Consider the following system: A + e- ⇌ B E0' = -0.5 V vs. SCE B -> C C + e- ⇌ D E0' = -1.0 V vs. SCE The half-life of B is 100 ms. Both charge-transfer reactions have large va... Consider the following system: A + e- ⇌ B E0' = -0.5 V vs. SCE B -> C C + e- ⇌ D E0' = -1.0 V vs. SCE The half-life of B is 100 ms. Both charge-transfer reactions have large values of k0. Draw the expected cyclic voltammograms for scans beginning at 0.0 V vs. SCE and reversing at -1.2 V. Show curves for rates of 50 mV/s, 1 V/s, and 20 V/s. Read more

Steps to Solve

1. Understanding the Electrochemical System

The system has two reversible electrochemical reactions (A/B and C/D) and one irreversible chemical reaction (B to C). The standard reduction potentials ($E^{0'}$) for A/B and C/D are -0.5 V and -1.0 V vs. SCE, respectively. The forward rate constants ($k^{0}$) for both redox couples are large, implying rapid electron transfer. The half-life of species B converting to C is 100 ms, meaning the kinetics of this conversion will influence the cyclic voltammograms, particularly at slower scan rates.

2. Analyzing the Scan Rates

We need to consider three different scan rates: 50 mV/s, 1 V/s, and 20 V/s. Each scan starts at 0.0 V and reverses at -1.2 V. The scan rate will dictate how much of species B converts to C during the time of the experiment

3. Cyclic Voltammetry at 50 mV/s

At a slow scan rate (50 mV/s), the experiment timescale is long. This gives species B ample time to convert to C via the chemical reaction $B \rightarrow C$. As the potential is scanned negatively from 0.0 V, a reduction peak for A to B will be observed near -0.5 V. However, due to the slow scan rate, most of the generated B will convert to C before the reverse scan. During the reverse scan (positive direction), the oxidation peak of B back to A will be significantly diminished because B has largely been consumed. Instead, because B converts to C, and C is reduced to D, a prominent reduction of C to D will occur at $E^{0'}$ = -1.0 V. The oxidation peak of D will be observed on the reverse scan. So we can expect to see peaks around -0.5 V and -1.0 V. The peak at -0.5 V will have a small return peak, and should diminish over time.

4. Cyclic Voltammetry at 1 V/s

At an intermediate scan rate (1 V/s), the timescale is shorter than the 50 mV/s case, but still allows a fraction of B to convert to C. As the potential is scanned negatively, the reduction peak for A to B emerges at -0.5 V. Now, a smaller but noticeable amount of B has time to convert to C. Thus, on the reverse scan, the oxidation peak for B back to A will be more significant compared to the 50 mV/s scan but still smaller than we'd expect for a fully reversible couple. We also observe the reduction of C to D near -1.0 V, though it's less pronounced than in the 50 mV/s case because less B has had time to convert to C. Because the reaction is reversible, we will also see the oxidation peak of D.

5. Cyclic Voltammetry at 20 V/s

At a fast scan rate (20 V/s), the experiment occurs quickly, minimizing the amount of B that can convert to C. The reduction peak for A to B will appear at -0.5 V. Because the scan is fast, only a minimal amount of B converts to C meaning on the reverse scan, a prominent oxidation peak for B back to A will be observed. The reduction and oxidation peaks for the A/B couple will be nearly symmetrical, indicating a reversible process. The effect on the C/D couple will be small.

6. Sketching the Voltammograms

Based on the above analysis, we can sketch the cyclic voltammograms. The x-axis is the potential (V vs. SCE), and the y-axis is the current. The positioning and size of the peaks for the electrochemical reactions A + e- ⇌ B and C + e- ⇌ D need to be illustrated according to the specified rates.