In a cyclic voltammetry experiment, you double the scan rate and observe that the peak current increases by a factor of approximately 1.41 (√2). What does this tell you about the process?
AThe reaction is irreversible because peak current depends on scan rate
BThe electrode is becoming saturated at faster scan rates
CThe process is diffusion-controlled, since peak current scales with the square root of scan rate (Randles–Ševčík behavior)
DThe analyte concentration decreased during the experiment
For a diffusion-controlled process, the Randles–Ševčík equation gives ip ∝ √ν. Doubling the scan rate increases √ν by √2 ≈ 1.41, which is exactly what was observed. This behavior arises because at faster scan rates the diffusion layer is thinner — reactant has less time to be depleted near the surface — so more current flows. Scan-rate dependence itself does not indicate irreversibility; irreversibility is diagnosed by peak separation exceeding 59/n mV.
Question 2 Multiple Choice
What is the primary purpose of the deposition step in anodic stripping voltammetry (ASV)?
ATo clean and activate the electrode surface before analysis
BTo measure background capacitive current so it can be subtracted during the stripping scan
CTo electroplate trace metal ions from a large solution volume onto the electrode, concentrating the analyte by orders of magnitude
DTo determine which metals are present by observing their reduction potentials during deposition
The deposition step is the source of ASV's extraordinary sensitivity. By holding the electrode at a very negative potential for several minutes, metal ions from a large volume of solution are electroplated onto a tiny electrode surface — concentrating parts-per-trillion analyte into a detectable quantity. When the potential is swept positively, each metal strips off at its characteristic potential, releasing minutes of accumulation in seconds. The signal amplification from this preconcentration step is what enables ppt detection limits impossible with direct CV.
Question 3 True / False
A cyclic voltammogram shows a perfectly symmetrical pair of peaks — the anodic and cathodic peaks are mirror images of each other. This symmetry proves that the redox reaction is electrochemically reversible.
TTrue
FFalse
Answer: False
Symmetry of peak shape is a necessary but not sufficient condition for reversibility. True electrochemical reversibility also requires that the peak separation (Ep,a − Ep,c) equals exactly 59/n mV at all scan rates. If the separation is larger than 59/n mV or increases with scan rate, electron transfer kinetics are sluggish (quasi-reversible or irreversible) even if the peaks look symmetric. Both criteria must be satisfied: correct peak separation AND scan-rate independence of that separation.
Question 4 True / False
Differential pulse voltammetry achieves lower detection limits than simple cyclic voltammetry primarily because it reduces the capacitive (charging) current relative to the faradaic (reaction) current.
TTrue
FFalse
Answer: True
The fundamental noise floor in electroanalytical measurements is set by capacitive current — the current that flows when the electrode double layer charges and discharges as the potential is scanned. This current is not related to the analyte and cannot be reduced by increasing sample concentration. Differential pulse methods sample current at specific points in a pulse cycle (end of pulse, when capacitive current has decayed exponentially but faradaic current remains) and subtract pairs of measurements to cancel the capacitive background. The result is a dramatic improvement in signal-to-noise for the analyte peak.
Question 5 Short Answer
In cyclic voltammetry, why does the current peak and then decline as the potential sweep continues past the reduction potential? What physical process drives this peak shape?
Think about your answer, then reveal below.
Model answer: As the potential reaches the reduction potential, analyte molecules at the electrode surface are reduced rapidly, generating a large current. But those surface molecules are quickly consumed and must be replenished by diffusion from the bulk solution. Diffusion is slow compared to the electrochemical reaction, so a depletion zone (diffusion layer) builds up near the electrode surface. As the diffusion layer thickens, fresh analyte must travel ever-increasing distances to reach the electrode, and the rate of arrival (and thus the current) falls. The peak marks the point where the reduction rate transitions from being limited by electrode kinetics to being limited by diffusion transport.
The peak shape is entirely a consequence of mass transport. Students often assume more negative potential should drive more current indefinitely, but the current is bounded by how fast analyte can arrive at the electrode surface. This is why peak current depends on √ν (diffusion rate scales with scan rate) and why peak current is proportional to concentration (more analyte diffuses to the surface per unit time).