UPSC Combined Geo-Scientist, Mains · 2025 · Mains Paper II
Q1(j). Equilibrium constant from standard potentials
Question
Given that E°(Cr³⁺/Cr) = −0.41 V and E°(Ce³⁺/Ce) = −2.33 V, calculate the value of the equilibrium constant for the following reaction at room temperature (25 °C): Ce(s) + Cr³⁺(aq) ⇌ Ce³⁺(aq) + Cr(s).
Answer in brief
E°cell = 1.92 V, log K = 97.37, K ≈ 2.3 × 10⁹⁷.
Key points
- E°cell = E°cathode − E°anode = −0.41 − (−2.33) = 1.92 V.
- ln K = nFE°/RT = (3)(96485)(1.92)/(8.314)(298.15) = 224.2.
- log K = 224.2/2.3026 = 97.37 ⇒ K = 1097.37 ≈ 2.3 × 10⁹⁷.
- Reading: K is enormous, so Ce reduces Cr³⁺ to the metal completely.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversequilibrium constant from emfstandard electrode potentialΔG° = −nFE°Nernst equationcerium
All parts of Question 1
- 1(a) Why Z > 1 for H₂ and He at all pressures5 marks
- 1(b) Corresponding state of H₂S for N₂ at 25 °C and 1 atm5 marks
- 1(c) Band structure of n-type and p-type semiconductors5 marks
- 1(d) Symmetry elements of orthorhombic Mg₂SiO₄5 marks
- 1(e) Fall of G with temperature at constant pressure5 marks
- 1(f) Z > 1 and f > P for the gas P(Vm − b) = RT5 marks
- 1(g) Overall equation and the role of each species5 marks
- 1(h) log k against pH for specific H⁺ catalysis5 marks
- 1(i) KCl salt bridge for a silver–lead cell5 marks
- 1(j) Equilibrium constant from standard potentials5 marks
- 1(k) Conditions for an acceptable wave function5 marks
- 1(l) No solution with E = 0 for the particle in a box5 marks
- 1(m) Discrete levels and the resonance condition5 marks
- 1(n) Rotational activity of CO and CO₂5 marks
- 1(o) Measuring light intensity with a chemical actinometer5 marks
- 1(p) Why watch hands glow in the dark5 marks