(a) Calculate the e.m.f. of the following electrochemical cell at 25 °C : Pt/H₂₍₁ ₐₜₘ₎|H⁺₍C=0.01 M₎‖Cu²⁺₍C=0.1 M₎|Cu(s) (10 marks)

(b) (i) A certain closed cell foam used as an insulating material is initially filled with polyatomic gas of molecular

Question

(a) Calculate the e.m.f. of the following electrochemical cell at 25 °C : Pt/H₂₍₁ ₐₜₘ₎|H⁺₍C=0.01 M₎‖Cu²⁺₍C=0.1 M₎|Cu(s) (10 marks)

(b) (i) A certain closed cell foam used as an insulating material is initially filled with polyatomic gas of molecular weight ~ 60. Later, the gas diffuses out of the foam and is replaced by dry air (mean molecular weight ~ 30). Assuming that insulating property arises largely from the thermal conductivity of the gas, explain the factors which influence the thermal conductivity of the gas. For each factor, make an argument whether insulating ability increases or decreases. What is the overall effect upon the insulating ability? (10 marks)
(ii) The critical temperature and pressure for NO gas are 177 K and 64 atm, respectively, and for CCl₄, they are 550 K and 45 atm, respectively. Which gas has the smaller values of the van der Waals' constants, a and b? Which is the most nearly ideal in behaviour at 300 K and 10 atm? (10 marks)

(c) Explain the phase diagram of phenol-water system by highlighting the importance of tie lines. (20 marks)

UPSC Answer Check · Accepted Answer

(a) e.m.f. of the cell
Cell: Pt | H₂(1 atm) | H⁺(0.01 M) ‖ Cu²⁺(0.1 M) | Cu. Anode (oxidation): H₂ → 2H⁺ + 2e⁻ (E° = 0, SHE); cathode (reduction): Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V). Net: H₂ + Cu²⁺ → 2H⁺ + Cu, n = 2.
E°cell = 0.34 − 0 = 0.34 V.
Nernst: E = E° − (0.0591/2) log([H⁺]²/[Cu²⁺]) (p_H₂ = 1).
= 0.34 − 0.02955 log[(0.01)²/0.1] = 0.34 − 0.02955 log(10⁻³) = 0.34 + 0.0887 = 0.429 V.
The positive sign confirms that the cell reaction is spontaneous as written.

(b)(i) Thermal conductivity of the gas in foam
By kinetic theory, κ = (1/3) C_V,m(N/V) c̄ λ, which leads to κ ∝ c̄ C_V,m/σ, independent of pressure. Factors:
1. Mean speed c̄ ∝ (T/M)^½: lighter air (M ≈ 30) molecules move about 1.4 times faster, so conductivity increases and insulation falls.
2. Heat capacity per molecule: polyatomic gas has more rotational and vibrational modes, so it stores and carries more energy per collision; replacing by diatomic air lowers κ, so insulation rises.
3. Collision cross-section σ (mean free path λ = kT/√2σp): the bigger polyatomic molecules have larger σ and a shorter λ, so low κ; smaller air molecules lengthen λ and raise κ.
4. Number density cancels with λ, so pressure has no effect.
Overall, factors 1 and 3 outweigh 2, so the conductivity of air is higher than that of the heavy polyatomic gas and the insulating ability of the foam decreases with ageing.

(b)(ii) van der Waals constants for NO and CCl₄
a = 27R²T_c²/64P_c and b = RT_c/8P_c.
NO: a = 27(0.08206)²(177)²/(64 × 64) = 1.39 L² atm mol⁻²; b = 0.08206 × 177/(8 × 64) = 0.0284 L mol⁻¹.
CCl₄: a = 27(0.08206)²(550)²/(64 × 45) = 19.1 L² atm mol⁻²; b = 0.08206 × 550/(8 × 45) = 0.125 L mol⁻¹.
NO has the smaller a and b (weaker attraction, smaller molecules).
At 300 K and 10 atm, NO has T_r = 300/177 = 1.7 and P_r = 10/64 = 0.16, so it is well above its critical temperature at low reduced pressure and nearly ideal (Z ≈ 1). CCl₄ is at T_r = 0.55, below its T_c, and is actually a liquid under these conditions, hence far from ideal.

(c) Phenol–water phase diagram and tie lines
Phenol and water are partially miscible. In the temperature–composition diagram, the mutual solubility curve forms a dome. Below the dome, two conjugate liquid layers exist: water-rich (phenol dissolved in water) and phenol-rich (water dissolved in phenol). At 20 °C they contain about 8% and 72% phenol by weight. Outside the dome, a single homogeneous solution exists.
As temperature rises, mutual solubility increases and the two compositions converge, meeting at the upper critical solution (consolute) temperature, 66.8 °C, at about 34% phenol by weight. Above it, the liquids mix in all proportions. Phenol–water shows no lower critical solution temperature.
A tie line is a horizontal line at a fixed temperature across the two-phase region, joining the compositions of the two conjugate phases. Its ends give the compositions of the layers, and the position of the overall composition on it gives their relative amounts by the lever rule (moles of water-rich phase × distance to its end = moles of phenol-rich phase × distance to the other end). Tie lines shorten with temperature and shrink to a point at the consolute point. Addition of impurities raises the critical temperature, so this system can be used to test purity.

(a) Calculate the e.m.f. of the following electrochemical cell at 25 °C : Pt/H₂₍₁ ₐₜₘ₎|H⁺₍C=0.01 M₎‖Cu²⁺₍C=0.1 M₎|Cu(s) (10…

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