Indian Forest Service (Mains), Chemistry optional · 2025 · Paper I
Q1(e). Fast and slow electrode systems
Question
Draw and explain steady-state current–potential curves for (i) a fast system and (ii) a slow system, with the same open-circuit potential and the same diffusion limiting currents.
Answer in brief
Both curves cross zero current at the same equilibrium potential and approach the same limiting plateaux; the fast system has a steeper slope near equilibrium and needs less overpotential.
Key points
- Open circuit means zero net current, although equal anodic and cathodic exchange currents still flow microscopically.
- A fast electron-transfer system has large exchange-current density j₀; Butler–Volmer linearisation gives j ≃ j₀nFη/(RT), so polarisation resistance is small.
- The slow system has smaller j₀; more overpotential is needed to drive a given current and its transition between plateaux is broader.
- At large overpotential, diffusion limits the current; since both systems have identical diffusion limits, their anodic and cathodic asymptotes coincide.
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.
Coverscurrent–potential curveexchange current densityButler–Volmer equationoverpotentiallimiting currentreversible and irreversible electrode
All parts of Question 1
- 1(a) Equivalent bonds in nitrate8 marks
- 1(b) Formula from occupied crystal voids8 marks
- 1(c) Formation free energy of uranium hydride8 marks
- 1(d) Pressure dependence of solid Gibbs energy8 marks
- 1(e) Fast and slow electrode systems8 marks