Indian Forest Service (Mains), Chemistry optional · 2021 · Paper I
Q1(e). Equilibrium constant of a Zn–Ag cell
Question
Calculate the equilibrium constant of the cell reaction 2Ag⁺ + Zn → 2Ag + Zn²⁺ occurring in the zinc–silver cell at 25 °C when [Zn²⁺] = 0.10 M and [Ag⁺] = 10 M. The e.m.f. of the cell is found to be 1.62 V.
Answer in brief
With concentrations used as activities, E° = 1.5313 V and K ≈ 6.0 × 10⁵¹.
Key points
- Two electrons are transferred; pure metal activities are unity and Q = a(Zn²⁺)/a(Ag⁺)² ≈ 0.10/10² = 10⁻³.
- From E = E° − (RT/2F) ln Q, E° = 1.62 + [8.314(298.15)/2(96500)] ln(10⁻³) = 1.53128 V.
- At equilibrium E = 0 and Q = K, so ln K = 2FE°/RT = 119.225; hence K ≈ 6.0 × 10⁵¹ (dimensionless).
- The reaction is strongly product-favoured; at 10 M activity coefficients matter, so this is an ideal-concentration estimate.
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.
CoversNernst equationequilibrium constant from EMFZn–Ag cellstandard cell potentialln K = nFE°/RT
All parts of Question 1
- 1(a) Frenkel disorder in silver bromide8 marks
- 1(b) Changing the enthalpy reference8 marks
- 1(c) C–F and C–I potential-energy curves8 marks
- 1(d) Desilverisation of lead8 marks
- 1(e) Equilibrium constant of a Zn–Ag cell8 marks