CBSE Class 12 Chemistry Electrochemistry Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. What is the effect of temperature on the electrical conduction of (i) metallic conductor (ii) electrolytic conductor ?
Answer:
With increase of temperature, the electrical conduction of metals decreases whereas that of electrolytes increases.

Teacher's Note:
1. In metals, hotter kernels vibrate more and obstruct electron flow; in electrolytes, heat lowers viscosity and inter-ionic attraction, so ions move faster.

Question 2. How do metallic and ionic substances differ in conducting electricity ? (CBSE 2009)
Answer:
In metallic substances, flow of electricity is due to flow of electrons while in ionic substances in the aqueous solution or in molten state, the flow of electricity is due to flow of ions.

Teacher's Note:
1. Metallic conduction involves no chemical change, while ionic conduction is always accompanied by chemical changes at the electrodes.

Question 3. Give the relationship between molar conductivity and specific conductivity. (CBSE 2008, 2011, 2012, Karnataka Board 2012)
Answer:
Λ_m = κ × \( \frac{1000}{M} \), where κ is specific conductivity in S cm⁻¹ and M is the molar concentration in mol L⁻¹, Λ_m is in S cm² mol⁻¹
or Λ_m = \( \frac{\kappa}{M} \) where κ is in S m⁻¹, M is in mol m⁻³, Λ_m is in S m² mol⁻¹.

Teacher's Note:
1. The factor 1000 only converts litres to cm³; drop it when you work fully in SI units.

Question 4. Express the relation among conductivity of the solution in the cell, the cell constant and resistance of solution in the cell. (CBSE 2011, 2012)
Answer:
Conductivity (κ) = \( \frac{1}{\text{Resistance (R)}} \) × Cell constant.

Teacher's Note:
1. Cell constant = l/A, so κ = (1/R)(l/A); its unit is cm⁻¹ or m⁻¹.

Question 5. Give the relationship between equivalent and molar conductance of a given solution ?
Answer:
Λ_m = κ × \( \frac{1000}{\text{Molarity}} \) and Λ_eq = κ × \( \frac{1000}{\text{Normality}} \)
∴ \( \frac{\Lambda_m}{\Lambda_{eq}} = \frac{\text{Normality}}{\text{Molarity}} \)

Teacher's Note:
1. Since Normality = n-factor × Molarity, Λ_m = n-factor × Λ_eq.

Question 6. What are the units of molar conductivity ?
Answer:
Ohm⁻¹ cm² mol⁻¹ or S cm² mol⁻¹.

Teacher's Note:
1. In SI units it is S m² mol⁻¹; 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.

Question 7. What is the relationship between specific conductance and equivalent conductance ?
Answer:
Λ_eq = \( \frac{\kappa\times 1000}{N} \), where κ is specific conductance in S cm⁻¹, N is normality in g eq L⁻¹, Λ_eq is in S cm² (g eq)⁻¹.

Teacher's Note:
1. This has the same form as the molar conductivity relation, with normality in place of molarity.

Question 8. How is cell constant calculated from conductance values ?
Answer:
Cell constant = specific conductance/observed conductance.

Teacher's Note:
1. In practice, a KCl solution of known conductivity is used to find the cell constant before measuring an unknown solution.

Question 9. State Kohlrausch’s law of independent migration of ions. (Assam Board 2013, CBSE 2022)
Answer:
Kohlrausch’s law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.
For an electrolyte A_xB_y, Λ°_m = xλ°₊ + yλ°₋, where λ°₊ and λ°₋ are the limiting molar conductivities of the cation and anion.

Teacher's Note:
1. Remember to multiply each ionic conductivity by the number of such ions in the formula, e.g. Λ°(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻).

Question 10. Write an expression to relate molar conductivity of an electrolyte to its degree of dissociation.
Answer:
α = \( \frac{\Lambda_m^c}{\Lambda_m^0} \) where Λ_m^c = molar conductivity at given concentration, c and Λ_m⁰ = molar conductivity at infinite dilution.

Teacher's Note:
1. This relation is used for weak electrolytes; α then gives the dissociation constant as K_a = cα²/(1 − α).

Question 11. What is the direction of flow of conventional current in a galvanic cell ?
Answer:
Conventional current flows from cathode to anode. It is opposite to the direction of flow of electrons.

Teacher's Note:
1. In the external circuit electrons flow from anode (negative) to cathode (positive), so current flows the other way.

Question 12. Write two applications of electrochemical series. (UP Board 2012)
Answer:
Two applications of electrochemical series are :
(i) To compare oxidizing or reducing powers of different elements.
(ii) To predict the spontaneity of any redox reaction.

Teacher's Note:
1. Other applications you can quote: calculating E°_cell, and predicting whether a metal displaces hydrogen from dilute acids.

Short Answer Questions (2 Marks)

Question 13. Write down the reactions that occur at the anode and cathode of H₂–O₂ fuel cell and the overall reaction.
Answer:
In the H₂–O₂ fuel cell, H₂ and O₂ are bubbled through porous carbon electrodes into concentrated aqueous KOH solution.
Anode: 2H₂ (g) + 4OH⁻ (aq) → 4H₂O (l) + 4e⁻
Cathode: O₂ (g) + 2H₂O (l) + 4e⁻ → 4OH⁻ (aq)
Overall reaction: 2H₂ (g) + O₂ (g) → 2H₂O (l)

Teacher's Note:
1. Check that electrons cancel (4 on each side) and OH⁻ cancels, leaving only water as the product.

Question 14. Give reason:
(i) Rusting of iron pipe can be prevented by joining it with a piece of magnesium.
(ii) Conductivity of an electrolyte solution decreases with the decrease in concentration.
Answer:
(i) It is due to cathodic protection in which magnesium metal is oxidized in preference to iron and acts as the anode.
(ii) It is due to the fact that on dilution, number of ions per unit volume decreases.

Teacher's Note:
1. Mg is used because its reduction potential (−2.37 V) is much lower than that of iron (−0.44 V); the Mg block is replaced from time to time.

Question 15. Corrosion is an electrochemical phenomenon. The oxygen in moist air reacts as follows:
O₂ (g) + 2H₂O (l) → 4OH⁻ (aq)
Write down the possible reactions for corrosion of zinc occuring at anode, cathode and overall reaction to form a white layer of zinc hydroxide. [CBSE Sample Paper 2022-23]
Answer:
Anode: [Zn (s) → Zn²⁺ (aq) + 2e⁻] × 2
Cathode: O₂ (g) + 2H₂O (l) + 4e⁻ → 4OH⁻ (aq)
Overall: 2Zn (s) + O₂ (g) + 2H₂O (l) → 2Zn²⁺ (aq) + 4OH⁻ (aq)
or 2Zn (s) + O₂ (g) + 2H₂O (l) → 2Zn(OH)₂ (ppt)

Teacher's Note:
1. Multiply the anode reaction by 2 so that 4 electrons are given and taken before adding.

Question 16. Give reasons for the following:
(i) Copper displaces silver from silver nitrate solution.
(ii) Iron pipes are usually coated with zinc. [Karnataka Board 2012]
Answer:
(i) Copper lies above silver in the electrochemical series, i.e., copper is more active than silver.
(ii) This is done to protect iron pipes from rusting because zinc is more electropositive than iron.

Teacher's Note:
1. E°(Cu²⁺/Cu) = +0.34 V is lower than E°(Ag⁺/Ag) = +0.80 V, so Cu reduces Ag⁺; zinc coating also protects by being oxidised first if scratched.

Question 17. Define electrochemical equivalent. How is it related to the equivalent weight of the element ?
Answer:
Electrochemical equivalent (Z) of a substance is the mass of the substance deposited or liberated when one coulomb of electricity is passed through its solution (or one ampere current is passed for one second).
Since 96500 C (1 F) deposit one gram equivalent of the substance, Z = \( \frac{\text{Equivalent weight (E)}}{96500} \) or E = Z × 96500.

Teacher's Note:
1. Use Z in Faraday’s first law, w = Z I t; for Cu (E = 31.75), Z = 3.29 × 10⁻⁴ g C⁻¹.

Question 18. Why is it necessary to platinize the electrodes of a conductivity cell before it is used for conductance measurement ?
Answer:
The electrodes are coated with platinum black (finely divided platinum) to increase the effective surface area of the electrodes and to reduce the polarization effect (accumulation of electrolysis products on the electrodes), which would otherwise change the resistance being measured.

Teacher's Note:
1. Alternating current is used for the same reason: it prevents electrolysis products from building up at the electrodes.

Question 19. What is cell constant ? How is it determined ?
Answer:
Cell constant (G*) is the ratio of the distance between the electrodes (l) to the area of cross-section of the electrodes (A), i.e., G* = l/A. Its unit is cm⁻¹.
It is determined by measuring the resistance (R) of the cell filled with a standard solution of KCl whose conductivity (κ) at that temperature is accurately known. Then G* = κ × R.

Teacher's Note:
1. Measuring l and A directly is not accurate, which is why a KCl solution of known conductivity is used.

Question 20. What is conductivity water ? How is it obtained ?
Answer:
Pure water which has very low conductivity (about 10⁻⁷ S cm⁻¹ or less) and is used for preparing solutions for conductivity measurements is called conductivity water.
It is obtained by distilling ordinary distilled water a number of times in a glass or quartz apparatus, to which a little KMnO₄ and KOH are added (to oxidise organic impurities), and collecting it in a quartz vessel while preventing contact with air (CO₂).

Teacher's Note:
1. Ordinary distilled water contains dissolved CO₂ and traces of ions, which would add their own conductance to that of the solute.

Question 21. With the help of a graph explain why it is not possible to determine Λ°_m for a weak electrolyte by extrapolating the concentration-molar conductance curve as for strong electrolyte.

Answer:
For a strong electrolyte (e.g., KCl), Λ_m increases slowly with dilution and the plot of Λ_m against √C is a straight line. On extrapolating it to zero concentration, the intercept gives Λ°_m.
For a weak electrolyte (e.g., CH₃COOH), Λ_m increases very sharply at low concentrations because its degree of dissociation increases steeply on dilution. The curve becomes almost parallel to the Λ_m axis near zero concentration and does not cut the axis at a definite point. Hence, it cannot be extrapolated to get Λ°_m. (It is obtained using Kohlrausch’s law.)

Teacher's Note:
1. For a weak electrolyte Λ°_m is found indirectly, e.g. Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl).

Question 22. Define molar conductivity for the solution of an electrolyte. How does it vary with concentration ? (CBSE 2022)
Answer:
Molar conductivity of a solution is the conductance of the volume of solution containing one mole of the electrolyte kept between two electrodes with area of cross-section A and distance of unit length. Λ_m = \( \frac{\kappa\times 1000}{M} \) S cm² mol⁻¹.
Molar conductivity increases with decrease in concentration. For strong electrolytes, it increases slowly (due to decrease in inter-ionic attractions), while for weak electrolytes, it increases sharply at low concentrations (due to increase in degree of dissociation).

Teacher's Note:
1. Conductivity (κ) falls on dilution but molar conductivity rises; keep the two trends clearly apart in your answer.

Question 23. How is the molar conductivity of a weak electrolyte at infinite dilution determined ?
Answer:
It cannot be determined by extrapolation of the Λ_m vs √C curve. It is determined by applying Kohlrausch’s law, using the Λ°_m values of suitable strong electrolytes. For example, for acetic acid :
Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl)
because [λ°(CH₃COO⁻) + λ°(Na⁺)] + [λ°(H⁺) + λ°(Cl⁻)] − [λ°(Na⁺) + λ°(Cl⁻)] = λ°(H⁺) + λ°(CH₃COO⁻).

Teacher's Note:
1. Choose the strong electrolytes so that the unwanted ions (here Na⁺ and Cl⁻) cancel out.

Question 24. Define molar conductivity. Express the relationship between degree of dissociation of an electrolyte and its molar conductivities.
Answer:
Molar conductivity (Λ_m) is the conducting power of all the ions produced by dissolving one mole of the electrolyte in solution. Λ_m = \( \frac{\kappa\times 1000}{M} \) S cm² mol⁻¹.
Degree of dissociation, α = \( \frac{\Lambda_m^c}{\Lambda_m^\circ} \), where Λ_m^c is molar conductivity at concentration c and Λ°_m is the limiting molar conductivity.

Teacher's Note:
1. The relation α = Λ_m/Λ°_m is meaningful only for weak electrolytes.

Question 25. Explain the function of salt bridge in an electrochemical cell.
Answer:
(i) It completes the inner circuit by allowing the movement of ions from one solution to the other without mixing of the two solutions.
(ii) It maintains the electrical neutrality of the solutions in the two half-cells. (Without it, positive charge would accumulate around the anode and negative charge around the cathode and the flow of electrons would stop.)
(iii) It minimises the liquid junction potential.

Teacher's Note:
1. The salt bridge contains an inert electrolyte like KCl or KNO₃ in agar-agar gel, whose ions do not react with the solutions.

Short Answer Questions (3 Marks)

Teacher's Note:
1. An oxidant can oxidise a species only if its reduction potential is higher than that of the species’ couple.

Question 27. Predict the products of electrolysis of an aqueous solution of CuCl₂ with platinum electrodes.
Given: E°_Cu²⁺/Cu = +0.34 V, E°_½Cl2/Cl⁻ = +1.36 V, E°_{H⁺/H2 (g), Pt} = 0.00 V, E°_½O2/H2O = +1.23 V [CBSE 2020]
Answer:
CuCl₂ (aq) → Cu²⁺ (aq) + 2Cl⁻ (aq); H₂O (l) → H⁺ (aq) + OH⁻ (aq)
Probable reactions at the cathode:
Cu²⁺ (aq) + 2e⁻ → Cu (s), E°_Red = +0.34 V
H⁺ (aq) + e⁻ → \( \frac{1}{2} \)H₂, E°_Red = 0.0
As E°_Red for first reaction is higher, first reaction will occur, i.e., Cu will be deposited on the cathode.
Probable reactions at the anode:
2Cl⁻ (aq) → Cl₂ (g) + 2e⁻, E°_ox = −1.36 V
H₂O (l) → \( \frac{1}{2} \)O₂ (g) + 2H⁺ (aq) + 2e⁻, E°_ox = −1.23 V
As E°_ox for second reaction is higher, O₂ should be liberated on the basis of E° values alone. But oxidation of water needs a high overpotential, so in practice Cl⁻ ions are oxidised and Cl₂ is liberated at the anode.

Teacher's Note:
1. Overpotential of oxygen is the standard reason why Cl₂, not O₂, is obtained at the anode from chloride solutions (as in the electrolysis of brine).

Question 28. What is electrolysis ? Discuss its mechanism by taking the electrolysis of lead bromide as an example.
Answer:
The process of decomposition of an electrolyte by passing electric current through its aqueous solution or molten state is called electrolysis.
Mechanism : When an electrolyte is dissolved in water or melted, it dissociates into ions. On passing current, the cations move towards the cathode and the anions move towards the anode. At the cathode, the cations gain electrons (reduction) and at the anode, the anions lose electrons (oxidation). Thus they are discharged as neutral atoms/molecules.
For molten lead bromide : PbBr₂ → Pb²⁺ + 2Br⁻
At cathode : Pb²⁺ + 2e⁻ → Pb (lead metal is deposited)
At anode : 2Br⁻ → Br₂ + 2e⁻ (bromine is liberated)

Teacher's Note:
1. In electrolysis, reduction always occurs at the cathode and oxidation at the anode, just as in a galvanic cell; only the signs of the electrodes are reversed.

Question 29. What is electrolysis ? Write the reactions occurring at the two electrodes during electrolysis of aqueous solution of sodium chloride. (Hr. Board 2011)
Answer:
The process of decomposition of an electrolyte by passing electric current through its aqueous solution or molten state is called electrolysis.
In aqueous NaCl solution : NaCl → Na⁺ + Cl⁻
At cathode : 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) (water is reduced in preference to Na⁺)
At anode : 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Overall : 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g)

Teacher's Note:
1. Na⁺ is not discharged from aqueous solution because E°(Na⁺/Na) = −2.71 V is far lower than that for reduction of water (−0.83 V).

Teacher's Note:
1. The temperature effect is the most commonly asked difference; remember it is opposite for the two types.

Question 31. Define the following and write the formula and unit of each : (a) Conductivity (b) Molar conductivity (c) Cell constant (MP Board 2012, J & K Board 2012)
Answer:
(a) Conductivity (κ) is the conductance of a solution of 1 cm length and having 1 cm² area of cross-section, i.e., conductance of 1 cm³ of the solution. κ = \( \frac{1}{R} \) × \( \frac{l}{A} \). Unit : S cm⁻¹ (SI unit S m⁻¹).
(b) Molar conductivity (Λ_m) is the conducting power of all the ions produced by dissolving one mole of the electrolyte in solution. Λ_m = \( \frac{\kappa\times 1000}{M} \). Unit : S cm² mol⁻¹ (SI unit S m² mol⁻¹).
(c) Cell constant (G*) is the ratio of the distance between the electrodes (l) to the area of cross-section of the electrodes (A). G* = \( \frac{l}{A} \) = \( \frac{\kappa}{G} \) = κ × R. Unit : cm⁻¹ (SI unit m⁻¹).

Teacher's Note:
1. Writing the unit correctly fetches marks; molar conductivity has cm² (not cm⁻¹) because conductivity is multiplied by the volume containing one mole.

Question 32. Describe the construction and working of calomel electrode. (Maharashtra Board 2012)
Answer:
Construction : It consists of a glass tube having a side tube. At the bottom of the tube, mercury is placed, which is covered with a paste of mercury and mercurous chloride (calomel, Hg₂Cl₂). Above this paste is a solution of KCl of known concentration (saturated, 1 M or 0.1 M). A platinum wire sealed in a glass tube dips into mercury to make electrical contact. The side tube acts as a salt bridge to connect it with the other electrode.
Representation : Hg, Hg₂Cl₂(s) | KCl(sol)
Working : When it acts as cathode, the reaction is Hg₂Cl₂(s) + 2e⁻ → 2Hg(l) + 2Cl⁻(aq)
When it acts as anode, the reverse reaction takes place : 2Hg(l) + 2Cl⁻(aq) → Hg₂Cl₂(s) + 2e⁻
Its potential depends only upon the concentration of Cl⁻ ions. The reduction potential of saturated calomel electrode at 298 K is 0.2415 V (1 M KCl : 0.2800 V ; 0.1 M KCl : 0.3338 V).

Teacher's Note:
1. The calomel electrode is used as a secondary reference electrode because the hydrogen electrode is difficult to set up and maintain.

Question 33. How does molar conductivity vary with concentration for (i) weak electrolyte (ii) strong electrolyte ? Give reasons for these variations

Answer:
Molar conductivity increases with decrease in concentration (i.e., on dilution) for both types of electrolytes.
(i) Weak electrolyte : Λ_m increases steeply on dilution, especially at low concentrations. This is because the degree of dissociation increases with dilution, so the number of ions produced from one mole of the electrolyte increases.
(ii) Strong electrolyte : Λ_m increases slowly with dilution and follows the equation Λ_m = Λ°_m − A√C. A strong electrolyte is completely ionised at all concentrations; on dilution, the inter-ionic attractions decrease, so the mobility of the ions increases.

Teacher's Note:
1. Do not say a strong electrolyte ionises more on dilution; it is already fully ionised, and only the ionic mobility rises.

Question 34. Define Kohlrausch’s law. How can it be used to find the degree of dissociation of a weak electrolyte ?
Answer:
Kohlrausch’s law : The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte, i.e., Λ°_m = ν₊λ°₊ + ν₋λ°₋.
Degree of dissociation : Λ°_m of the weak electrolyte is calculated from the λ° values of its ions (or from Λ°_m of suitable strong electrolytes), and its molar conductivity at the given concentration, Λ_m^c, is measured. Then
α = \( \frac{\Lambda_m^c}{\Lambda_m^\circ} \)
For example, for acetic acid : Λ°_m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) and α = Λ_m^c/Λ°_m. From α, K_a = \( \frac{c\alpha^2}{1-\alpha} \).

Teacher's Note:
1. α must come out less than 1 for a weak electrolyte; a value above 1 means the two conductivities were swapped.

Question 35. What is an electrochemical cell ? Describe the working of Daniell cell ? (Hr. Board 2011)
Answer:
A device in which chemical energy of a spontaneous redox reaction is converted into electrical energy is called an electrochemical (galvanic) cell.
Daniell cell : It consists of a zinc rod dipping in ZnSO₄ solution and a copper rod dipping in CuSO₄ solution, the two solutions being connected by a salt bridge. When the two rods are connected by a wire, the following reactions take place :
At anode (oxidation) : Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode (reduction) : Cu²⁺(aq) + 2e⁻ → Cu(s)
Net reaction : Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Electrons flow from zinc to copper through the external wire, and current flows from copper to zinc. The salt bridge completes the inner circuit. The cell is represented as Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu and its emf is 1.10 V when the concentrations of Zn²⁺ and Cu²⁺ are 1 M.

Teacher's Note:
1. In a galvanic cell the anode is negative and the cathode positive; zinc dissolves (rod gets thinner) and copper deposits on the copper rod.

Long Answer Questions (5 Marks)

Teacher's Note:
1. Increasing [Zn²⁺] (a product) increases log Q, so E_cell falls; for CaCl₂ remember the factor 2 for the two Cl⁻ ions.

Question 37. Conductivity of 0.00241 M acetic acid is 7.896 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity and if Λ°_m for acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?
Answer:
Λ^c_m = \( \frac{\kappa \times 1000}{M} \)
= \( \frac{7.896 \times 10^{-5}\ \text{S cm}^{-1} \times 1000\ \text{cm}^3\ \text{L}^{-1}}{0.00241\ \text{mol L}^{-1}} \)
= 32.76 S cm² mol⁻¹
α = \( \frac{\Lambda_m^c}{\Lambda_m^\circ} \) = \( \frac{32.76}{390.5} \) = 8.39 × 10⁻²
K_a = \( \frac{C\alpha^2}{1-\alpha} \) = \( \frac{0.00241 \times (8.39 \times 10^{-2})^2}{1 - 8.39 \times 10^{-2}} \)
= 1.85 × 10⁻⁵

Teacher's Note:
1. Use the molar conductivity at the given concentration (not Λ°_m) in the numerator of α.

Question 38. Three electrolytic cells A, B, C containing solution of ZnSO₄, AgNO₃ and CuSO₄ respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and of zinc were deposited?
Answer:
Ag⁺ + e⁻ → Ag(s)
108 g of silver is deposited by 96500 C.
1.45 g silver is deposited by = \( \frac{96500 \times 1.45}{108} \) = 1295.6 C
Q = I × t
1295.6 = 1.5 × t
t = \( \frac{1295.6}{1.5} \) = 863 s
In cell A, the electrode reaction is Zn²⁺ + 2e⁻ → Zn
2F of electricity deposit Zn = 65.3 g
1295.6 C of electricity deposit Zn = \( \frac{65.3 \times 1295.6}{2 \times 96500} \) = 0.438 g
In cell C, the electrode reaction is Cu²⁺ + 2e⁻ → Cu(s)
2F of electricity deposit Cu = 63.5 g
1295.6 C of electricity deposit Cu = \( \frac{63.5 \times 1295.6}{2 \times 96500} \) = 0.426 g

Teacher's Note:
1. Cells in series carry the same charge; find Q from the silver cell first, then use it for Zn and Cu.

Question 39. (a) Define Kohlrausch’s law.
(b) Suggest a way to determine the Λ°_m for CH₃COOH.
(c) The Λ°_m for sodium acetate, HCl, NaCl are 91.0, 425.9 and 126.4 S cm² mol⁻¹ respectively at 298 K. Calculate Λ°_m for CH₃COOH.
Answer:
(a) The molar conductivity at infinite dilution for a given salt can be expressed as the sum of the individual contributions from the ions of the electrolyte.
(b) Λ°(CH₃COOH) = ?
λ°(CH₃COO⁻) + λ°(H⁺) = λ°(CH₃COO⁻) + λ°(Na⁺) + λ°(H⁺) + λ°(Cl⁻) − λ°(Na⁺) − λ°(Cl⁻)
Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl)
(c) Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl)
= 91.0 + 425.9 − 126.4
= 390.5 S cm² mol⁻¹

Teacher's Note:
1. Λ°_m of a weak electrolyte cannot be found by extrapolating the Λ_m vs √C graph, which is why Kohlrausch’s law is used.

Question 40. (a) Define weak and strong electrolytes.
(b) The E° values corresponding to the following two reduction electrode processes are:
(i) Cu⁺/Cu = 0.52 V   (ii) Cu²⁺/Cu⁺ = 0.16 V
Formulate the galvanic cell for their combination. Calculate the cell potential and ΔG° for the cell reaction.
Answer:
(a) Weak electrolyte: The substance which is partially ionised in solution is known as weak electrolyte. Example: NH₄OH.
Strong electrolyte: The substance which is completely ionised in solution is known as strong electrolyte. Example: NaCl.
(b) Cathode: Cu⁺ + e⁻ → Cu
Anode: Cu⁺ → Cu²⁺ + e⁻
Overall cell reaction: 2Cu⁺ → Cu + Cu²⁺
Pt | Cu⁺(aq), Cu²⁺(aq) || Cu⁺(aq) | Cu
E°_cell = 0.52 − 0.16 = 0.36 V
ΔG° = −nFE°_cell
= −1 × 96500 × 0.36
= −34740 J mol⁻¹

Teacher's Note:
1. A positive E°_cell here shows Cu⁺ disproportionates in aqueous solution.