Cyclic Voltammetry Basics

Introduction

Cyclic voltammetry (CV) is often the first experiment an electrochemist runs on a new compound. In a few minutes it reveals the potentials at which a species is oxidised or reduced, whether the electron transfer is fast or slow, and whether the product is stable. Its characteristic "duck-shaped" trace combines everything learned about electrode kinetics, diffusion layers and limiting currents into a single picture.

Core explanation

The experiment. A three-electrode cell is used: a working electrode where the reaction of interest occurs, a reference electrode that fixes the potential scale and a counter electrode that completes the circuit. A potentiostat sweeps the working-electrode potential linearly from a starting value to a switching potential and back again, forming a triangular waveform. The scan rate v typically ranges from 10 mV s⁻¹ to several V s⁻¹. Current is plotted against potential. The solution is unstirred and contains excess supporting electrolyte, so the analyte moves by diffusion.

Why a peak, not a plateau? Consider a solution of O swept in the cathodic direction. As the potential passes the formal potential, reduction accelerates and current rises. Soon the surface concentration of O approaches zero. Because the solution is still, the diffusion layer then grows outwards with time, the concentration gradient becomes shallower and the current falls. The competition between rising rate and a thickening depleted zone produces a cathodic peak , followed by a decaying tail.

The return sweep. After the switching potential, the sweep reverses. R accumulated near the surface is now re-oxidised as the potential passes back through E°′, giving an anodic peak of opposite sign. The two peaks together form the typical CV shape.

Diagnosing a reversible couple. For a fast, chemically stable couple (Nernstian behaviour) at 25 °C: - peak separation ΔE p = E pa − E pc ≈ 59/n mV (about 57 mV in practice), independent of scan rate; - peak current ratio i pa/i pc ≈ 1; - E₁/₂ = (E pa + E pc)/2 ≈ E°′; - peak current proportional to √v.

Quasi-reversible and irreversible couples. If electron transfer is slow compared with the timescale of the sweep, extra overpotential is needed, and peaks move apart as scan rate increases. When ΔE p grows well beyond 59/n mV, the couple is quasi-reversible; if the return peak barely appears, it is irreversible.

Chemical follow-up reactions. If the product R decomposes before the return scan, less R is available to re-oxidise, so i pa/i pc < 1. Faster scans give R less time to decompose and restore the return peak, allowing the decomposition rate to be estimated.

The Randles–Ševčík equation. For a reversible couple at 25 °C:

i p = (2.69 × 10⁵) n^(3/2) A D^(1/2) c v^(1/2)

with i p in amperes, A in cm², D in cm² s⁻¹, c in mol cm⁻³ and v in V s⁻¹. The √v dependence confirms diffusion control; a current proportional to v instead suggests a species adsorbed on the surface.

Formulae

Reversible peak separation: ΔE p ≈ 2.22RT/nF ≈ 57/n mV at 25 °C (often quoted as 59/n mV). E₁/₂ = (E pa + E pc)/2. Randles–Ševčík: i p = 2.69 × 10⁵ n^(3/2) A D^(1/2) c v^(1/2). Capacitive background current: i c = C dl A v.

Step-by-step reasoning

To interpret a cyclic voltammogram:

1. Locate anodic and cathodic peak potentials and currents, measured from the extrapolated baseline. 2. Calculate ΔE p and compare it with 59/n mV. 3. Calculate the peak current ratio to test for chemical stability of the product. 4. Repeat at several scan rates; plot i p against √v. 5. Check whether ΔE p changes with v to decide between reversible and quasi-reversible behaviour.

Visual explanation

Imagine a current–potential graph. The trace starts near zero on the right, sweeps left and swells into a sharp downward peak (cathodic, using the IUPAC convention of plotting reduction current as negative), then tails off. On the return sweep to the right it rises into an upward peak slightly displaced to more positive potential. The loop resembles a stretched, tilted figure with two humps offset by about 60 mV.

Real-world analogy

Imagine harvesting apples by walking along an orchard row. At first you pick faster and faster as you reach the heavy trees, but soon you strip the nearby branches and must reach further, so your picking rate falls. Coming back along the same row, you encounter the baskets you left earlier and can process them again, producing a second burst of activity.

Real-world example

Ferrocene is widely used as an internal standard in non-aqueous electrochemistry because its one-electron oxidation to the ferrocenium ion is fast and chemically reversible. Its CV shows peaks roughly 60 to 70 mV apart with a current ratio near one, giving a reliable reference point for comparing the redox potentials of new molecules, such as battery electrolytes or organic semiconductors.

Why?

Why does peak current depend on √v rather than v? Faster sweeps reach the peak potential sooner, when the diffusion layer is thinner and the gradient steeper. Diffusion-layer thickness scales as √(Dt), and the time spent near E°′ scales as 1/v, so the gradient, and hence the peak current, scales as √v.

Common misconception

"The cathodic peak occurs exactly at the formal potential." For a reversible couple the cathodic peak lies about 29/n mV beyond E°′ and the anodic peak about 29/n mV on the other side. The formal potential lies at the midpoint, E₁/₂, not at either peak.

Worked example

Question: A one-electron couple shows E pa = +0.470 V and E pc = +0.400 V at 100 mV s⁻¹; peak separation stays constant between 50 and 500 mV s⁻¹ and i pa/i pc = 1.0. Estimate E°′ and comment on reversibility.

Reasoning: E₁/₂ = (0.470 + 0.400)/2 = 0.435 V. ΔE p = 70 mV, close to 59 mV, and independent of scan rate; the current ratio is 1.

Answer: E°′ ≈ +0.435 V; the couple is essentially reversible with a stable product (the small excess separation is probably due to uncompensated resistance).

Quick check

1. What does a peak current ratio i pa/i pc much less than one suggest about a reduction product? Answer: The product is being consumed by a chemical reaction before it can be re-oxidised on the return sweep.

Exam focus

Know the three-electrode set-up, why peaks form, the reversibility criteria (ΔE p ≈ 59/n mV, ratio ≈ 1, i p ∝ √v) and how to calculate E₁/₂. Be ready to interpret a CV that changes with scan rate.

Advanced insight

The Nicholson method uses the growth of ΔE p with scan rate to extract the standard rate constant k° of a quasi-reversible couple. It works through a dimensionless parameter comparing k° with the rate of diffusion on the timescale of the scan. Uncompensated solution resistance also increases ΔE p, so careful work applies IR compensation before assigning slow kinetics.

Summary

Cyclic voltammetry sweeps potential in a triangle and records current. Peaks form because the reaction accelerates while the depleted diffusion layer grows. Reversible couples show ΔE p ≈ 59/n mV, equal peak currents and i p ∝ √v, with E°′ at the midpoint. Wider or scan-dependent separations signal slow electron transfer, and a missing return peak signals follow-up chemistry.

Practice questions

1. Why is the solution not stirred during cyclic voltammetry? Answer: So that transport occurs by diffusion alone, giving well-defined peak shapes that theory can interpret. 2. Predict ΔE p at 25 °C for a reversible two-electron couple. Answer: About 59/2 ≈ 30 mV. 3. If the scan rate is increased from 25 to 100 mV s⁻¹, by what factor does a diffusion-controlled peak current increase? Answer: By √4 = 2. 4. A redox-active film attached to the electrode gives a peak current proportional to scan rate. Explain. Answer: No diffusion is needed; the fixed amount of surface species reacts faster at higher scan rate, so current is proportional to v. 5. What role does the reference electrode play in the experiment? Answer: It provides a stable potential against which the working-electrode potential is controlled and measured, while carrying negligible current.