_cathode − E°_anode = E°_Ag⁺/Ag − E°_Ni²⁺/Ni
= 0.80 V − (−0.25 V)
E°_cell = 1.05 V
(ii) \( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} \)
Here, n = 2, E°_cell = 1.05 V, [Ni²⁺] = 0.1 M, [Ag⁺] = 1.0 M
\( E_{\text{cell}} = 1.05\ \text{V} - \frac{0.059}{2}\log\frac{(0.1)}{(1)^2} \)
E_cell = 1.05 V − 0.0295 log 10⁻¹
= 1.05 + 0.0295 V = 1.0795 V

Teacher's Note:
1. Lowering the product ion concentration (Ni²⁺) makes E_cell larger than E°_cell – a quick sense check of the sign.

Question 27. The emf of a cell corresponding to the reaction
Zn(s) + 2H⁺(aq) → Zn²⁺(0.1 M) + H₂ (g, 1 atm) is 0.28 volt at 25°C.
Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode.
E°_Zn²⁺/Zn = −0.76 V, E°_H⁺/H2 = 0 V [HOTS]
Answer:
Half-cell reactions:
At anode: Zn → Zn²⁺ + 2e⁻
At cathode: 2H⁺ + 2e⁻ → H₂
Cell reaction: Zn + 2H⁺ → Zn²⁺ + H₂
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log\frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2} \)
\( = \left(E^\circ_{\text{H}^+/\text{H}_2} - E^\circ_{\text{Zn}^{2+}/\text{Zn}}\right) - \frac{0.0591}{2}\log\frac{0.1}{[\text{H}^+]^2} \)
= [0 − (−0.76)] − 0.02955 [log 10⁻¹ − 2 log (H⁺)]
0.28 = 0.76 − 0.02955 (−1 + 2 pH)   [∵ pH = −log (H⁺)]
2 pH − 1 = 16.244
pH = 8.62

Teacher's Note:
1. −2 log[H⁺] becomes +2 pH; most errors in this question come from that sign change.

Question 28. From the following molar conductivities at infinite dilution, calculate Λ°_m for NH₄OH.
Λ°_m for Ba(OH)₂ = 457.6 Ω⁻¹ cm² mol⁻¹
Λ°_m for BaCl₂ = 240.6 Ω⁻¹ cm² mol⁻¹
Λ°_m for NH₄Cl = 129.8 Ω⁻¹ cm² mol⁻¹
Answer:
Λ°_m(NH4OH) = λ°_NH4⁺ + λ°_OH⁻
= (λ°_NH4⁺ + λ°_Cl⁻) + \( \frac{1}{2} \)(λ°_Ba²⁺ + 2λ°_OH⁻) − \( \frac{1}{2} \)(λ°_Ba²⁺ + 2λ°_Cl⁻)
= Λ°_m(NH4Cl) + \( \frac{1}{2} \)[Λ°_m(Ba(OH)2)] − \( \frac{1}{2} \)[Λ°_m(BaCl2)]
= 129.8 + \( \frac{1}{2} \) × 457.6 − \( \frac{1}{2} \) × 240.6
= 238.3 ohm⁻¹ cm² mol⁻¹

Teacher's Note:
1. Write each Λ° in terms of ions first, then choose the combination that leaves only NH₄⁺ and OH⁻.

Question 29. Conductivity of 2.5 × 10⁻⁴ M methanoic acid is 5.25 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity and degree of dissociation.
Given: λ°(H⁺) = 349.5 S cm² mol⁻¹ and λ°(HCOO⁻) = 50.5 S cm² mol⁻¹. [CBSE Allahabad 2015]
Answer:
\( \Lambda_m^c = \frac{\kappa\times1000}{M} \)
\( \Lambda_m^c = \frac{5.25\times10^{-5}\ \text{S cm}^{-1}\times1000\ \text{cm}^3\ \text{L}^{-1}}{2.5\times10^{-4}\ \text{mol L}^{-1}} \)
Λ^c_m = 210 S cm² mol⁻¹
Λ°_m(HCOOH) = Λ°_HCOO⁻ + Λ°_H⁺ = 50.5 S cm² mol⁻¹ + 349.5 S cm² mol⁻¹ = 400 S cm² mol⁻¹
Degree of dissociation, \( \alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ} = \frac{210\ \text{S cm}^2\text{mol}^{-1}}{400\ \text{S cm}^2\text{mol}^{-1}} = 0.525 \)
or α = 52.5%

Teacher's Note:
1. Two-step pattern: κ → Λ_m using the 1000 factor, then α = Λ_m/Λ°_m using Kohlrausch’s law.

Question 30. (i) A current of 1.50 A was passed through an electrolytic cell containing AgNO₃ solution with inert electrodes. The weight of Ag deposited was 1.50 g. How long did the current flow?
(ii) Write the reactions taking place at the anode and cathode in the above cell.
(iii) Give reactions taking place at the two electrodes if these are made up of Ag.
Answer:
(i) Ag⁺ + e⁻ → Ag
Quantity of charge required to deposit 108 g of silver = 96500 C
∴ Quantity of charge required to deposit 1.50 g of silver = \( \frac{96500}{108} \) × 1.50 = 1340.28 C
\( t = \frac{Q}{I} \)
∴ Time taken = \( \frac{1340.28}{1.50} \) = 893.52 s
(ii) Inert electrodes
Anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
Cathode: Ag⁺(aq) + e⁻ → Ag(s)
(iii) Ag electrodes
Anode: Ag(s) → Ag⁺(aq) + e⁻
Cathode: Ag⁺(aq) + e⁻ → Ag(s)

Teacher's Note:
1. With an active (Ag) anode the metal itself is oxidised; with an inert anode, water (not NO₃⁻) is oxidised.

Question 31. (i) Calculate Δ_rG° for the reaction
Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Given: E°_cell = +2.71 V, 1 F = 96500 C mol⁻¹
(ii) Name the type of cell which was used in Apollo space programme for providing electrical power. [CBSE (AI) 2014]
Answer:
(i) Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Δ_rG° = −nFE°_cell
Here, n = 2, F = 96500 C mol⁻¹ and E°_cell = 2.71 V
∴ Δ_rG° = −2 × 96500 C mol⁻¹ × 2.71 V = −523030 J mol⁻¹
= −523.03 kJ mol⁻¹
(ii) H₂–O₂ fuel cell

Teacher's Note:
1. 1 C × 1 V = 1 J, so −nFE° comes out directly in J mol⁻¹; divide by 1000 for kJ.

Question 32. Calculate the molar conductivity and degree of dissociation.
Conductivity of 2.5 × 10⁻⁴ M methanoic acid is 5.25 × 10⁻⁵ S cm⁻¹.
Given : Λ^°(H⁺) = 349.5 S cm² mol⁻¹ and Λ^°(HCOO⁻) = 50.5 S cm² mol⁻¹. [CBSE OD 2015]
Answer:
Λ_m = \( \frac{1000 \times \kappa}{M} \) S cm² mol⁻¹
Λ_m = \( \frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}} \) S cm² mol⁻¹
= 210 S cm² mol⁻¹
Λ^°_m(HCOOH) = λ^°(HCOO⁻) + λ^°(H⁺)
= (50.5 + 349.5) S cm² mol⁻¹ = 400 S cm² mol⁻¹
α = Λ_m/Λ^°_m
α = 210/400 = 0.525

Teacher's Note:
1. Find Λ_m from κ first and Λ^°_m from Kohlrausch law second; α is their ratio.

Question 33. (i) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
(ii) The products of electrolysis of aqueous NaCl at the respective electrodes are:
Cathode: H₂
Anode: Cl₂ and not O₂. Explain. [CBSE SQP 2017]
Answer:
(i) ‘B’ is a strong electrolyte.
A strong electrolyte is already dissociated into ions, but on dilution interionic forces are overcome, ions are free to move. So there is slight increase in molar conductivity on dilution.
(ii) On anode water should get oxidised in preference to Cl⁻, but due to overvoltage/overpotential Cl⁻ is oxidised in preference to water.

Teacher's Note:
1. A small rise in Λ_m on dilution means a strong electrolyte; a very large rise means a weak electrolyte whose degree of dissociation increases.

Question 34. (i) What are fuel cells ? Explain the electrode reactions involved in the working of H₂ – O₂ fuel cell.
(ii) Represent the galvanic cell in which the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) takes place. [CBSE Comptt. Delhi/OD 2013]
Answer:
(i) The cells which convert chemical energy of a fuel directly into electrical energy is known as fuel cells.
The electrode reactions are :
Anode : [H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻] × 2
Cathode : O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Net reaction : 2H₂(g) + O₂(g) → 2H₂O(l)
(ii)

May be represented as Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)

Teacher's Note:
1. Multiply the anode reaction by 2 so that 4 electrons cancel with the cathode reaction before writing the net reaction.

Question 35. Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery ? [CBSE Comptt. OD 2012]
Answer:
The cell reactions when the battery is in use are given below :
Anode : Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
Cathode : PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
i.e., Overall cell reaction consisting of cathode and anode reactions is :
Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
On charging the battery, the electrode reactions are reverse of those that occur during discharge.

Teacher's Note:
1. Students often get confused between the reactions at cathode and anode; remember that H₂SO₄ is used up on discharge, so its density falls.

Long Answer Questions (5 Marks)

Question 36. Consider the figure given below and answer the questions (i) to (vi):
(i) Write the direction of electron flow.
(ii) Is silver plate the anode or cathode?
(iii) What will happen if salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will concentration of Zn²⁺ ions and Ag⁺ ions be affected when the cell functions?
(vi) How will the concentration of Zn²⁺ ions and Ag⁺ ions be affected after the cell becomes ‘dead’? [NCERT Exemplar] [HOTS]

Answer:
(i) Electrons move from Zn to Ag.
(ii) Ag is the cathode.
(iii) Cell will stop functioning.
(iv) When E_cell = 0.
(v) Concentration of Zn²⁺ ions will increase and concentration of Ag⁺ ions will decrease.
(vi) When E_cell = 0, equilibrium is reached and concentration of Zn²⁺ ions and Ag⁺ ions will not change.

Teacher's Note:
1. Zinc has the lower E°, so it is the anode; the metal with the higher E° (Ag) is always the cathode.

Question 37. Calculate e.m.f and ΔG for the following cell at 298 K:
Mg(s) | Mg²⁺ (0.01 M) || Ag⁺ (0.0001 M) | Ag(s)
Given: [E°_Mg²⁺/Mg = −2.37 V, E°_Ag⁺/Ag = +0.80 V] [CBSE Guwahati 2015]
Answer:
At anode: Mg → Mg²⁺ + 2e⁻
At cathode: [Ag⁺ + e⁻ → Ag] × 2
Mg + 2Ag⁺ → Mg²⁺ + 2Ag; n = 2
E°_cell = E°_cathode − E°_anode = E°_Ag⁺/Ag − E°_Mg²⁺/Mg = 0.80 V − (−2.37 V) = 3.17 V
Substituting E°_cell = 3.17 V, n = 2, [Mg²⁺] = 1 × 10⁻² M, [Ag⁺] = 1 × 10⁻⁴ M in Nernst equation for above cell reaction,
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} \), we get
\( E_{\text{cell}} = 3.17 - \frac{0.059}{2}\log\frac{10^{-2}}{(10^{-4})^2} \)
E_cell = 3.17 − 0.0295 log 10⁶ = 3.17 − 0.177 V
E_cell = 2.993 V
Substituting n = 2, F = 96500 C mol⁻¹, E_cell = 2.993 V in the expression, ΔG = −nFE_cell we get
ΔG = −2 × 96500 C mol⁻¹ × 2.993 V
ΔG = −577649 J mol⁻¹ = −577.649 kJ mol⁻¹

Teacher's Note:
1. Use E_cell (not E°_cell) for ΔG when non-standard concentrations are given.

Question 38. (i) Calculate the standard free energy change for the following reaction at 25°C.
Au(s) + Ca²⁺(1 M) → Au³⁺(1 M) + Ca(s)
E°_Au³⁺/Au = +1.50 V, E°_Ca²⁺/Ca = −2.87 V
Predict whether the reaction will be spontaneous or not at 25°C. Which of the above two half cells will act as an oxidising agent and which one will be a reducing agent?
(ii) Tarnished silver contains Ag₂S. Can this tarnish be removed by placing tarnished silver ware in an aluminium pan containing an inert electrolytic solution such as NaCl? The standard electrode potential for half reaction:
Ag₂S(s) + 2e⁻ → 2Ag(s) + S²⁻ is −0.71 V
and for Al³⁺ + 3e⁻ → Al(s) is −1.66 V [HOTS]
Answer:
(i) Balanced reaction: 2Au(s) + 3Ca²⁺(1 M) → 2Au³⁺(1 M) + 3Ca(s); n = 6
E°_cell = E°_Ca²⁺/Ca − E°_Au³⁺/Au = (−2.87 V) − (1.50 V) = −4.37 V
Δ_rG° = −nFE°_cell = −6 × 96500 × (−4.37 V)
= +2530.230 kJ/mol
Since Δ_rG° is positive, therefore, reaction is non-spontaneous.
Au³⁺/Au half cell will be an oxidising agent while Ca²⁺/Ca half cell will be a reducing agent.
(ii) E°_cell for reaction of tarnished silver ware with aluminium pan is
(−0.71 V) − (−1.66 V) i.e., +0.95 V
Tarnished silver ware, therefore, can be cleaned by placing it in an aluminium pan as E°_cell is positive.

Teacher's Note:
1. Always balance electrons first: Au → Au³⁺ loses 3 and Ca²⁺ → Ca gains 2, so n = 6.

Teacher's Note:
1. “Centinormal” means 0.01 N; for a monoacidic base like NH₄OH this equals 0.01 M.

Question 40. (i) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.
(ii) Calculate Λ°_m for acetic acid.
Given that Λ°_m (HCl) = 426 S cm² mol⁻¹, Λ°_m (NaCl) = 126 S cm² mol⁻¹, Λ°_m (CH₃COONa) = 91 S cm² mol⁻¹ [CBSE Delhi 2010]
Answer:
(i) Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
Expression for the molar conductivity of acetic acid: Λ°_m(CH₃COOH) = Λ°_CH3COO⁻ + Λ°_H⁺
(ii) Λ°_m(CH₃COOH) = Λ°_CH3COO⁻ + Λ°_H⁺
= Λ°_CH3COO⁻ + Λ°_Na⁺ + Λ°_H⁺ + Λ°_Cl⁻ − (Λ°_Na⁺ + Λ°_Cl⁻)
= Λ°_m(CH3COONa) + Λ°_m(HCl) − Λ°_m(NaCl)
= (91 + 426 − 126) S cm² mol⁻¹
= 391 S cm² mol⁻¹

Teacher's Note:
1. Add the two salts that contain CH₃COO⁻ and H⁺, then subtract the one that removes the extra Na⁺ and Cl⁻.