CBSE Class 12 Chemistry Electrochemistry Questions Answers

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ELECTROCHEMISTRY - ONE MARK EACH

1. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________. [1 Mark]

Answer: Cell emf (electromotive force).

Teacher's Note:
a) Emf is the potential difference measured when no current is drawn, so it is the maximum potential the cell can supply.

2. Under what condition an electrochemical cell can behave like an electrolytic cell ? [1 Mark]

Answer: When an external opposing potential greater than the cell emf (E_ext > E_cell) is applied, the cell behaves as an electrolytic cell.

Teacher's Note:
a) The reverse reaction then takes place and the cell consumes electrical energy.

3. What is the quantity of charge in faraday is required to obtain one mole of aluminum from Al₂O₃ ? [1 Mark]

Answer: 3 F (3 × 96500 C).

Teacher's Note:
a) Al³⁺ + 3e^- → Al, so each mole of aluminium needs three moles of electrons.

4. How the cell constant of a conductivity cell changes with change of electrolyte, concentration and temperature? [1 Mark]

Answer: It remains unchanged.

Teacher's Note:
a) Cell constant depends only on the distance between the electrodes and the area of the electrodes, not on the solution.

5. What will happen at anode during the electrolysis of aqueous solution of CuSO₄ in the presence of Cu electrodes? [1 Mark]

Answer: Copper will dissolve at the anode (Cu → Cu²⁺ + 2e^-).

Teacher's Note:
a) With copper electrodes, the anode itself is oxidised, so the concentration of Cu²⁺ in the solution stays constant.

6. Under what condition is E_Cell = 0 or Δ_rG = 0 ? [1 Mark]

Answer: When the cell reaction reaches equilibrium.

Teacher's Note:
a) At equilibrium there is no net driving force, so the cell emf and the free energy change both become zero.

OR

Give the condition for Daniel Cell in which there is no flow of electrons or current. [1 Mark]

Answer: When the external opposing potential is equal to the emf of the Daniell cell, no current flows.

Teacher's Note:
a) This is the same equilibrium condition, seen from the external circuit.

7. Why is alternating current used for measuring resistance of an electrolytic solution? [1 Mark]

Answer: Alternating current prevents electrolysis, so the concentration of ions in the solution remains constant.

Teacher's Note:
a) With direct current, the ions would be discharged at the electrodes and the solution would change composition.

8. How will the pH of brine (aq. NaCl solution) be affected when it is electrolyzed ? [1 Mark]

Answer: The pH of the solution rises, because NaOH is formed at the cathode.

Teacher's Note:
a) Hydroxide ions are produced at the cathode as hydrogen gas is released.

9. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why? [1 Mark]

Answer: Ions are not involved in the overall cell reaction of the mercury cell.

Teacher's Note:
a) The reactants and products are solids, so their concentrations do not change during use.

10. Mention the purpose of salt-bridge placed between two half-cells of a galvanic cell? [1 Mark]

Answer: The salt bridge completes the electrical circuit and keeps the two half-cells electrically neutral.

Teacher's Note:
a) Ions flow through the salt bridge to balance the charge build-up in each half-cell.

11. Two metals A and B have electrode potential values of - 0.25V and 0.80V respectively. Which of these will liberate hydrogen gas from dilute H₂SO₄? [1 Mark]

Answer: Metal A (E° = -0.25 V).

Teacher's Note:
a) Only metals with a negative standard electrode potential, lower than that of H⁺/H₂, can displace hydrogen from acids.

12. What is the effect of temperature on molar conductivity? [1 Mark]

Answer: Molar conductivity increases with rise in temperature.

Teacher's Note:
a) Higher temperature lowers the viscosity of the solution and makes the ions move faster.

13. What is the role of ZnCl₂ in the dry cell? [1 Mark]

Answer: ZnCl₂ absorbs the ammonia produced in the reaction by forming the complex [Zn(NH₃)₄]²⁺.

Teacher's Note:
a) Removing ammonia prevents polarisation of the carbon electrode, so the cell keeps working steadily.

14. Why is the equilibrium constant K, related to only E° cell and not E_cell ? [1 Mark]

Answer: Because E_cell is zero at equilibrium.

Teacher's Note:
a) The equilibrium constant is calculated from the standard emf, which is measured under standard conditions.

15. Rusting of iron is quicker in saline water than in ordinary water. Why is it so? [1 Mark]

Answer: Saline water contains more ions, so it conducts electricity better and the corrosion cell works faster.

Teacher's Note:
a) A higher conductivity lets electrons and ions move more easily between the anodic and cathodic areas on the iron surface.

16. Why rusting of iron prevented in alkaline medium? [1 Mark]

Answer: In alkaline medium, atmospheric oxygen is unable to accept the electrons released by the oxidation of Fe, so rusting is prevented.

Teacher's Note:
a) Rusting needs the cathodic reduction of oxygen, which is hindered in an alkaline medium.

17. Which will have greater molar conductivity and why? 1 mole KCl dissolved in 200 cc of the solution OR 1 mole KCl dissolved in 500 cc of the solution. [1 Mark]

Answer: 1 mole KCl dissolved in 500 cc of the solution, because of the greater mobility of ions in the more dilute solution.

Teacher's Note:
a) Dilution reduces the interionic attraction, so the ions move more freely and the molar conductivity increases.

18. Why Lead storage battery as a secondary cell can be recharged? [1 Mark]

Answer: PbSO₄ formed during discharge is a sticky solid which adheres to the electrodes, so it can be converted back to Pb and PbO₂ on charging.

Teacher's Note:
a) Because the product stays on the electrodes, the reaction can be reversed by passing current in the opposite direction.

19. Write the name of a chemical substance which is used to prevent corrosion. [1 Mark]

Answer: Biphenyl.

Teacher's Note:
a) Zinc coating (galvanisation) is also commonly used to prevent rusting of iron.

20. Write the unit of Faraday constant. [1 Mark]

Answer: Coulomb per mol (C mol^-1).

Teacher's Note:
a) One faraday is the charge of one mole of electrons, about 96500 C mol^-1.

21. Can you store CuSO₄ solution in Zn pot ? [1 Mark]

Answer: No, Zn is more reactive than Cu.

Teacher's Note:
a) Zinc displaces copper from the solution (Zn + CuSO₄ → ZnSO₄ + Cu), so the pot would be damaged.

22. Write the name of electrolyte used in mercury cell. [1 Mark]

Answer: A paste of ZnO and KOH.

Teacher's Note:
a) The mercury cell uses a paste of zinc oxide and potassium hydroxide as the electrolyte.

23. What does the negative value of E°_cell indicate? [1 Mark]

Answer: The reaction is non-spontaneous.

Teacher's Note:
a) The reverse reaction is the spontaneous one under standard conditions.

24. What flows in the internal circuit of the Galvanic cell? [1 Mark]

Answer: Ions.

Teacher's Note:
a) Cations move towards the cathode and anions towards the anode inside the cell.

25. What is the EMF of the cell when the cell reaction attains equilibrium? [1 Mark]

Answer: E_cell = 0 V.

Teacher's Note:
a) At equilibrium the cell can no longer deliver any current.

26. Why does an aqueous solution of NaCl in electrolysis give H₂ gas at cathode not sodium metal? [1 Mark]

Answer: The reduction potential of H⁺/H₂ is higher than that of Na⁺/Na, so H⁺ ions are discharged in preference to Na⁺.

Teacher's Note:
a) Water is reduced at the cathode because its reduction potential is less negative than that of sodium.

27. Which type of metal can be used in cathodic protection of iron against rusting? [1 Mark]

Answer: Metals more reactive than iron, such as zinc (Zn) or magnesium (Mg).

Teacher's Note:
a) The more reactive metal gets oxidised instead of the iron, which is the principle of sacrificial protection.

28. Why does the conductivity of solution decreases on dilution? [1 Mark]

Answer: Because there are fewer ions per unit volume.

Teacher's Note:
a) Conductivity depends on the number of ions carrying charge in each cubic centimetre of solution.

29. Except hydrogen, write the name of two chemical species which are used in fuel cell. [1 Mark]

Answer: Methane and methanol.

Teacher's Note:
a) Both are oxidised at the anode of a fuel cell while oxygen is reduced at the cathode.

30. How many coulombs are required for conversion of 1 mol FeO into Fe₂O₃? [1 Mark]

Answer: 1 faraday, that is about 96500 C.

Teacher's Note:
a) Each Fe²⁺ changes to Fe³⁺ by losing one electron, so one mole of FeO needs one mole of electrons.

31. State the factors which influence the value of cell potential. [1 Mark]

Answer: Concentration of the electrolyte, temperature and the number of electrons exchanged.

Teacher's Note:
a) These three factors appear in the Nernst equation.

32. What is the relationship between the free energy change and EMF of the cell? [1 Mark]

Answer: Δ_rG° = -nFE°_cell.

Teacher's Note:
a) A positive cell emf gives a negative free energy change, so the reaction is spontaneous.

33. What are the products of electrolysis of molten NaCl and aqueous solution of NaCl? [1 Mark]

Answer: Molten NaCl gives sodium (Na) and chlorine (Cl₂). Aqueous NaCl gives hydrogen (H₂) and chlorine (Cl₂).

Teacher's Note:
a) In aqueous solution, NaOH is also formed in the solution along with hydrogen.

34. What is the role of ZnCl₂ in dry cell? [1 Mark]

Answer: Zn²⁺ ions combine with NH₃ to form the complex [Zn(NH₃)₄]²⁺.

Teacher's Note:
a) The printed key writes [Zn(NH₃)₂]²⁺, but the common complex formed is the tetraammine zinc(II) ion.

35. Rusting of iron is quicker in saline than ordinary water. Why? [1 Mark]

Answer: Due to the presence of salt, which increases the ions in the water.

Teacher's Note:
a) More ions mean higher conductivity, which speeds up the electrochemical corrosion of iron.

36. Write one use of Kohlrausch's law. [1 Mark]

Answer: To find the limiting molar conductivity of a weak electrolyte.

Teacher's Note:
a) A weak electrolyte cannot be measured at infinite dilution, so its limiting value is found from strong electrolytes.

37. What is cell constant? [1 Mark]

Answer: Cell constant = l/A, the ratio of the distance between the electrodes (l) to the area of cross-section of the electrodes (A).

Teacher's Note:
a) Its unit is cm^-1.

38. What is the relationship between Molar conductivity of electrolytic solution to its degree of dissociation? [1 Mark]

Answer: α = Λ_m / Λ°_m, where Λ_m is the molar conductivity and Λ°_m is the limiting molar conductivity.

Teacher's Note:
a) The printed key writes Λ^c_m, which is a typing slip for the molar conductivity Λ_m.

39. Which type of cell is used in wrist watch? [1 Mark]

Answer: Mercury cell (button cell).

Teacher's Note:
a) The constant potential of the mercury cell suits the steady power needed by a watch.

40. Which acid does not react with rust? [1 Mark]

Answer: Carbonic acid (H₂CO₃).

Teacher's Note:
a) The printed key shows only the letter O, so the answer is taken from chemistry: strong acids such as HCl and H₂SO₄ dissolve rust.
b) Carbonic acid is too weak to dissolve iron(III) oxide in any useful amount.

THREE MARKS QUESTIONS

1. Explain construction and working of standard Hydrogen electrode? [3 Marks]

Answer:
1. The standard hydrogen electrode consists of a platinum electrode coated with platinum black, dipped in an acidic solution of 1 M H⁺ ions.
2. Pure hydrogen gas at 1 bar pressure is bubbled over the electrode, and its potential is set to 0.00 V by convention.
3. At the electrode, the reaction is H₂(1 bar) ⇌ 2H⁺(1 M) + 2e^-; it acts as the anode when it is oxidised and as the cathode when 2H⁺ + 2e^- → H₂ occurs.

Teacher's Note:
a) Both the gas pressure and the ion concentration must be unity for the electrode potential to be exactly zero.
b) A platinum black coating gives a large surface area for the reaction.
c) Students often forget the 1 bar pressure condition, which is part of the standard state.

2. What is an electrochemical series? How does it predict the feasibility of a certain redox reaction? [3 Marks]

Answer:
1. An electrochemical series is the arrangement of metals and ions in order of their standard electrode potentials.
2. A redox reaction is feasible when the standard emf of the cell is positive.
3. The species with the lower reduction potential acts as the anode (it is oxidised), and the one with the higher reduction potential acts as the cathode (it is reduced).

Teacher's Note:
a) The series runs from the most negative (most reactive) to the most positive (least reactive) electrode potential.
b) A metal higher in the series displaces any metal below it from its salt solution.
c) The reverse reaction is not feasible because its emf would be negative.

3. The conductivity of an aqueous solution of NaCl in a cell is 92.1 cm^-1 the resistance offered by this cell is 247.8. Calculate the cell constant? [3 Marks]

Answer:
1. Cell constant = conductivity × resistance.
2. Taking the conductivity as 0.0921 S cm^-1, cell constant = 0.0921 × 247.8 = 22.8 cm^-1.

Teacher's Note:
a) The printed value 92.1 gives a cell constant of about 22,800 cm^-1, which is not physically possible for a laboratory cell.
b) Conductivity has the unit S cm^-1, not cm^-1, so the printed unit is also a typing error.
c) Cell constants in practice are usually between 0.1 and 10 cm^-1, which supports the reading 0.0921.

4. The measured resistance of a cell containing 7.5 x 10^-3 M solution of KCl at 25⁰C was 1005 Ω. Calculate (a) Specific conductance and (b) Molar conductance of the solution. Cell Constant = 1.25 cm^-1. [3 Marks]

Answer:
1. Specific conductance κ = cell constant / resistance = \( \frac{1.25}{1005} \) = 1.244 × 10^-3 S cm^-1 (a).
2. Molar conductance Λ_m = κ × 1000 / C = \( \frac{1.244 \times 10^{-3} \times 1000}{7.5 \times 10^{-3}} \) = 165.8 S cm² mol^-1 (b).

Teacher's Note:
a) Specific conductance is the conductance of one centimetre cube of the solution.
b) Multiplying by 1000 converts molarity in mol L^-1 to mol cm^-3, which is why the factor is used.
c) Keep the powers of ten carefully, since a slip changes the answer by a factor of a thousand.

5. Enlist the factors affecting corrosion? [3 Marks]

Answer:
1. Nature of the metal: more reactive metals corrode faster.
2. Presence of electrolytes such as salts, which increase conductivity.
3. Presence of moisture, oxygen and acidic gases such as CO₂.

Teacher's Note:
a) A metal with a lower electrode potential is oxidised more easily.
b) Temperature and pH also change the rate, since higher temperature speeds up reactions.
c) Contact between two dissimilar metals sets up a galvanic cell that speeds up corrosion.

TWO MARKS EACH

1. Solutions of two electrolytes 'A' and 'B' are diluted. The Λ_m of 'B' increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. [2 Marks]

Answer: Electrolyte 'B' is the strong electrolyte. On dilution its molar conductivity rises only slightly (1.5 times), because the number of ions stays the same and only the interionic attraction decreases.

Teacher's Note:
a) A weak electrolyte increases its molar conductivity sharply on dilution because its degree of dissociation rises.
b) Compare the size of the increase, not just the value of molar conductivity.

2. When acidulated water (dil. H₂SO₄ solution) is electrolysed, will the pH of the solution be affected? Justify your answer. [2 Marks]

Answer: The pH of the solution will not be affected, as [H⁺] remains constant. Only water is decomposed (2H₂O → 2H₂ + O₂), while H₂SO₄ stays in the solution, so [H⁺] rise as the volume shrinks.

Teacher's Note:
a) At the anode, 2H₂O → O₂ + 4H⁺ + 4e^-, and at the cathode, 4H⁺ + 4e^- → 2H₂.
b) The solution volume is taken as constant, so [H⁺] and the pH remain unchanged.

3. What advantage do the fuel cells have over primary and secondary batteries? [2 Marks]

Answer:
1. A fuel cell runs continuously as long as the reactants are supplied and the products are removed.
2. Unlike primary batteries it does not go dead, and unlike secondary batteries it needs no long recharging.

Teacher's Note:
a) Primary batteries stop when their reactants are used up.
b) Fuel cells are also efficient and produce little pollution.

4. How does the density of the electrolyte change when the lead storage battery is discharged? [2 Marks]

Answer: The density of the electrolyte decreases, because sulphuric acid is consumed and water is formed during discharge.

Teacher's Note:
a) Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O.
b) Density is a simple way to check the state of charge of the battery.

5. Why on dilution the Λ_m of CH₃COOH increases drastically, while that of CH₃COONa increases gradually? [2 Marks]

Answer: CH₃COOH is a weak electrolyte, so on dilution its degree of dissociation rises sharply and the number of ions increases. For the strong electrolyte CH₃COONa the number of ions is already complete, and only the interionic attraction decreases.

Teacher's Note:
a) The sharp rise in a weak electrolyte comes from more ions being formed.
b) The gradual rise in a strong electrolyte comes from ions moving more freely.

6. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell? [2 Marks]

Answer:
1. Δ_rG = -nFE_cell.
2. The maximum work is obtained when the cell is operated reversibly.

Teacher's Note:
a) Under standard conditions, with unit concentration of all reacting species, the formula gives the standard free energy change.
b) Here n is the number of electrons transferred and F is the faraday constant.

7. Define corrosion. Write chemical formula of rust. [2 Marks]

Answer:
1. Corrosion is the process of formation of oxides, sulphides, carbonates, hydroxides, etc. on the surface of a metal by its reaction with air and water around it.
2. Chemical formula of rust is Fe₂O₃.xH₂O.

Teacher's Note:
a) Rust is hydrated iron(III) oxide, and the value of x varies.

8. Can you store copper sulphate solutions in a zinc pot? [2 Marks]

Answer: No. Zinc is more reactive than copper, so Zn displaces Cu²⁺ from the solution (Zn + Cu²⁺ → Zn²⁺ + Cu).

Teacher's Note:
a) This is a displacement reaction, and the zinc pot would be damaged.

9. Write the cell reaction which occur in the lead storage battery (a) when the battery is in use (b) when the battery is on charging. [2 Marks]

Answer:
(a) When in use (discharging): Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
(b) When on charging: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄

Teacher's Note:
a) At the anode during discharge, Pb + SO₄^2- → PbSO₄ + 2e^-.
b) At the cathode during discharge, PbO₂ + 4H⁺ + SO₄^2- + 2e^- → PbSO₄ + 2H₂O.
c) Charging is the reverse of this overall reaction.

Half-reaction scheme (as in the PDF):
(a) When battery is in use:
Oxidation reaction: Pb + SO₄^2- → PbSO₄ + 2e^-
Reduction reaction: PbO₂ + SO₄^2- + 4H⁺ + 2e^- → PbSO₄ + 2H₂O
Cell reaction: Pb + PbO₂ + 2SO₄^2- + 4H⁺ → 2PbSO₄ + 2H₂O
(b) When the battery is on charging:
Reduction reaction: PbSO₄ + 2e^- → Pb + SO₄^2-
Oxidation reaction: PbSO₄ + 2H₂O → PbO₂ + SO₄^2- + 4H⁺ + 2e^-
Cell reaction: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2SO₄^2- + 4H⁺

10. Write the product of electrolysis of aqueous copper sulphate by using platinum electrode. [2 Marks]

Answer:
1. Copper is deposited at the cathode (Cu²⁺ + 2e^- → Cu).
2. Oxygen gas is evolved at the anode (2H₂O → O₂ + 4H⁺ + 4e^-).

Electrolysis of aqueous CuSO₄ using Pt electrode (scheme as in the PDF):
CuSO₄ → Cu²⁺ + SO₄^2-; H₂O ⇌ H⁺ + OH^-
Anode [SO₄^2-, OH^-]: 4OH^- → 2H₂O + O₂ + 4e^-
Cathode [Cu²⁺, H⁺]: Cu²⁺ + 2e^- → Cu

Teacher's Note:
a) Platinum is inert, so it does not dissolve, and water is oxidised instead of the sulphate ion.
b) The solution becomes more acidic as H⁺ ions are formed at the anode.

11. Consider the standard electrode potential K⁺/K = -2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Cr³⁺/Cr = -0.74 V. Arrange these metals in their increasing order of reducing power. [2 Marks]

Answer: Ag < Hg < Cr < K

Teacher's Note:
a) A more negative standard potential means a stronger reducing agent.
b) Potassium is the strongest reducing agent, and silver is the weakest of the four.

12. The conductivity of 0.20 M solution of HCl at 298 K is 0.0248 S cm^-1. Calculate molar conductivity. [2 Marks]

Answer:
1. Λ_m = κ × 1000 / C
2. Λ_m = \( \frac{0.0248 \times 1000}{0.20} \) = 124 S cm² mol^-1

Teacher's Note:
a) The concentration must be in mol L^-1 when the factor 1000 is used.
b) A value of 124 S cm² mol^-1 is typical for a strong acid such as HCl.

13. How much charge will be required for 1 mol Cu²⁺ to Cu and for 1 mol MnO₄^- to Mn²⁺? [2 Marks]

Answer:
1. Cu²⁺ + 2e^- → Cu, so 2 F (2 × 96500 C = 1.93 × 10⁵ C) is required.
2. MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O, so 5 F (5 × 96500 C = 4.825 × 10⁵ C) is required.

Teacher's Note:
a) The number of faradays equals the number of electrons transferred per mole.
b) Permanganate needs more charge because five electrons are gained per ion.

14. Write the Variation of conductivity and Molar conductivity with dilution. [2 Marks]

Answer:
1. On dilution, conductivity decreases, because the number of ions per unit volume falls.
2. On dilution, molar conductivity increases, because the volume containing one mole of electrolyte increases.

Teacher's Note:
a) Conductivity is for a unit volume, while molar conductivity is for one mole of electrolyte.

15. Λ⁰_m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm²/mol respectively. Calculate Λ⁰_m for acetic acid. [2 Marks]

Answer:
1. Λ°_m(CH₃COOH) = Λ°_m(HCl) + Λ°_m(NaAc) - Λ°_m(NaCl)
2. = 425.9 + 91.0 - 126.4 = 390.5 S cm² mol^-1

Teacher's Note:
a) Kohlrausch's law lets the limiting value of a weak acid be found from strong electrolytes.
b) The NaCl value is subtracted because it contributes Na⁺ and Cl^- that cancel out.

16. A solution of Ni(NO₃)₂ is electrolyzed between Pt electrode using current of 5 amp. For 20 min. What mass of Ni will be deposited at cathode? [2 Marks]

Answer:
1. Q = I × t = 5 × 20 × 60 = 6000 C
2. Ni²⁺ + 2e^- → Ni, so mass = \( \frac{58.7 \times 6000}{2 \times 96500} \) = 1.825 g

Teacher's Note:
a) Use the molar mass of nickel, 58.7 g mol^-1, and two electrons per nickel atom.
b) Time must be converted to seconds before multiplying by the current.

17. State and explain Faraday laws of electrolysis. [2 Marks]

Answer:
1. First law: The mass of a substance deposited or liberated at an electrode is proportional to the quantity of electricity passed through the electrolyte.
2. Second law: For the same quantity of electricity, the masses of different substances deposited are proportional to their chemical equivalent weights. Mathematically, W = Z × Q = Z × I × t.

Teacher's Note:
a) Z is the electrochemical equivalent of the substance.
b) The printed key asks for a derivation, which is simply W = Z × Q.

18. Write the cell reaction of Lead storage battery. [2 Marks]

Answer:
1. Overall during discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
2. During charging the reaction is reversed.

As given in the PDF: at anode, PbSO₄ + 2H₂O → PbO₂ + SO₄^2- + 4H⁺ + 2e^-; at cathode, PbSO₄ + 2e^- → Pb + SO₄^2-.

Teacher's Note:
a) Lead sulphate forms on both electrodes during discharge.
b) The printed key gives only the charging half-reactions, which are the reverse of the discharge reactions.

19. Which cell is more efficient than others? Why? [2 Marks]

Answer:
1. The fuel cell is the most efficient of these cells.
2. It converts chemical energy directly into electrical energy, works continuously and produces little pollution.

Teacher's Note:
a) Fuel cells are not limited by stored reactants, unlike dry or mercury cells.

20. [2 Marks] (the question is not printed in the PDF)

Answer: A more reactive metal is used so as to follow the principle of sacrificial protection.

THREE MARKS EACH

1. Calculate the EMF of the cell in which the following reaction take place: Ni + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag. Given that E°_cell = 1.05 V. [3 Marks]

Answer:
1. E_cell = E°_cell - \( \frac{0.059}{n} \) log \( \frac{[Ni^{2+}]}{[Ag^{+}]^2} \)
2. n = 2, so E_cell = 1.05 - \( \frac{0.059}{2} \) log \( \frac{0.160}{(0.002)^2} \)
3. log (0.160 / 4 × 10^-6) = log (4 × 10⁴) = 4.602
4. E_cell = 1.05 - 0.0295 × 4.602 = 1.05 - 0.136 = 0.914 V

Teacher's Note:
a) Square the silver ion concentration, because the coefficient of Ag⁺ in the reaction is 2.
b) A lower Ni²⁺ or higher Ag⁺ concentration would change the emf.

2. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire? [3 Marks]

Answer:
1. Q = I × t = 0.5 × 2 × 60 × 60 = 3600 C
2. Charge of 96487 C corresponds to 6.022 × 10²³ electrons, so number of electrons = \( \frac{3600 \times 6.022 \times 10^{23}}{96487} \) = 2.25 × 10²²

Teacher's Note:
a) Electron count is the charge divided by the charge of one electron (1.6 × 10^-19 C).
b) Two hours must be turned into 7200 seconds before the charge is found.
c) Check the answer order of magnitude: a few hundred coulombs give about 10²² electrons.

3. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. [3 Marks]

Answer:
1. pH = 10, so [H⁺] = 10^-10 M.
2. E = -0.059 × pH = -0.059 × 10 = -0.59 V

Teacher's Note:
a) The potential of a hydrogen electrode at 298 K equals minus 0.0591 times the pH.
b) A negative potential shows that hydrogen is a weaker oxidising agent in basic solution.
c) Keep the sign negative, since the standard potential is zero.

4. 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 λ°_H+ = 346.6 S cm² mol^-1 and λ°_HCOO- = 54.6 S cm² mol^-1. [3 Marks]

Answer:
1. Λ°_m = λ°_H+ + λ°_HCOO- = 346.6 + 54.6 = 401.2 S cm² mol^-1
2. α = Λ_m / Λ°_m = \( \frac{46.1}{401.2} \) = 0.115
3. K_a = \( \frac{c\alpha^2}{1-\alpha} \) = \( \frac{0.025 \times (0.115)^2}{1 - 0.115} \) = 3.73 × 10^-4 mol L^-1

Teacher's Note:
a) The printed key used 349.6 and 546, which are typing slips; the given values are 346.6 and 54.6.
b) The degree of dissociation is small, so the acid is weak.
c) Using 0.114 instead of 0.115 would give a slightly different constant, so keep three figures throughout.

5. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire? [3 Marks]

Answer: This is the same question as question 2 of this set. Q = 3600 C, and the number of electrons is 2.25 × 10²².

Teacher's Note:
a) The working is the same as in question 2: charge = current × time, then divide by the charge of one electron.
b) Printed twice in the source, so both copies carry the same answer.

6. Calculate Λ°_m for CaCl₂ and MgSO₄ from the data given in the table of Book. [3 Marks]

Answer:
1. Λ°_m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl^-) = 119.0 + 2 × 76.3 = 271.6 S cm² mol^-1
2. Λ°_m(MgSO₄) = λ°(Mg²⁺) + λ°(SO₄^2-) = 106.0 + 160.0 = 266.0 S cm² mol^-1

Teacher's Note:
a) The ion values used are the standard ionic conductivities at 298 K.
b) Two chloride ions are needed because CaCl₂ gives two Cl^- ions.

7. The Conductivity of 0.001028 mol L^-1 acetic acid is 4.95 × 10^-5 S cm^-1. Calculate its dissociation constant if Λ°_m for acetic acid is 390.5 S cm² mol^-1. [3 Marks]

Answer:
1. Λ_m = κ × 1000 / C = \( \frac{4.95 \times 10^{-5} \times 1000}{0.001028} \) = 48.15 S cm² mol^-1
2. α = \( \frac{48.15}{390.5} \) = 0.1233
3. K_a = \( \frac{c\alpha^2}{1-\alpha} \) = \( \frac{0.001028 \times (0.1233)^2}{1 - 0.1233} \) = 1.78 × 10^-5 mol L^-1

Teacher's Note:
a) Convert conductivity to molar conductivity before finding the degree of dissociation.
b) Acetic acid is a weak acid, and its constant is of the order 10^-5.
c) The value agrees with the standard dissociation constant of acetic acid.

8. A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? [3 Marks]

Answer:
1. Q = I × t = 1.5 × 10 × 60 = 900 C
2. Cu²⁺ + 2e^- → Cu, so 2 × 96500 C deposit 63 g of Cu
3. Mass = \( \frac{63 \times 900}{2 \times 96500} \) = 0.2938 g

Teacher's Note:
a) The molar mass of copper is taken as 63 g mol^-1 as in the PDF, giving 0.2938 g; with the more precise value 63.5 g mol^-1, the mass is 0.296 g.
b) Charge is the product of current and time in seconds.
c) Copper is deposited, not dissolved, because it is the cathode.

9. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity. [3 Marks]

Answer:
1. Λ_m = κ × 1000 / C = \( \frac{0.0248 \times 1000}{0.20} \)
2. Λ_m = 124 S cm² mol^-1

Teacher's Note:
a) This is the same working as the two-mark question on HCl, since both solutions have the same conductivity and concentration.
b) Molar conductivity is in S cm² mol^-1, so the units must match.

10. Write the Nernst equation and find emf of the following cells at 298 K: Mg(s) | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu(s) [3 Marks]

Answer:
1. Nernst equation: E_cell = E°_cell - \( \frac{0.0591}{n} \) log \( \frac{[Mg^{2+}]}{[Cu^{2+}]} \)
2. E°_cell = E°_Cu - E°_Mg = 0.34 - (-2.37) = 2.71 V, and n = 2
3. E_cell = 2.71 - \( \frac{0.059}{2} \) log \( \frac{0.001}{0.0001} \) = 2.71 - 0.0295 × 1 = 2.6805 V

Teacher's Note:
a) The reaction is Mg + Cu²⁺ → Mg²⁺ + Cu, with two electrons transferred.
b) The log term equals 1 because the concentration ratio is 10.

FIVE MARKS EACH

1. (a) A Leclanche cell is also called dry cell. Why? (b) Why is the voltage of a mercury cell constant during its working? (c) Name two metals that can be used for cathodic protection of iron? (d) What do you mean by primary and secondary battery? [1+1+1+2]

Answer:

(a) Why is a Leclanche cell called a dry cell?

Answer: The Leclanche cell has a zinc container as anode and a carbon rod as cathode. Its electrolyte is a moist paste of MnO₂, ZnCl₂, NH₄Cl and carbon black. Because there is no free liquid in the cell, it is called a dry cell.

Teacher's Note:
a) The paste is moist but contains no free liquid, which makes the cell easy to handle.
b) Carbon black improves the conductivity of the paste.

(b) Why is the voltage of a mercury cell constant during its working?

Answer: All the reactants and products are in solid or liquid state, so their concentrations do not change during use.

Teacher's Note:
a) Since the concentrations are fixed, the electrode potential stays the same over the life of the cell.
b) Mercury(II) oxide and zinc are used in solid form.

(c) Name two metals that can be used for cathodic protection of iron.

Answer: Zinc and magnesium.

Teacher's Note:
a) Both metals are more reactive than iron and act as sacrificial anodes.

(d) What do you mean by primary and secondary battery?

Answer:
1. A primary battery cannot be used again once its reactants are used up.
2. A secondary battery can be recharged by passing current in the opposite direction and reused.

Teacher's Note:
a) A dry cell is a primary battery, while a lead storage battery is a secondary battery.
b) Secondary batteries are useful where repeated use is needed.
c) Recharging reverses the overall cell reaction.

2. (a) What do you understand by strong and weak electrolytes? (b) State Faraday's Laws of electrolysis? (c) Silver is deposited on a metallic vessel by passing a current of 0.2 amps. for 3 hrs. Calculate the weight of silver deposited. (At mass of silver = 108 amu, F = 96500 C) [1+2+2]

Answer:

(a) What do you understand by strong and weak electrolytes?

Answer: A strong electrolyte ionises completely in solution (for example NaCl, CaCl₂). A weak electrolyte ionises only partially in solution (for example CH₃COOH, NH₄OH).

Teacher's Note:
a) Strong electrolytes have a high degree of dissociation, close to one.
b) Weak electrolytes reach an equilibrium between ions and undissociated molecules.

(b) State Faraday's Laws of electrolysis.

Answer:
1. First law: The amount of chemical change at an electrode is proportional to the quantity of electricity passed through the electrolyte.
2. Second law: The masses of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights.

Teacher's Note:
a) The two laws together give W = Z × Q.
b) Chemical equivalent weight is the molar mass divided by the number of electrons transferred.

(c) Silver is deposited on a metallic vessel by passing a current of 0.2 amps. for 3 hrs. Calculate the weight of silver deposited.

Answer:
1. Q = I × t = 0.2 × 3 × 60 × 60 = 2160 C
2. Ag⁺ + e^- → Ag, so mass = \( \frac{108 \times 2160}{96500} \) = 2.417 g

Teacher's Note:
a) Silver needs one electron per atom, so the molar mass is used directly.
b) Hours must be converted to seconds before multiplying by the current.
c) Silver deposited on the vessel is the cathode reaction.

3. (a) Define the term resistivity and give its SI unit. (b) What are the factors on which conductivity of an electrolyte depend? (c) The molar conductivity of 0.1 M CH₃COOH solution is 4.6 cm² mol^-1. What is the conductivity and resistivity of the solution? [1+2+2]

Answer:

(a) Define the term resistivity and give its SI unit.

Answer: Resistivity is the resistance of a substance of unit length and unit area of cross-section. Its SI unit is ohm metre (Ω m).

Teacher's Note:
a) Resistivity is the reciprocal of conductivity.

(b) What are the factors on which conductivity of an electrolyte depend?

Answer:
1. Nature of the electrolyte and the size of the ions it produces.
2. Nature of the solvent and its viscosity, the concentration of the electrolyte, and the temperature.

Teacher's Note:
a) Larger ions carry charge more slowly, which lowers conductivity.
b) Higher viscosity slows the movement of ions.

(c) The molar conductivity of 0.1 M CH₃COOH solution is 4.6 cm² mol^-1. What is the conductivity and resistivity of the solution?

Answer:
1. κ = Λ_m × C / 1000 = 4.6 × 0.1 / 1000 = 4.6 × 10^-4 S cm^-1
2. Resistivity ρ = 1 / κ = \( \frac{1}{4.6 \times 10^{-4}} \) = 2.17 × 10³ Ω cm, which is 21.7 Ω m

Teacher's Note:
a) Divide by 1000 to convert the concentration from mol L^-1 to mol cm^-3.
b) Resistivity in ohm metre is found by dividing the value in ohm centimetre by 100.

4. (a) State the factors that affect the value of electrode potential? (b) Write Nernst equation for a Al-ZnSO₄ cell? (c) write the chemistry of rusting of iron [1+2+2]

Answer:

(a) State the factors that affect the value of electrode potential.

Answer: Concentration of the electrolyte and temperature.

Teacher's Note:
a) Higher ion concentration on the reduction side increases the electrode potential.
b) Temperature changes the value of the term RT/nF in the Nernst equation.

(b) Write Nernst equation for a Al-ZnSO₄ cell.

Answer:
1. Cell: Al | Al³⁺ || Zn²⁺ | Zn
2. Cell reaction: 2Al + 3Zn²⁺ → 2Al³⁺ + 3Zn
3. E_cell = E°_cell - \( \frac{0.059}{6} \) log \( \frac{[Al^{3+}]^2}{[Zn^{2+}]^3} \)

Teacher's Note:
a) Six electrons are transferred in the balanced cell reaction, so n = 6.
b) The concentrations are squared and cubed to match the coefficients.

(c) Write the chemistry of rusting of iron.

Answer:
1. Acidic medium is formed when atmospheric CO₂ dissolves in water: CO₂ + H₂O → H₂CO₃.
2. At the anode, iron is oxidised: Fe → Fe²⁺ + 2e^-.
3. At the cathode, oxygen is reduced with H⁺ ions: O₂ + 4H⁺ + 4e^- → 2H₂O.
4. Atmospheric oxygen oxidises Fe²⁺ to Fe³⁺, which forms hydrated ferric oxide: 2Fe²⁺ + \( \frac{1}{2} \) O₂ + 2H₂O → Fe₂O₃ + 4H⁺.
5. Ferric oxide combines with water to form rust: Fe₂O₃ + xH₂O → Fe₂O₃.xH₂O (rust).

Teacher's Note:
a) Rust forms in the presence of both water and oxygen, not in either one alone.
b) The electrons released at the anode travel through the iron to the cathode region.
c) Acid from dissolved CO₂ makes the process faster.

5. (a) Can an electrochemical cell act as electrolytic cell? How? (b) Explain construction and working of standard Hydrogen electrode? (c) What is an electrochemical series? How does it predict the feasibility of a certain redox reaction? [1+2+2]

Answer:

(a) Can an electrochemical cell act as electrolytic cell? How?

Answer: Yes. When an external opposing potential greater than the cell emf is applied, the cell reaction is reversed and the cell behaves as an electrolytic cell.

Teacher's Note:
a) The electrical energy supplied drives the non-spontaneous reaction.
b) The electrodes change roles: the anode of the galvanic cell becomes the cathode.

(b) Explain construction and working of standard Hydrogen electrode.

Answer:
1. Construction: A platinum electrode coated with platinum black is dipped in a solution of H⁺ ions with concentration 1 M. Pure hydrogen gas at 1 bar is bubbled over it.
2. Working: The electrode can act as the anode, H₂ → 2H⁺ + 2e^-, or as the cathode, 2H⁺ + 2e^- → H₂. Its standard electrode potential is taken as 0.00 V.

If it acts as cathode: the maximum bubbling of hydrogen gas from the solution will evolve.
If it acts as anode: the minimum bubbling of hydrogen gas from the solution will evolve.
E°(2H⁺ | H₂) = 0

Teacher's Note:
a) The printed key's statements about bubbling at the cathode and anode are unclear, so the reactions above are given instead.
b) Platinum black provides a large surface on which the gas and ions can react.

(c) What is an electrochemical series? How does it predict the feasibility of a certain redox reaction?

Answer:
1. An electrochemical series is the arrangement of electrodes or redox couples in increasing order of their standard electrode potentials.
2. A redox reaction is feasible when the standard emf is positive; the couple with the lower reduction potential is oxidised and the couple with the higher potential is reduced.

Teacher's Note:
a) The reverse of a non-feasible reaction is the one that occurs spontaneously.
b) This is the same idea used in question 2 of the three-mark group.

Answer of one mark:-

1. Cell emf

2. When E_ext> _Ecell

3. 3F

4. Remain unchanged for a cell

7. Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant. Otherwise if DC is used the ions will get discharged and electrolysis will occur

10. Neutralize the two half cell.

11. Metal - A

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