CBSE Class 12 Chemistry Electrochemistry Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Depict the galvanic cell in which the cell reaction is Cu + 2Ag⁺ → 2Ag + Cu²⁺. [NCERT Exemplar] [HOTS]
Answer:
Cu | Cu²⁺(conc.) || Ag⁺(conc.) | Ag

Teacher's Note:
1. The metal that loses electrons (Cu) goes on the left; the stoichiometric coefficient 2 does not appear in the cell notation.

Question 2. What is the reference electrode in determining the standard electrode potential?
Answer:
Normal hydrogen electrode (NHE).

Teacher's Note:
1. Its potential is taken as exactly 0 V at all temperatures, with H₂ at 1 bar and [H⁺] = 1 M.

Question 3. Why is it not possible to measure the single electrode potential?
Answer:
Oxidation or reduction cannot take place alone. Moreover, electrode potential is a relative tendency and can be measured with respect to a reference electrode only.

Teacher's Note:
1. This is why every tabulated E° value is really the emf of a cell with the SHE as the other electrode.

Question 4. Define electrochemical series.
Answer:
The arrangement of elements in the increasing or decreasing order of their standard reduction potential is called electrochemical series.

Teacher's Note:
1. Higher in the series (more positive E°) means a stronger oxidising agent; lower means a stronger reducing agent.

Question 5. Write Nernst equation for single electrode potential.
Answer:
For the electrode reaction Mⁿ⁺ + ne⁻ → M
\( E_{\text{M}^{n+}/\text{M}} = E^\circ_{\text{M}^{n+}/\text{M}} - \frac{2.303RT}{nF}\log\frac{[\text{M}]}{[\text{M}^{n+}]} = E^\circ_{\text{M}^{n+}/\text{M}} - \frac{2.303RT}{nF}\log\frac{1}{[\text{M}^{n+}]} \)

Teacher's Note:
1. The concentration of a pure solid metal is taken as 1, which is why [M] disappears from the second form.

Question 6. Write Nernst equation for the general cell reaction aA + bB → xX + yY (with n electrons transferred).
Answer:
If n moles of electrons are transferred, Nernst equation is \( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln\frac{[\text{X}]^x[\text{Y}]^y}{[\text{A}]^a[\text{B}]^b} \).

Teacher's Note:
1. Products go in the numerator and reactants in the denominator, each raised to its stoichiometric coefficient.

Question 7. Write Nernst equation for the reaction
2Cr + 3Fe²⁺ → 2Cr³⁺ + 3Fe
Answer:
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{6F}\ln\frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} \)   (∵ n = 6)

Teacher's Note:
1. Find n from the balanced equation: 2 Cr atoms each lose 3 electrons, so n = 6; solids Cr and Fe are left out of the quotient.

Question 8. How can the reduction potential of an electrode be increased?
Answer:
Mⁿ⁺ + ne⁻ → M,
\( E_{\text{M}^{n+}/\text{M}} = E^\circ_{\text{M}^{n+}/\text{M}} - \frac{RT}{nF}\ln\frac{1}{[\text{M}^{n+}]} = E^\circ_{\text{M}^{n+}/\text{M}} + \frac{RT}{nF}\ln[\text{M}^{n+}] \)
Thus, electrode potential can be increased by increasing the metal ion concentration.

Teacher's Note:
1. Only E changes with concentration; the standard value E° is a constant at a given temperature.

Question 9. What flows in the internal circuit of a galvanic cell?
Answer:
Ions.

Teacher's Note:
1. Electrons flow in the external circuit (wire); ions carry the current inside the cell and through the salt bridge.

Question 10. What does the negative value of E°_cell indicate?
Answer:
Negative E°_cell value means Δ_rG° will be +ve, and the cell will not work.

Teacher's Note:
1. Use Δ_rG° = −nFE°_cell: the signs of Δ_rG° and E°_cell are always opposite.

Question 11. Why is alternating current used for measuring resistance of an electrolytic solution? [NCERT Exemplar]
Answer:
Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant.

Teacher's Note:
1. With DC, products would form at the electrodes and polarisation would make the measured resistance wrong.

Question 12. What is meant by cell constant?
Answer:
Cell constant is the ratio of distance (l) between electrodes and area of cross-section (A). It is denoted by \( \frac{l}{A} \). Its unit is cm⁻¹.

Teacher's Note:
1. In practice it is found from κ = G* / R using a KCl solution of known conductivity.

Short Answer Questions (2 Marks)

Question 13. The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10⁻⁴ S cm⁻¹. Calculate molar conductivity (Λ_m) of the solution. [CBSE Comptt. Delhi/OD 2018]
Answer:
Λ_m = \( \frac{1000\kappa}{C} \)
Λ_m = \( \frac{1.65\times10^{-4}\ \text{S cm}^{-1}\times1000\ \text{cm}^{3}\text{L}^{-1}}{0.01\ \text{mol L}^{-1}} \)
= 16.5 S cm² mol⁻¹

Teacher's Note:
1. The factor 1000 converts litres to cm³; it is needed only when κ is in S cm⁻¹ and C in mol L⁻¹.

Question 14. 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.
Answer:
Λ_m = \( \frac{1000\kappa}{C} \)
Λ_m = \( \frac{1000\ \text{cm}^{3}\text{L}^{-1}\times4.1\times10^{-5}\ \text{S cm}^{-1}}{10^{-3}\ \text{mol L}^{-1}} \)
= 41 S cm² mol⁻¹
α = \( \frac{\Lambda_m^{c}}{\Lambda_m^{\circ}} \)
α = \( \frac{41}{390.5} \) = 0.105

Teacher's Note:
1. Degree of dissociation has no unit, and for a weak acid like CH₃COOH it must come out less than 1 – a quick check on your arithmetic.

Question 15. Why on dilution the Λ_m of CH₃COOH increases drastically, while that of CH₃COONa increases gradually?
Answer:
In case of CH₃COOH which is a weak electrolyte, the number of ions increases on dilution due to an increase in degree of dissociation resulting in drastic increase in Λ_m.
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺
In the case of CH₃COONa which is a strong electrolyte, the number of ions remains the same but the inter-ionic attraction decreases resulting in gradual increase in Λ_m.

Teacher's Note:
1. Key contrast: weak electrolyte – more ions are formed on dilution; strong electrolyte – same ions, only less inter-ionic attraction.

Question 16. 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.
Answer:
‘B’ is a strong electrolyte.
B is a strong electrolyte which is completely dissociated into ions, but on dilution interionic forces overcome and ions are free to move. So there is slight increase in molar conductivity on dilution.

Teacher's Note:
1. A small rise in molar conductivity on dilution signals a strong electrolyte; a many-fold rise signals a weak one whose dissociation is increasing.

Question 17. The resistivity of a 0.8M solution of electrolyte is 5 × 10⁻³ Ωcm. Calculate its molar conductivity.
Answer:
Resistivity (ρ) = 5 × 10⁻³ Ω cm
Conductivity of solution (κ) = \( \frac{1}{\text{Resistivity}} \) = \( \frac{1}{5\times10^{-3}\ \Omega\,\text{cm}} \)
= 0.2 × 10³ Ω⁻¹cm⁻¹ = 200 Ω⁻¹cm⁻¹
Molar conductivity, Λ_m = \( \frac{1000\times\kappa}{M} \) = \( \frac{1000\times200}{0.8} \) = 2.5 × 10⁵ Ω⁻¹cm²mol⁻¹

Teacher's Note:
1. Do not stop at κ = 200 Ω⁻¹ cm⁻¹; the question asks for molar conductivity, so divide by concentration as well.

Question 18. In a galvanic cell, the following cell reaction occurs:
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s); E°_cell = +1.56 V
(i) Is the direction of flow of electrons from zinc to silver or silver to zinc?
(ii) How will concentration of Zn²⁺ ions and Ag⁺ ions be affected when the cell functions? [CBSE Foreign Set-1, 2, 3 2017]
Answer:
(i) Zinc to silver.
(ii) Concentration of Zn²⁺ ions will increase and Ag⁺ ions will decrease.

Teacher's Note:
1. Electrons in the external circuit always flow from the anode (the species oxidised, here Zn) to the cathode.

Question 19. 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]
Answer:
2Cr(s) + 3Fe²⁺(aq) → 3Fe(s) + 2Cr³⁺(aq)
Where, n = 6
E_cell = E°_cell − \( \frac{2.303RT}{nF} \)\( \log\frac{[\text{Cr}^{3+}]^{2}}{[\text{Fe}^{2+}]^{3}} \)
E_cell = 0.30 V − \( \frac{0.059}{6} \) V \( \log\frac{[10^{-1}]^{2}}{[10^{-2}]^{3}} \)
E_cell = 0.26 V

Teacher's Note:
1. Find n from the balanced overall equation (2Cr give 6 electrons), and raise each concentration to its stoichiometric power.

Teacher's Note:
1. An oxidising agent is itself reduced, so a larger (more positive) reduction potential means a stronger oxidising agent.

Question 21. Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes:
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 electrode potential values, which reaction is feasible at cathode and why?
Answer:
Ag⁺(aq) + e⁻ → Ag(s)
Because it has higher reduction potential.
Detailed Answer: As reaction with higher value of standard electrode potential occurs at cathode, Ag gets reduced. So, the reaction occurring at cathode is
Ag⁺(aq) + e⁻ → Ag(s)

Teacher's Note:
1. At the cathode the species with the higher (more positive) reduction potential is discharged first; at the anode it is the one most easily oxidised.

Question 22. Calculate E°_cell for the following reaction at 298 K:
2Cr(s) + 3Fe²⁺ (0.01M) → 2Cr³⁺(0.01M) + 3Fe(s)
Given: E_cell = 0.261 V
Answer:
Nernst Equation:
E_cell = E°_cell − \( \frac{0.059}{n} \)\( \log\frac{[\text{Prod.}]}{[\text{React.}]} \)
E°_cell = E_cell + \( \frac{0.059}{n} \)\( \log\frac{[\text{Prod.}]}{[\text{React.}]} \)
= 0.261 + \( \frac{0.059}{6} \)\( \log\frac{[\text{Cr}^{3+}]^{2}}{[\text{Fe}^{2+}]^{3}} \)
= 0.261 + \( \frac{0.059}{6} \)\( \log\frac{(10^{-2})^{2}}{(10^{-2})^{3}} \)
= 0.261 + \( \frac{0.059}{6} \) log 10²
= 0.261 + \( \frac{0.059\times2}{6} \)
= 0.261 + 0.01966
= 0.28068 V ≈ 0.281 V

Teacher's Note:
1. When E_cell is given and E°_cell is asked, rearrange the Nernst equation first – the log term is then added, not subtracted.

Question 23. Iron displaces copper from copper sulphate solution but Pt does not why?
Answer:
Oxidation potential of Fe is more than oxidation potential of Cu. So, Fe displaces Cu from copper sulphate while oxidation potential of Pt is less than Cu. Due to this reason, Pt cannot displace Cu from copper sulphate.

Teacher's Note:
1. In terms of standard reduction potentials: Fe (−0.44 V) is below Cu (+0.34 V) in the series and can reduce Cu²⁺; Pt (about +1.2 V) lies above Cu and cannot.

Question 24. Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. [CBSE Outside Delhi Set-1, 2017]
Answer:
Mercury cell.
Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
Cathode: HgO + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Detailed Answer: Mercury cell is generally used in hearing aids.
At anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
At cathode: HgO(s) + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Overall reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

Teacher's Note:
1. The mercury cell gives a constant voltage (about 1.35 V) throughout its life because no ions appear in the overall reaction.

Question 25. From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell.
Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life? [CBSE Delhi Set-1, 2 & 3 2016]
Answer:
(i) Mercury cell
(ii) Fuel cell
(iii) Lead storage cell
(iv) Dry cell

Teacher's Note:
1. The dry cell has a short life because the acidic NH₄Cl corrodes the zinc container even when the cell is not in use.

Short Answer Questions (3 Marks)

Question 26. Depict the galvanic cell in which the reaction Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2Ag_(s) takes place. Further show:
(i) Which of the electrodes is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:
The reaction is Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2Ag_(s)
The cell can be represented as Zn|Zn²⁺_(aq)||Ag⁺_(aq)|Ag

(i) The zinc electrode is negatively charged (anode) as it pushes the electrons into the external circuit.
(ii) Ions are the current carriers within the cell.
(iii) The reactions occurring at the two electrodes are:
At zinc electrode (anode): Zn_(s) → Zn²⁺_(aq) + 2e⁻
At silver electrode (cathode): Ag⁺_(aq) + e⁻ → Ag_(s)

Teacher's Note:
1. Electrons carry the current in the external wire, while ions carry it inside the cell and through the salt bridge.
2. In a galvanic cell the anode is negative; in an electrolytic cell it is positive.

Question 27. What is the difference between a chemical cell and a concentration cell?
Answer:
A chemical cell is a galvanic cell in which electrical energy is produced due to chemical changes occurring within the cell and no transfer of matter takes place. It involves the use of two different electrodes dipped in solutions of different electrolytes.
A concentration cell is a galvanic cell in which electrical energy is produced due to a physical change involving transfer of matter from one part of the cell to the other. It involves the use of the same electrodes dipped in solutions of the same electrolyte with different concentrations (or electrodes of different concentrations dipped in the same solution of the electrolyte).

Teacher's Note:
1. A concentration cell has E°_cell = 0, so its whole EMF comes from the concentration term of the Nernst equation.

Question 28. A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell.
Given: E°_Ag⁺/Ag = +0.80 V, E°_Cu²⁺/Cu = +0.34 V
Answer:
The given cell may be represented as Cu_(s)|Cu²⁺ (0.10 M)||Ag⁺ (C)|Ag_(s)
E°_cell = E°_cathode − E°_anode = 0.80 V − 0.34 V = 0.46 V
E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Cu^{2+}]}{[Ag^{+}]^{2}} \)
or 0.422 V = 0.46 V − \( \frac{0.0591}{2} \) log \( \frac{0.1}{[Ag^{+}]^{2}} \)
−0.038 V = −0.0295 log \( \frac{0.1}{[Ag^{+}]^{2}} \)
or log \( \frac{0.1}{[Ag^{+}]^{2}} \) = \( \frac{-0.038}{-0.0295} \) = 1.288
or \( \frac{0.1}{[Ag^{+}]^{2}} \) = antilog 1.288 = 19.41
∴ [Ag⁺]² = \( \frac{0.1}{19.41} \) = 5.1519 × 10⁻³
[Ag⁺] = 7.1 × 10⁻² M

Teacher's Note:
1. Here the Nernst equation is used backwards: the measured E_cell is known and the unknown concentration is found.
2. Remember to take the square root at the end, because [Ag⁺] appears squared.

Question 29. The resistance of 100 cm³ aqueous solution of 0.025 M CuSO₄ is 520 ohm at 298 K. Calculate the molar conductivity if the cell constant of the conductivity cell is 153.7 m⁻¹.
Answer:
Given: V = 100 cm³, M = 0.025 M, R = 520 ohm
G* = 153.7 m⁻¹ = 1.537 cm⁻¹, Λ_m = ?
κ = G* × \( \frac{1}{R} \) = 1.537 cm⁻¹ × \( \frac{1}{520\ \text{ohm}} \) = 2.95 × 10⁻³ ohm⁻¹ cm⁻¹
Again, Λ_m = \( \frac{\kappa\times10^{3}}{M} \) = \( \frac{2.95\times10^{-3}\times10^{3}}{0.025} \)
Λ_m = 118.0 ohm⁻¹ cm² mol⁻¹

Teacher's Note:
1. The volume of solution (100 cm³) is not needed; it is extra data meant to test whether you know the formula.
2. Convert the cell constant from m⁻¹ to cm⁻¹ (divide by 100) before using it with the × 1000 formula.

Question 30. When a certain conductance cell was filled with 0.1 M KCl, it had a resistance of 85 ohms at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohms. Calculate the molar conductance of the electrolyte at this concentration.
[Specific conductance of 0.1 M KCl = 1.29 × 10⁻² ohm⁻¹ cm⁻¹]
Answer:
κ = 1.29 × 10⁻² ohm⁻¹ cm⁻¹ = 1.29 S m⁻¹
κ = \( \frac{1}{R} \) × Cell constant
⇒ Cell constant = κ × R = 1.29 S m⁻¹ × 85 Ω = 109.65 m⁻¹
For the second solution,
κ = \( \frac{1}{R} \) × Cell constant = \( \frac{1}{96\ \Omega} \) × 109.65 m⁻¹ = 1.142 Ω⁻¹ m⁻¹
Λ_m = \( \frac{\kappa\times1000}{M} \) = \( \frac{1.142\times10^{-2}\ \Omega^{-1}\text{cm}^{-1}\times1000\ \text{cm}^{3}}{0.052\ \text{mol}} \)
= 219.62 S cm² mol⁻¹

Teacher's Note:
1. Use the standard KCl solution to find the cell constant first; it stays the same when the solution is changed.
2. Keep the units consistent: 1.142 Ω⁻¹ m⁻¹ = 1.142 × 10⁻² Ω⁻¹ cm⁻¹ before multiplying by 1000.

Question 31. The following curve is obtained when molar conductivity (Λ_m) is plotted against the square root of concentration, C^1/2 for two electrolytes A and B:
(i) How do you account for the increase in the molar conductivity of the electrolyte A on dilution?
(ii) As seen from the graph, the value of limiting molar conductivity (Λ°_m) for electrolyte B cannot be obtained graphically. How can this value be obtained?
(iii) Define limiting molar conductivity.

Answer:
(i) As seen from the graph, electrolyte A is a strong electrolyte which is completely ionised in solution. With dilution, the ions are far apart from each other and hence the molar conductivity increases.
(ii) To determine the value of limiting molar conductivity for electrolyte B, indirect method based upon Kohlrausch’s law of independent migration of ions is used.
(iii) When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

Teacher's Note:
1. A straight-line plot that can be extrapolated to C = 0 identifies a strong electrolyte; a steep curve that rises sharply near zero is a weak electrolyte.

Teacher's Note:
1. The table shows both trends at once: κ falls from 106.74 to 23.15 (×10⁻⁴) while Λ_m rises from 106.7 to 115.8 as the solution is diluted.

Question 33. The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 × 10³ ohm. Calculate its molar conductivity. [CBSE Foreign Set-1, 2, 3 2017]
Answer:
A = πr²
= 3.14 × 0.5 × 0.5 cm² = 0.785 cm²
l = 45.5 cm
G* = l/A = 45.5 cm/0.785 cm² = 57.96 cm⁻¹
κ = G*/R = 57.96 cm⁻¹/4.55 × 10³ Ω = 1.27 × 10⁻² S cm⁻¹
Λ_m = κ × 1000/C
= [1.27 × 10⁻² S cm⁻¹] × 1000/0.05 mol/cm³
= 254.77 S cm² mol⁻¹

Teacher's Note:
1. Use the radius (0.5 cm), not the diameter, in πr², and keep everything in cm – converting only some quantities to metres is a common slip.

Question 34. (i) State the law which helps to determine the limiting molar conductivity of weak electrolyte.
(ii) Calculate limiting molar conductivity of CaSO₄ (limiting molar conductivity of calcium and sulphate ions are 119.0 and 160.0 S cm² mol⁻¹ respectively).
Answer:
Kohlrausch law of independent migration of ions:
(i) The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contribution of the anions and cations of the electrolyte.
(ii) Λ°_m(CaSO4) = Λ°_Ca²⁺ + Λ°_SO4²⁻
= 119.0 S cm² mol⁻¹ + 160.0 S cm² mol⁻¹
= 279.0 S cm² mol⁻¹

Teacher's Note:
1. Write the law in words as well as the equation – marks are given for the statement, not just the formula.

Question 35. Consider the following reaction:
Cu(s) + 2Ag⁺(aq) → 2Ag(s) + Cu²⁺(aq)
(i) Depict the galvanic cell in which the given reaction takes place.
(ii) Give the direction of flow of current.
(iii) Write the half-cell reactions taking place at cathode and anode. [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) Cu(s) | Cu²⁺(aq) || Ag⁺ (aq) | Ag(s)
(ii) Current will flow from silver to copper electrode in the external circuit.
(iii) Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)
Anode: Cu(s) → Cu²⁺ (aq) + 2e⁻

Teacher's Note:
1. Current flows opposite to electrons: electrons go Cu → Ag in the wire, so conventional current goes Ag → Cu.

Long Answer Questions (5 Marks)

Question 36. The molar conductivities of acetic acid at 298 K at the concentrations of 0.1 M and 0.001 M are 5.20 and 49.2 S cm²mol⁻¹ respectively. Calculate the degree of dissociation of acetic acid at these concentration. Given λ^∞_(H⁺) and λ^∞_(CH3COO⁻) are 349.8 and 40.9 ohm⁻¹cm²mol⁻¹ respectively.
Answer:
Given
λ^∞_(H⁺) = 349.8 ohm⁻¹cm²mol⁻¹
λ^∞_(CH3COO⁻) = 40.9 ohm⁻¹cm²mol⁻¹
Λ^∞_m(CH3COOH) = λ^∞_(CH3COO⁻) + λ^∞_(H⁺)
= 40.9 + 349.8 = 390.7 ohm⁻¹cm²mol⁻¹
At C = 0.1 M
Degree of dissociation, α = \( \frac{\Lambda_m}{\Lambda_m^{\infty}} \)
= \( \frac{5.20}{390.7} \) = 0.013 i.e. 1.3%
At C = 0.001 M
α = \( \frac{\Lambda_m}{\Lambda_m^{\infty}} \)
= \( \frac{49.2}{390.7} \) = 0.126 i.e. 12.6%

Teacher's Note:
1. α = Λ_m/Λ°_m applies to weak electrolytes; note how α increases about ten times when the solution is diluted 100 times.

Teacher's Note:
1. Multiply κ by 1000 only when it is in S cm⁻¹ and concentration is in mol L⁻¹; this gives Λ_m in S cm² mol⁻¹.

Question 38. (a) Calculate e.m.f. of the following cell:
Zn(s)/Zn²⁺ (0.1 M) || (0.01 M) Ag⁺/Ag(s)
Given: E°_Zn²⁺/Zn = −0.76 V, E°_Ag⁺/Ag = +0.80 V
[Given: log 10 = 1]
(b) X and Y are two electrolytes. On dilution molar conductivity of ‘X’ increases 2.5 times while that Y increases 25 times. Which of the two is a weak electrolyte and why? [CBSE Outside Delhi Set-1, 2020]
Answer:
(a) Zn(s)/Zn²⁺ (0.1 M) || (0.01 M) Ag⁺/Ag(s)
E°_Zn²⁺/Zn = −0.76 V
E°_Ag⁺/Ag = +0.80 V   emf = ?
E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[\text{Anode}]}{[\text{Cathode}]} \)
E°_cell = E°_cathode − E°_anode
= E°_Ag⁺/Ag − E°_Zn²⁺/Zn
= 0.80 − (−0.76) = 1.56 V
E_cell = 1.56 − \( \frac{0.0591}{2} \) log \( \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \)
= 1.56 − \( \frac{0.0591}{2} \) log \( \frac{[0.1]}{[0.01]^2} \)
= 1.56 − 0.0295 log 1000
= 1.56 − 3 (0.0295) = 1.56 − 0.09 = 1.4715 V
(b) Y is a weak electrolyte as on dilution complete dissociation of weak electrolyte takes place and thus a sharp increase in molar conductivity while in strong electrolyte it has already dissociated completely. So on dilution molar conductivity does not rise much.

Teacher's Note:
1. In the Nernst term, raise each concentration to the power of its coefficient in the balanced cell reaction: here [Ag⁺] is squared.
2. A large jump in Λ_m on dilution means more ionisation, which is the signature of a weak electrolyte.

Question 39. (a) Represent the cell in which the following reaction takes place:
2Al(s) + 3Ni²⁺(0.1M) → 2Al³⁺(0.01M) + 3Ni(s)
Calculate its emf if E°_cell = 1.41V.
(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity (Λ°_m) for weak electrolyte? [CBSE Outside Delhi Set-2, 2019]
Answer:
(a) Al(s) | Al³⁺(0.01M) || Ni²⁺ (0.1 M) | Ni(s)
E_cell = E°_cell − \( \frac{0.059}{6} \) log \( \frac{[\text{Al}^{3+}]^2}{[\text{Ni}^{2+}]^3} \)
E_cell = 1.41 V − \( \frac{0.059}{6} \) log \( \frac{[0.01]^2}{[0.1]^3} \)
E_cell = 1.4198 V or E_cell = 1.42 V
(b) Λ_m decreases with increase in concentration for both strong & weak electrolyte. Λ°_m can be obtained for weak electrolyte by applying Kohlrausch law
Λ°_m = ν₊λ°₊ + ν₋λ°₋
Detailed Answer:
(a) 2Al(s) + 3Ni²⁺(0.1M) → 2Al³⁺(0.01M) + 3Ni(s)
Al(s) | Al³⁺ || Ni²⁺ | Ni(s) → Cell reaction
E_cell = E°_cell − \( \frac{0.0591}{6} \) log \( \frac{[0.01]^2}{[0.1]^3} \)
= 1.41 − \( \frac{0.0591}{6} \) log 0.1
= 1.41 + \( \frac{0.0591}{6} \) log 10
= 1.41 + \( \frac{0.0591}{6} \)
= 1.41 + 0.00985 = 1.42 V
(b) When the concentration of weak electrolyte becomes very low, its degree of ionization rises. This increase leads to increase in the number of ions in the solution. Thus, the molar conductivity rises sharply of a weak electrolyte at low concentration. The molar conductivity of strong electrolyte decreases a bit with an increase in concentration. This is due to increase in interionic attraction due to higher number of ions per unit volume. On dilution, ions move apart, weakening interionic attractions and thus conductance increases.
Limiting molar conductivity for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions.

Teacher's Note:
1. Here 6 electrons are transferred (2Al → 2Al³⁺ + 6e⁻), so n = 6 in the Nernst equation, not 2 or 3.

Question 40. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
Sn (s) | Sn²⁺ (0.004 M) || H⁺ (0.020 M) | H₂ (g) (1 bar) | Pt (s)
(Given: E°_Sn²⁺/Sn = −0.14V)
(b) Give reasons:
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH₃COOH decreases on dilution. [CBSE Delhi/Outside Delhi, 2018]
Answer:
(a) Sn + 2H⁺ → Sn²⁺ + H₂
E_cell = E°_cell − \( \frac{0.059}{2} \) log \( \frac{[\text{Sn}^{2+}]}{[\text{H}^+]^2} \)
E_cell = [0 − (−0.14)] − 0.0295 log \( \frac{(0.004)}{(0.02)^2} \)
= 0.14 − 0.0295 log 10 = 0.11 V / 0.1105 V
(b) (i) Due to over potential/ over voltage of O₂
(ii) The number of ions per unit volume decreases.
Detailed Answer:
(a) Sn(s) | Sn²⁺(0.004 M) || H⁺(0.020 M) | H₂(g)(1 bar) | Pt(s)
E°_cell = E°_(H⁺/H2) − E°_(Sn²⁺/Sn)
= 0.00 − (− 0.14) = +0.14 V
Sn(s) → Sn²⁺(aq) + 2e⁻
2H⁺(aq) + 2e⁻ → H₂(g)
Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g)
E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[\text{Sn}^{2+}]}{[\text{H}^+]^2} \)
= 0.14 − \( \frac{0.0591}{2} \) log \( \frac{(4\times10^{-3})}{(2\times10^{-2})^2} \)
= 0.14 − 0.0295 log 10
= 0.14 − 0.0295 = 0.1105 V
(b) (i) NaCl → Na⁺ + Cl⁻
H₂O ⇌ H⁺ + OH⁻
The value of E° of O₂ is higher than Cl₂ but O₂ is evolved from H₂O only when the higher voltage is applied. So, because of this Cl₂ is evolved instead of O₂.
(ii) Conductivity varies with the change in the concentration of the electrolyte. The number of ions per unit volume decreases on dilution. So, conductivity decreases with decrease in concentration. Therefore, conductivity of CH₃COOH decreases on dilution.

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