Lectures 5–6 Practice: Energy Barriers, Butler–Volmer Kinetics and Catalytic Electrodes

Eleven short problems that examine Lectures 5 and 6 together, in the formats you may meet in the assessment: sorting, true or false, matching, step-by-step and multi-step calculations, comparing, finding a mistake and analysing data. Problems 9 and 10 link back to Lectures 1 and 3, and Problem 11 is a Tafel analysis. They start with the ideas and build up to the calculations, so begin with Problem 1. Each answer is checked as you go, and the working is shown once each step is correct.

Was this lesson useful?

All feedback is anonymous. No names or logins are collected.

Equation sheet

Overpotential: η = E − E_eq (E_eq = E° at 1 M)
Butler–Volmer: i = i₀[e^αnFη/RT − e^{−(1−α)nFη/RT}]
With α = 0.5 the exponents are ±nη/(2 × 0.02569 V)
Barriers: ΔG‡_ox = ΔG‡₀ − αnFη; ΔG‡_red = ΔG‡₀ + (1 − α)nFη
Rate ratio: k_ox/k_red = e^nFη/RT
Tafel (large |η|): |η| = b log(|i|/i₀), b = 2.303RT/(αnF) = 0.118 V/n per decade
Current = current density × area; q = it; moles = q/(zF)
E°_cell = E°(cathode) − E°(anode); ΔG° = −nFE°
Electrolyser voltage = 1.23 V + η_anode + |η_cathode|

Key terms, if this is new to you

Activation barrier, ΔG‡: The free energy that must be supplied to reach the transition state. The overpotential lowers it in one direction and raises it in the other.
Exchange current density, i₀: The current density in each direction at equilibrium. A large i₀ means fast electron transfer: it plays the part of the Arrhenius factor A.
Transfer coefficient, α: Also called the symmetry factor: the position of the transition state between reactants and products. In this course α = 0.5.
Anodic and cathodic current: An anodic current (oxidation) is positive; a cathodic current (reduction) is negative.
Catalytic electrode: An inert electrode, such as platinum, that transfers electrons to or from species in solution without being consumed.
Tafel plot and Tafel slope: A plot of log |i| against η. Far from equilibrium it is a straight line with slope 1/b; extrapolated to η = 0 it gives log i₀.
HER and OER: The hydrogen evolution reaction at the cathode and the oxygen evolution reaction at the anode of a water electrolyser.

About the values

F = 96,485 C mol⁻¹; R = 8.314 J K⁻¹ mol⁻¹; T = 298.15 K, so RT/F = 0.02569 V.
α = 0.5 throughout, and n is the number of electrons in the half-reaction, as in the lecture examples.
Exchange current densities are those in the Lecture 5 and 6 tables. Molar masses: Ag 107.87, CH₃OH 32.04 g mol⁻¹.
By convention an anodic (oxidation) current is positive and a cathodic (reduction) current negative.
Small rounding differences are accepted, so you do not need to match every decimal place.
Your answers are saved in this browser, so you can leave and come back later.