(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

The three parts all turn on departure from ideal behaviour, but each requires a different physical mechanism.

Electrochemical cell. The anode reaction is H₂ → 2H⁺ + 2e⁻ and the cathode reaction is Cu²⁺ + 2e⁻ → Cu, so the cell reaction is H₂ + Cu²⁺ → 2H⁺ + Cu with n = 2. E°cell = E°Cu²⁺/Cu − E°H⁺/H₂ = 0.34 − 0 = 0.34 V. At 25 °C, E = E° − (0.0591/2) log Q, where Q = [H⁺]² / ([Cu²⁺] P_H₂) = (0.01)² / (0.1 × 1) = 10⁻³. Hence E = 0.34 − (0.0591/2)(−3) = 0.34 + 0.0887 = 0.4287 V. The non-standard concentrations make the cell more spontaneous than the standard cell.

Gas thermal conductivity. For a dilute gas at fixed T and P, kinetic theory gives κ ∝ C_v,m v_mean λ, where C_v,m is molar heat capacity, v_mean ∝ 1/√M, and λ is mean free path. If collision diameters are similar, λ is mainly pressure-dependent, so the molecular-weight part is κ ∝ C_v,m/√M; if C_v,m is comparable, this reduces to κ ∝ 1/√M. First, replacing a gas of M ≈ 60 by dry air of M ≈ 30 increases v_mean by √2, raising κ and reducing insulation. Second, the original polyatomic gas has more rotational and vibrational degrees of freedom, hence a larger C_v,m than diatomic air; this factor alone would raise κ, but air’s lower C_v,m is a partial offset. Third, the heavier polyatomic gas usually has a larger collision cross-section, so its λ is shorter and κ is lower; air’s smaller molecules give a longer λ and higher κ. Overall, the lower molecular weight and longer mean free path of air dominate, so the foam becomes more thermally conductive and its insulating ability decreases.

Van der Waals comparison. For a real gas, a = 27R²Tc²/(64Pc) and b = RTc/(8Pc). NO has Tc = 177 K and Pc = 64 atm; CCl₄ has Tc = 550 K and Pc = 45 atm. Since a ∝ Tc²/Pc and b ∝ Tc/Pc, NO has the smaller values of both a and b. At 300 K and 10 atm, NO has Tr = 300/177 ≈ 1.70 and Pr = 10/64 ≈ 0.16, whereas CCl₄ has Tr ≈ 0.55 and Pr ≈ 0.22. NO is well above its critical temperature and at low reduced pressure, so attractions and molecular volume matter less. Therefore NO is the most nearly ideal gas at 300 K and 10 atm.

Phenol-water phase diagram. Phenol and water are partially miscible below an upper consolute temperature, about 66 °C at about 34 wt% phenol. In a temperature-composition diagram, the two-liquid region is a lens bounded by two solubility curves: the left curve gives the phenol-poor aqueous phase and the right curve the phenol-rich phase. Above 66 °C the curves meet and the mixture is one homogeneous liquid. Below it, a composition inside the lens separates into two conjugate liquids. Tie lines are horizontal lines at constant temperature joining the equilibrium compositions of these two phases. They are important because they determine the actual phase compositions for any overall composition in the two-phase region, and the lever rule uses their segment lengths to calculate the relative amounts of the aqueous and phenol-rich phases. This is essential for designing phenol extraction, recovery and waste-water treatment, where one must know how much phenol remains in each phase and how many contacts are required.

Thus the cell shows concentration-driven non-ideality, the gases show van der Waals corrections, and the liquid mixture shows partial miscibility. Recognising these mechanisms is directly useful in Indian chemical engineering and materials applications, from hot-climate building insulation to phenol recovery in petrochemical and coal-tar units.

(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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