An electric current is a flow of charge, so the current at an electrode measures the rate of the reaction taking place there. At the equilibrium potential, oxidation and reduction continue at equal rates and no net current flows; an overpotential makes one direction faster than the other. Work through the three stages in order.
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Predict, then see: starting from 0.50 A for 1 hour on zinc,
Or explore freely:
0.50 A
60 min
Predict, then see:
Or explore freely:
0.000 V
At η = 0 both arrows are the same size: the reaction has not stopped, it runs equally fast both ways.
Compare η = −0.05 V with η = −0.10 V: doubling η multiplies the current by about seven.
Remember
Current is the rate of flow of charge: i = q/t, so q = it. 1 A = 1 C s−1.
Moles of electrons = q/F, with F = 96,485 C mol−1. Moles of metal = q/(zF), where z is the number of electrons per metal ion (2 for Zn2+); mass = moles × M.
At E = Eeq, oxidation and reduction continue at equal and opposite rates: dynamic equilibrium, no net current.
Overpotential: η = E − Eeq. η < 0 makes reduction faster (net cathodic current); η > 0 makes oxidation faster (net anodic current).
Both rates change exponentially with η, in opposite directions, so the net current grows much faster than η itself.
In stage 2 the partial current densities are drawn as ia = i0 e+η/0.0257 V and ic = −i0 e−η/0.0257 V, with i0 = 1 mA cm−2: the symmetric case of the Butler–Volmer equation for a two-electron couple, which Lecture 5 derives. i0, the current in each direction at equilibrium, is the exchange current density.