CBSE Class 12 Chemistry Electrochemistry Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Express the relation between conductivity and molar conductivity of a solution held in a cell.
Answer:
Λ_m = \( \frac{\kappa\times1000}{M} \) (in CGS units);   Λ_m = \( \frac{\kappa\times10^{-3}}{M} \) (in SI units)
where κ is the conductivity, M is the molar concentration and Λ_m is the molar conductivity.

Teacher's Note:
1. Keep track of which unit system you are in – the factor is 1000 with κ in S cm⁻¹, or 10⁻³ with κ in S m⁻¹.

Question 2. Limiting molar conductivity of an electrolyte cannot be determined experimentally for weak electrolytes. Why?
Answer:
In a weak electrolyte, the conductivity of the solution increases very slowly with dilution and goes on increasing up to infinity. Therefore, it cannot be measured experimentally – the Λ_m vs √C plot is not linear and cannot be extrapolated to C = 0.

Teacher's Note:
1. This is exactly why Kohlrausch’s law exists – it gives an indirect route to Λ°m for weak electrolytes.

Question 3. Following reactions occur at the cathode during the electrolysis of aqueous silver chloride solution:
Ag⁺_(aq) + e⁻ → Ag_(s), E° = +0.80 V
H⁺_(aq) + e⁻ → ½H_2(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?
Answer:
The species that gets reduced at the cathode is the one having the higher value of standard reduction potential.
Hence, the reaction that will occur at the cathode is Ag⁺_(aq) + e⁻ → Ag_(s).

Teacher's Note:
1. Comparing E° values this way is the standard method for predicting which species is reduced first at the cathode.

Question 4. Give reason: Molar conductivity of CH₃COOH increases on dilution.
Answer:
Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases.
It has been found that the decrease in κ on dilution of a solution is more than compensated by the increase in its volume.

Teacher's Note:
1. For a weak electrolyte like CH3COOH, dilution also pushes the dissociation equilibrium further forward, which adds to this effect.

Question 5. Give reason: On the basis of E° values, O₂ gas should be liberated at the anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
Answer:
The reaction at the anode with the lower value of E° is preferred, i.e., O₂ gas should be liberated, but on account of the overpotential of oxygen at the anode, the preferred reaction is
Cl⁻_(aq) → ½Cl_2(g) + e⁻, i.e., Cl₂ gas is liberated at the anode in the electrolysis of aqueous NaCl.

Teacher's Note:
1. Overpotential is an extra, kinetic barrier that can override the plain thermodynamic (E°-based) prediction at an electrode.

Question 6. What is the necessity to use a salt bridge in a Galvanic cell?
Answer:
The salt bridge allows the movement of ions from one solution to the other without mixing of the two solutions. Moreover, it helps to maintain the electrical neutrality of the solutions in the two half cells.

Teacher's Note:
1. Without a salt bridge, charge would build up in each half-cell and the current would quickly stop.

Question 7. What is the use of the platinum foil in the hydrogen electrode?
Answer:
It is used for the inflow and outflow of electrons.

Teacher's Note:
1. Platinum is chemically inert here, so it only conducts electrons without itself taking part in the redox reaction.

Question 8. Out of HCl and NaCl, which do you expect will have a greater value for Λ_m and why?
Answer:
HCl will have a greater value of Λ_m because H⁺ ions are smaller than Na⁺ ions and hence H⁺ ions have greater ionic mobility than Na⁺ ions.

Teacher's Note:
1. H+ (and OH-) move abnormally fast in water through a special relay (Grotthuss) mechanism, on top of their small size.

Question 9. State Kohlrausch’s law of independent migration of ions. Write its one application.
Answer:
It 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.
Kohlrausch’s law helps in the calculation of the degree of dissociation of weak electrolytes like acetic acid.

Teacher's Note:
1. Besides degree of dissociation, the same law is also used to find Λ°m of a weak electrolyte and the solubility of a sparingly soluble salt.

Question 10. What would happen if no salt bridge were used in an electrochemical cell (like Zn–Cu cell)?
Answer:
The metal ions (Zn²⁺) formed by the loss of electrons will accumulate in one electrode and the negative ions (SO₄²⁻) will accumulate in the other. Thus, the solutions will develop charges and the current will stop flowing. Moreover, the inner circuit will not be completed.

Teacher's Note:
1. The salt bridge has two jobs: it completes the internal circuit and keeps both half-cells electrically neutral.

Question 11. Why does a galvanic cell become dead after some time?
Answer:
As the reaction proceeds, concentration of ions in anodic half keeps on increasing while in the cathodic half it keeps on decreasing. Hence, their electrode potentials also keeps on changing until they become equal and then e.m.f. of the cell becomes zero.

Teacher's Note:
1. A dead cell is a cell at equilibrium: E_cell = 0 and Δ_rG = 0.

Question 12. Consider a cell given below:
Cu | Cu²⁺ || Cl⁻ | Cl₂, Pt
Write the reactions that occur at anode and cathode. [NCERT Exemplar]
Answer:
Anode: Cu → Cu²⁺ + 2e⁻
Cathode: Cl₂ + 2e⁻ → 2Cl⁻

Teacher's Note:
1. In cell notation the left side is always the anode (oxidation) and the right side the cathode (reduction).

Short Answer Questions (2 Marks)

Question 13. Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation.
Answer:
The fraction of the total number of molecules present in solution as ions is known as the degree of dissociation.
Molar conductivity (λ_m) = αλ°_m, where λ°_m is the molar conductivity at infinite dilution.

Teacher's Note:
1. This same relation, rearranged as α = Λm/Λ°m, is used throughout the chapter to find degree of dissociation from measured conductivity.

Question 14. (i) Explain why fluorine is the strongest oxidising agent.
(ii) Lithium metal is the strongest reducing agent. Why?
Answer:
(i) Because fluorine has the highest reduction potential.
(ii) Lithium metal is the strongest reducing agent because Li has the lowest reduction potential, i.e., E°_Li⁺/Li = −3.05 V

Teacher's Note:
1. Strongest oxidiser = highest E°; strongest reducer = lowest (most negative) E° — the two ends of the electrochemical series.

Question 15. The standard electrode potential (E°) for the Daniell cell is +1.1 V. Calculate the Δ_rG° for the reaction,
Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s)
(1 F = 96500 C mol⁻¹)
Answer:
Here n = 2, E°_cell = 1.1 V, F = 96500 C mol⁻¹
Δ_rG° = −nFE°_cell
Δ_rG° = −2 × 1.1 × 96500 = −212300 J mol⁻¹ = −212.3 kJ mol⁻¹

Teacher's Note:
1. n = 2 comes from the balanced equation itself (Zn loses 2 electrons, Cu2+ gains 2) — always read n off the equation, not off the half-cell notation alone.

Question 16. What is the difference between electronic and electrolytic conductors?
Answer:
The substance which conducts electricity by ions present in solution is called an electrolytic conductor, e.g., NaCl solution.
Substances which conduct electricity in the solid state are called electronic conductors. These are made up of metals, e.g., Cu, Zn, Al. (Electrolytes are electrolytic conductors while electrodes are electronic conductors.)

Teacher's Note:
1. Electronic conductance falls with rising temperature (more scattering); electrolytic conductance rises with temperature (ions move faster, viscosity drops) — opposite trends.

Question 17. Define electrochemical cell. What happens if the external potential applied becomes greater than E°_cell of the electrochemical cell?
Answer:
The device which converts the chemical energy liberated during a chemical reaction to electrical energy is called an electrochemical cell.
If the external potential applied becomes greater than E°_cell of the electrochemical cell, then it behaves as an electrolytic cell and the direction of flow of current is reversed.

Teacher's Note:
1. This threshold behaviour is the same idea tested in several MCQs earlier in this chapter (the 1.1 V opposing-potential question).

Question 18. Why does a galvanic cell stop working after some time?
Answer:
With time, the concentrations of the electrolytic solutions change. Hence, their electrode potentials change.
When the electrode potentials of the two half-cells become equal, the cell stops working (EMF = 0, equilibrium reached).

Teacher's Note:
1. This is the everyday meaning of a "dead battery" — the cell hasn’t run out of material, it has simply reached electrochemical equilibrium.

Question 19. In the plot of molar conductivity (Λ_m) vs square root of concentration (c^1/2), the following curves are obtained for two electrolytes A and B.
Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of Λ_m to concentration approaching zero for electrolytes A and B?

Answer:
(i) Electrolyte A is a strong electrolyte while electrolyte B is a weak electrolyte.
(ii) For electrolyte A, the plot becomes linear near high dilution and thus can be extrapolated to zero concentration to get the molar conductivity at infinite dilution.
For weak electrolyte B, Λ_m increases steeply on dilution and extrapolation to zero concentration is not possible. Hence, molar conductivity at infinite dilution cannot be determined directly for B.

Teacher's Note:
1. The shape of this graph is the single most reliable visual test in the chapter for telling a strong electrolyte from a weak one.

Question 20. Two half-reactions of an electrochemical cell are given below:
MnO₄⁻_(aq) + 8H⁺_(aq) + 5e⁻ → Mn²⁺_(aq) + 4H₂O_(l), E° = +1.51 V
Sn²⁺_(aq) → Sn⁴⁺_(aq) + 2e⁻, E° = +0.15 V
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured.
Answer:
At anode: [Sn²⁺_(aq) → Sn⁴⁺_(aq) + 2e⁻] × 5
At cathode: [MnO₄⁻_(aq) + 8H⁺_(aq) + 5e⁻ → Mn²⁺_(aq) + 4H₂O_(l)] × 2
Net cell reaction: 2MnO₄⁻_(aq) + 5Sn²⁺_(aq) + 16H⁺_(aq) → 2Mn²⁺_(aq) + 5Sn⁴⁺_(aq) + 8H₂O_(l)
E°_cell = E°_cathode − E°_anode = 1.51 V − 0.15 V = 1.36 V
Since the cell potential is positive, the reaction is product favoured.

Teacher's Note:
1. Balance electrons lost = electrons gained first (LCM of 5 and 2 = 10) before adding the two half-reactions, so nothing cancels incorrectly.

Question 21. The conductivity of 0.001 M acetic acid is 4 × 10⁻⁵ S/cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm²/mol.
Answer:
C = 0.001 M, κ = 4 × 10⁻⁵ S cm⁻¹, Λ°_m = 390 S cm²/mol
Λ^c_m = \( \frac{\kappa\times1000}{C} \) = \( \frac{4\times10^{-5}\times1000}{0.001} \) = 40 S cm²/mol
α = \( \frac{\Lambda^{c}_{m}}{\Lambda^{\circ}_{m}} \) = \( \frac{40}{390} \) = 0.10256 ≈ 0.103
CH₃COOH ⇌ CH₃COO⁻ + H⁺
K_a = \( \frac{C\alpha^{2}}{1-\alpha} \) = \( \frac{0.001(0.103)^{2}}{1-0.103} \) = \( \frac{1.061\times10^{-5}}{0.897} \) = 1.18 × 10⁻⁵

Teacher's Note:
1. This is a two-stage calculation — find Λcm from κ first, then α from Λcm/Λ°m, before touching the Ka formula.

Question 22. Equilibrium constant (K_c) for the given cell reaction is 10. Calculate E°_cell.
A_(s) + B²⁺_(aq) ⇌ A²⁺_(aq) + B_(s)
Answer:
Here n = 2
Using the formula, E°_cell = \( \frac{0.059}{n} \) log K_c
E°_cell = \( \frac{0.059}{2} \) log 10
E°_cell = 0.0295 V

Teacher's Note:
1. log 10 = 1 exactly, which makes this a quick plug-in once n is identified correctly from the two-electron exchange shown.

Question 23. Given that the standard electrode potential (E°) of metals are:
K⁺/K = −2.93 V, Ag⁺/Ag = 0.80 V, Cu²⁺/Cu = 0.34 V, Cr³⁺/Cr = −0.74 V, Fe²⁺/Fe = −0.44 V
Arrange these metals in increasing order of their reducing power.
Answer:
The reducing power increases with decreasing value of electrode potential.
Hence, the order is Ag < Cu < Fe < Cr < K.

Teacher's Note:
1. The lower (more negative) the reduction potential, the more easily the metal loses electrons, so K is the strongest reducing agent and Ag the weakest.

Question 24. Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λ_m) is 39.05 S cm² mol⁻¹. Given Λ°_(H⁺) = 349.6 S cm² mol⁻¹ and Λ°_(CH3COO⁻) = 40.9 S cm² mol⁻¹. [CBSE Delhi Set-1, 2, 3 2017]
Answer:
Λ°_CH3COOH = Λ°_CH3COO⁻ + Λ°_H⁺
= 40.9 + 349.6 = 390.5 S cm²/mol
Now, α = Λ_m/Λ°_m
= 39.05/390.5 = 0.1

Teacher's Note:
1. Write the working formula in each step, then substitute values, and give units with the answer – marks are lost for skipping steps.

Short Answer Questions (3 Marks)

Question 25. Mention a few applications of the electrochemical series.
Answer:
(i) Ions with higher reduction potentials are strong oxidising agents while those with lower reduction potentials are strong reducing agents.
(ii) The electrode with higher electrode potential (E°) acts as the cathode while the one with lower electrode potential acts as the anode.
(iii) Predicting the feasibility of a redox reaction.
(iv) Predicting the capability of a metal to evolve H₂ gas from acid.

Teacher's Note:
1. Metals with negative E° (below hydrogen in the series) can liberate H₂ from dilute acids; copper and silver cannot.
2. A redox reaction is feasible when E°_cell worked out from the series comes out positive.

Question 26. A voltaic cell is set up at 25°C with the following half cells:
Al/Al³⁺ (0.001 M) and Ni/Ni²⁺ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(E°_Ni²⁺/Ni = −0.25 V and E°_Al³⁺/Al = −1.66 V; log 8 × 10⁻⁶ = −5.09)
Answer:
At anode: [Al_(s) → Al³⁺_(aq) + 3e⁻] × 2
At cathode: [Ni²⁺ + 2e⁻ → Ni_(s)] × 3
Cell reaction: 2Al_(s) + 3Ni²⁺_(aq) → 2Al³⁺_(aq) + 3Ni_(s)
Applying the Nernst equation to the above cell reaction,
E_cell = E°_cell − \( \frac{0.0591}{2\times3} \) log \( \frac{[Al^{3+}]^{2}}{[Ni^{2+}]^{3}} \)
Now, E°_cell = E°_Ni²⁺/Ni − E°_Al³⁺/Al = −0.25 − (−1.66) = 1.41 V
∴ E_cell = 1.41 − \( \frac{0.0591}{6} \) log \( \frac{(10^{-3})^{2}}{(0.5)^{3}} \)
= 1.41 − \( \frac{0.0591}{6} \) log (8 × 10⁻⁶)
= 1.41 − \( \frac{0.0591}{6} \) (−5.09)
= 1.41 + 0.050 = 1.46 V

Teacher's Note:
1. Balance the electrons first (6 here); that same n goes into the Nernst equation, and the coefficients become the powers on the concentrations.
2. The cell potential comes out above E° because the product ion Al³⁺ is dilute and the reactant ion Ni²⁺ is fairly concentrated.

Question 27. A cell is prepared by dipping a copper rod in 1 M copper sulphate solution and a zinc rod in 1 M zinc sulphate solution. The standard reduction potentials of copper and zinc are 0.34 V and −0.76 V respectively.
(i) What will be the cell reaction?
(ii) What will be the standard electromotive force of the cell?
(iii) Which electrode will be positive?
Answer:
(i) The cell reactions are:
Zn_(s) → Zn²⁺_(aq) + 2e⁻ (Anode)
Cu²⁺_(aq) + 2e⁻ → Cu_(s) (Cathode)
Net reaction: Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s)
(ii) E°_cell = E°_cathode − E°_anode = 0.34 V − (−0.76 V) = 1.10 V
(iii) The copper electrode, on which reduction takes place, will be positive.

Teacher's Note:
1. This is the Daniell cell. In a galvanic cell the cathode is the positive terminal, the reverse of an electrolytic cell.

Question 28. Resistance of a conductivity cell filled with 0.1 mol L⁻¹ KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L⁻¹ KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L⁻¹ KCl solution. The conductivity of 0.1 mol L⁻¹ KCl solution is 1.29 × 10⁻² Ω⁻¹ cm⁻¹.
Answer:
Resistance of 0.1 M KCl solution, R = 100 Ω
Conductivity, κ = 1.29 S m⁻¹
Cell constant, G* = κ × R = 1.29 × 100 = 129 m⁻¹
Resistance of 0.02 M KCl solution, R = 520 Ω
Conductivity, κ = \( \frac{\text{cell constant}}{R} \) = \( \frac{129\ \text{m}^{-1}}{520\ \Omega} \) = 0.248 S m⁻¹
Concentration, C = 0.02 mol L⁻¹ = 1000 × 0.02 mol m⁻³ = 20 mol m⁻³
Molar conductivity, Λ_m = \( \frac{\kappa}{C} \) = \( \frac{0.248\ \text{S m}^{-1}}{20\ \text{mol m}^{-3}} \) = 0.0124 S m² mol⁻¹

Teacher's Note:
1. 1.29 × 10⁻² S cm⁻¹ equals 1.29 S m⁻¹ (multiply by 100), which is why the SI value appears in the first step.
2. The cell constant depends only on the cell, so it is found with the standard KCl solution and then reused for the second solution.

Question 29. Calculate the potential for a half-cell containing
0.10 M K₂Cr₂O_7(aq), 0.20 M Cr³⁺_(aq) and 1.0 × 10⁻⁴ M H⁺_(aq). The half-cell reaction is
Cr₂O₇²⁻_(aq) + 14H⁺_(aq) + 6e⁻ → 2Cr³⁺_(aq) + 7H₂O_(l)
and the standard electrode potential is given as E° = 1.33 V.
Answer:
For the half cell reaction,
Cr₂O₇²⁻_(aq) + 14H⁺_(aq) + 6e⁻ → 2Cr³⁺_(aq) + 7H₂O_(l)
E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[Cr^{3+}]^{2}}{[Cr_2O_7^{2-}][H^{+}]^{14}} \)
Given, E°_cell = 1.33 V, n = 6, [Cr³⁺] = 0.2 M, [Cr₂O₇²⁻] = 0.1 M, [H⁺] = 1 × 10⁻⁴ M
E_cell = 1.33 − \( \frac{0.0591}{6} \) log \( \frac{(0.20)^{2}}{(0.1)(10^{-4})^{14}} \)
= 1.33 − \( \frac{0.0591}{6} \) log (4 × 10⁵⁵)
= 1.33 − \( \frac{0.0591}{6} \) [log 4 + 55 log 10]
= 1.33 − \( \frac{0.0591}{6} \) [0.602 + 55]
= 1.33 − 0.548 = 0.782 V

Teacher's Note:
1. The [H⁺] term is raised to the power 14, so a low acid concentration lowers the electrode potential a lot.
2. This is why dichromate is a much stronger oxidising agent in strongly acidic solution.

Question 30. Estimate the minimum potential difference needed to reduce Al₂O₃ at 500°C. The Gibbs energy change for the decomposition reaction, \( \frac{2}{3} \)Al₂O₃ → \( \frac{4}{3} \)Al + O₂ is 960 kJ. (F = 96500 C mol⁻¹)
Answer:
Al₂O₃ (2Al³⁺ + 3O²⁻) → 2Al + \( \frac{3}{2} \)O₂, n = 6e⁻
∴ \( \frac{2}{3} \)Al₂O₃ → \( \frac{4}{3} \)Al + O₂, n = \( \frac{2}{3} \) × 6e⁻ = 4e⁻
ΔG = 960 × 1000 = 960000 J
Now, ΔG = −nFE_cell
E_cell = −\( \frac{\Delta G}{nF} \) = \( \frac{-960000}{4\times96500} \) = −2.487 V
Minimum potential difference needed to reduce Al₂O₃ is 2.487 V (the negative sign shows that the reaction is non-spontaneous and the potential must be applied from outside).

Teacher's Note:
1. n must match the equation as written: 4/3 mol Al, each needing 3 electrons, gives n = 4.
2. A positive ΔG gives a negative E, which is why electrolysis needs at least this much applied voltage.

Question 31. For the cell reaction,
Ni_(s)|Ni²⁺_(aq)||Ag⁺_(aq)|Ag_(s)
Calculate the equilibrium constant at 25°C. How much maximum work would be obtained by operation of this cell?
E°_Ni²⁺/Ni = −0.25 V and E°_Ag⁺/Ag = 0.80 V
Answer:
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 = 1.05 V
E°_cell = \( \frac{0.0591}{n} \) log K_c
log K_c = \( \frac{E^{\circ}_{cell}\times n}{0.0591} \) = \( \frac{1.05\times2}{0.0591} \) = 35.53
K_c = antilog 35.53 = 3.38 × 10³⁵
Maximum work = −ΔG° = nFE°_cell = 2 × 96500 × 1.05 = 202650 J = 202.65 kJ

Teacher's Note:
1. The maximum useful work a cell can do equals −ΔG°, so it is found with the same nFE° expression.
2. A K of the order of 10³⁵ means the reaction goes essentially to completion.

Question 32. Calculate Δ_rG and log K_c for the following reaction.
Cd²⁺_(aq) + Zn_(s) → Zn²⁺_(aq) + Cd_(s)
Given: E°_Cd²⁺/Cd = −0.403 V; E°_Zn²⁺/Zn = −0.763 V
Answer:
E°_cell = E°_cathode − E°_anode = −0.403 − (−0.763) = 0.36 V
Δ_rG° = −nFE°_cell = −2 × 96500 × 0.36 = −69480 J = −69.48 kJ
Using the formula, log K_c = \( \frac{nE^{\circ}_{cell}}{0.059} \) = \( \frac{2\times0.36}{0.059} \) = 12.20
K_c = antilog 12.20 = 1.58 × 10¹²

Teacher's Note:
1. Zn has the lower E°, so it is the anode; Cd²⁺ is reduced at the cathode.
2. Both ΔG° and log K come from the same E°_cell, so work that out carefully first.

Question 33. The equivalent conductivity of 0.05 N solution of a monobasic acid is 15.8 mho cm² eq⁻¹. If equivalent conductivity of the acid at infinite dilution is 350 mho cm² eq⁻¹, calculate (a) degree of dissociation of acid (b) dissociation constant of acid.
Answer:
(a) Degree of dissociation, α = \( \frac{\Lambda_{eq}}{\Lambda^{\infty}_{eq}} \)
∴ α = \( \frac{15.8}{350} \) = 0.04514
(b) For a monobasic acid, HA ⇌ H⁺ + A⁻
K = \( \frac{C\alpha^{2}}{(1-\alpha)} \) = Cα²
As α << 1, hence (1 − α) ≈ 1
∴ K = 0.05 × (0.04514)² ⇒ K = 1.019 × 10⁻⁴

Teacher's Note:
1. The approximation (1 − α) ≈ 1 is safe here because α is only about 4.5%.
2. For a monobasic acid, normality equals molarity, so C = 0.05 mol L⁻¹.

Question 34. The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.5 × 10³ ohm. Calculate its resistivity, conductivity and molar conductivity.
Answer:
Given: Diameter = 1 cm, length = 50 cm, R = 5.5 × 10³ ohm, M = 0.05 M
ρ = ?, κ = ?, Λ_m = ?
Area of the column, a = πr² = 3.14 × (\( \frac{1}{2} \) cm)² = \( \frac{3.14}{4} \) cm²
Resistivity, ρ = R · \( \frac{a}{l} \) = 5.5 × 10³ ohm × \( \frac{3.14\ \text{cm}^{2}}{4\times50\ \text{cm}} \) = 86.35 ohm cm
Again, conductivity, κ = \( \frac{1}{\rho} \) = \( \frac{1}{86.35} \) = 1.158 × 10⁻² ohm⁻¹ cm⁻¹
and molar conductivity, Λ_m = κ · \( \frac{10^{3}}{M} \) = 1.158 × 10⁻² ohm⁻¹ cm⁻¹ × \( \frac{10^{3}}{5\times10^{-2}} \)
= 231.6 ohm⁻¹ cm² mol⁻¹

Teacher's Note:
1. Use the radius (0.5 cm), not the diameter, when finding the cross-sectional area.
2. The three quantities follow in a chain: R gives ρ, ρ gives κ, and κ gives Λ_m.

Long Answer Questions (5 Marks)

Question 35. E°_cell for the given redox reaction is 2.71 V.
Mg_(s) + Cu²⁺ (0.01 M) → Mg²⁺ (0.001 M) + Cu_(s)
Calculate E_cell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V
Answer:
E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[Mg^{2+}]}{[Cu^{2+}]} \)
= 2.71 − \( \frac{0.0591}{2} \) log \( \frac{0.001}{0.01} \) = 2.73955 V
(i) If the external opposing potential is less than 2.71 V, then current will flow from Cu to Mg.
(ii) If the external opposing potential is greater than 2.71 V, then current will flow in the opposite direction, i.e., from Mg to Cu.

Teacher's Note:
1. Current (conventional) flows opposite to electrons: electrons go from Mg to Cu in the external circuit, so current goes from Cu to Mg.
2. Once the opposing potential exceeds the cell potential, the cell works as an electrolytic cell and both directions reverse.

Question 36. (a) Calculate the standard emf of the cell in which the following reaction takes place at 25°C.
Cu_(s) + Cl_2(g) ⇌ Cu²⁺ + 2Cl⁻
E°_Cl2/Cl⁻ = +1.36 V, E°_Cu²⁺/Cu = +0.34 V
Also calculate the standard free energy change and equilibrium constant of the reaction.
(b) The emf of a galvanic cell composed of two hydrogen electrodes is 0.16 volt at 25°C. Calculate the pH of the anode solution if the cathode is in a solution with pH = 1.
Answer:
(a) The given cell may be represented as Cu_(s)|Cu²⁺||Cl₂|Cl⁻
(i) E°_cell = E°_c − E°_a = (+1.36 V) − (+0.34 V) = 1.02 V
(ii) Δ_rG° = −nFE° = −2 × 96500 C × 1.02 V = −196.86 kJ
(iii) E°_cell = \( \frac{0.0591}{n} \) log K
log K = \( \frac{2\times1.02}{0.0591} \) = 34.51; K = antilog (34.51)
K = 3.236 × 10³⁴
(b) The given cell may be represented as Pt, H₂ (1 atm)|H⁺ (pH = ?)||H⁺ (pH = 1)|H₂ (1 atm)
Using the formula, E_cell = \( \frac{0.0591}{1} \) log \( \frac{[H^{+}]_{c}}{[H^{+}]_{a}} \)
or 0.16 = 0.0591 [log [H⁺]_c − log [H⁺]_a]
or 0.16 = 0.0591 [pH_a − pH_c]
0.16 = 0.0591 [pH_a − 1]
or pH_a − 1 = \( \frac{0.16}{0.0591} \) = 2.70
or pH_a = 2.70 + 1 = 3.70

Teacher's Note:
1. ΔG° must carry a minus sign here because E°_cell is positive and the reaction is spontaneous.
2. In the hydrogen concentration cell, since −log[H⁺] = pH, the log ratio turns into a simple pH difference.

Question 37. (a) Calculate the cell emf and ΔG° for the cell reaction at 25°C for the cell:
Zn_(s)|Zn²⁺ (0.0004 M)||Cd²⁺ (0.2 M)|Cd_(s)
E° values at 25°C: Zn²⁺/Zn = −0.763 V; Cd²⁺/Cd = −0.403 V; F = 96500 C mol⁻¹; R = 8.314 J K⁻¹ mol⁻¹.
(b) If E° for the copper electrode is 0.34 V, how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions? How does emf for the copper electrode change when the concentration of Cu²⁺ ions in the solution is decreased?
Answer:
(a) E°_cell = E°_cathode − E°_anode = −0.403 − (−0.763) = 0.36 V
The net cell reaction is Zn_(s) + Cd²⁺_(aq) → Zn²⁺_(aq) + Cd_(s). Here, the value of n = 2.
E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Zn^{2+}]}{[Cd^{2+}]} \)
= 0.36 − \( \frac{0.0591}{2} \) log \( \frac{0.0004}{0.2} \)
= 0.36 − \( \frac{0.0591}{2} \) (−2.69) = 0.36 + 0.08 = 0.44 V
∴ ΔG = −nFE_cell = −2 × 96500 × 0.44 = −84920 J/mol
(b) Cu²⁺_(aq) + 2e⁻ → Cu_(s)
E_Cu²⁺/Cu = E°_Cu²⁺/Cu − \( \frac{0.059}{2} \) log \( \frac{[Cu]}{[Cu^{2+}]} \)
= 0.34 − \( \frac{0.059}{2} \) log \( \frac{1}{0.1} \) = 0.34 − \( \frac{0.059}{2} \) log 10
= 0.34 − \( \frac{0.059}{2} \) × (1) = 0.34 − 0.0295 = 0.3105 V
When the concentration of Cu²⁺ ions is decreased, the electrode potential of the copper electrode decreases.

Teacher's Note:
1. The value above uses E_cell (0.44 V), so it is ΔG at the given concentrations. The standard value ΔG° uses E°_cell (0.36 V) and comes to −69480 J/mol.
2. For a metal electrode, the metal is a pure solid with [Cu] = 1, so only [Cu²⁺] affects the potential.

Question 38. (a) Equivalent conductance of a 0.0128 N solution of acetic acid is 1.4 mho cm² eq⁻¹ and conductance at infinite dilution is 391 mho cm² eq⁻¹. Calculate the degree of dissociation and dissociation constant of acetic acid.
(b) The equivalent conductances of sodium acetate, sodium chloride and hydrochloric acid are 83, 127 and 426 mho cm² eq⁻¹ at 25°C respectively. Calculate the equivalent conductance of acetic acid solution.
Answer:
(a) Given: Λ_eq = 1.4 mho cm² eq⁻¹, Λ^∞_eq = 391 mho cm² eq⁻¹, α = ?, K_a = ?
Using the formula, α = \( \frac{\Lambda_{eq}}{\Lambda^{\infty}_{eq}} \) = \( \frac{1.4}{391} \) = 0.00358
K_a = \( \frac{\alpha^{2}C}{1-\alpha} \) = \( \frac{(0.00358)^{2}\times0.0128}{1-0.00358} \)
= \( \frac{1.64\times10^{-7}}{0.99642} \) = 1.64 × 10⁻⁷
(b) Given: Λ°_eq (CH₃COONa) = 83 mho cm² eq⁻¹, Λ°_eq (NaCl) = 127 mho cm² eq⁻¹, Λ°_eq (HCl) = 426 mho cm² eq⁻¹, Λ°_eq (CH₃COOH) = ?
Using Kohlrausch's law of independent migration of ions,
Λ°_eq (CH₃COOH) = Λ°_eq (CH₃COONa) + Λ°_eq (HCl) − Λ°_eq (NaCl)
or Λ°_eq (CH₃COOH) = 83 + 426 − 127 = 382 mho cm² eq⁻¹

Teacher's Note:
1. In part (b), adding sodium acetate and HCl and subtracting NaCl removes the Na⁺ and Cl⁻ contributions and leaves H⁺ + CH₃COO⁻.
2. The value in (b) is the limiting value at infinite dilution, which is what Kohlrausch's law gives.

Question 39. (a) The electrical resistance of a column of 0.05 M KOH solution of length 50 cm and area of cross-section 0.625 cm² is 5 × 10³ ohm. Calculate its resistivity, conductivity and molar conductivity.
(b) 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, OD Set-1, 2020]
Answer:
(a) Given: A = 0.625 cm², l = 50 cm, R = 5 × 10³ ohm, ρ = ?, M = 0.05 M, κ = ?, Λ_m = ?
Cell constant = \( \frac{l}{A} \) = \( \frac{50\text{ cm}}{0.625\text{ cm}^2} \) = 80 cm⁻¹
Resistivity = \( \frac{R}{\text{cell constant}} \) or \( \frac{R\times A}{l} \)
⇒ \( \frac{5\times10^3\text{ ohm}}{80\text{ cm}^{-1}} \) = 62.5 ohm cm
Conductivity = \( \frac{1}{\text{Resistance}} \) × \( \frac{l}{A} \)
= \( \frac{1}{5\times10^3} \) × 80 = 0.016 S cm⁻¹
Molar conductivity (Λ_m) = \( \frac{\kappa\times1000}{M} \)
Λ_m = \( \frac{0.016\times1000}{0.05} \) = 320 S cm² mol⁻¹
(b) Given: E°_Cu²⁺/Cu = +0.34 V, E°_{(½Cl2/Cl⁻)} = +1.36 V, E°_{H⁺/H2(g), Pt} = 0.00 V, E°_(O2/H2O) = +1.23 V
At cathode:
Cu²⁺(aq) + 2e⁻ → Cu(s); E° = 0.34 V
H⁺(aq) + e⁻ → ½H₂(g); E° = 0 V
The reaction with a higher value of E° takes place at the cathode, so deposition of copper will take place at the cathode.
At anode: The oxidation reactions are possible at the anode.
Cl⁻(aq) → ½Cl₂(g) + e⁻; E° = 1.36 V
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻; E° = +1.23 V
At the anode the reaction with a lower value of E° is preferred. But due to the over potential of oxygen, Cl⁻ gets oxidised at anode to produce Cl₂ gas.

Teacher's Note:
1. Resistivity ρ = RA/l and conductivity κ = 1/ρ; get the cell constant (l/A) first and both follow quickly.
2. In aqueous chloride electrolysis, Cl₂ (not O₂) forms at the anode because of the overpotential of oxygen.