Question

In experiment 1, a cyclic voltammogram at an HMDE was obtained from a $0.159-\mathrm{mM}$ solution of data-custom-editor="chemistry" ${\mathrm{Pb}}^{2+}$ at a scan rate of $2.9\mathrm{V}/\mathrm{s}$. In experiment 2, a second CV is to be obtained from a $4.26-\mathrm{mM}$ solution of data-custom-editor="chemistry" ${\mathrm{Cd}}^{2+}$ using the same HMDE. What must the scan rate be in experiment 2 to record the same peak current in both experiments if the diffusion coefficients of data-custom-editor="chemistry" ${\mathrm{Cd}}^{2+}\mathrm{and}\mathrm{}{\mathrm{Pb}}^{2+}$ are $0.72\times {10}^{-5}{\mathrm{cm}}^2{\mathrm{s}}^{-1}\mathrm{and}0.98{\mathrm{cm}}^2{\mathrm{s}}^{-1}$, respectively. Assume that the reductions of both cations are reversible at the HMDE.

Short answer

The scan rate for experiment 2 is $7.5{\mathrm{mVs}}^{-1}$.

Step-by-step solution

Step 1. Given information

$\text{Volume of}P{b}^{2+}=0.159\text{-m}\mathrm{M}$

Scan rate$=2.9V/s$

Compound 1 =HMDE

Volume of $C{d}^{2+}=4.26-mM$

Diffusion coefficients of $C{d}^{2+}$$=0.72\times {10}^{-5}{\mathrm{cm}}^2{\mathrm{s}}^{-1}$

Diffusion coefficients of $P{b}^{2+}$$\mathrm{=}0.98{\mathrm{cm}}^2{\mathrm{s}}^{-1}$

Step 2. Randles–Sevik equation 

In cyclic voltammetry, the Randles–Sevik equation describes the relationship between peak current and scan rate. The peak current depends on the concentration of the ions, the diffusion coefficient and the scan rate. The Randles-Sevik equation is as follows:

$i_p=2.686\times {10}^5{n}^{1/2}Ac{D}^{1/2}{v}^{1/2}$

Here,

$\begin{array}{c}{i}_p=\mathrm{peak}\mathrm{current}\left(\mathrm{A}\right)\\ A=t\mathrm{he}\mathrm{electrode}\mathrm{area}\left({\mathrm{cm}}^2\right)\\ D=\mathrm{diffusion}\mathrm{coefficient}\left({\mathrm{cm}}^2/\mathrm{s}\right)\\ c=\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{ion}\mathrm{and}\mathrm{measure}\mathrm{in}\mathrm{moleper}\mathrm{cubic}\mathrm{centimeter}\\ v=\mathrm{the}\mathrm{scan}\mathrm{rate}\mathrm{of}\mathrm{ions}\mathrm{and}\mathrm{measure}\mathrm{in}\mathrm{V}/\mathrm{s}\end{array}$

Step 3. Calculation

The peak current ${\left(i_p\right)}_1$for Experiment 1 can be written as:

${\left(i_p\right)}_1=2.686\times {10}^5{n}^{3/2}A{c}_1{D}_1^{1/2}{v}_1^{1/2}$

Put the values in above equation

localid="1645608737005" ${\left(i_p\right)}_1=2.686\times {10}^5{n}^{3/2}A(0.159\mathrm{mM}){\left(0.98\mathrm{cm}{\mathrm{s}}^{-1}\right)}^{1/2}{\left(2.9\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}$

The peak current ${\left(i_p\right)}_2$for Experiment 2 can be written as:

${\left(i_{\mathrm{P}}\right)}_2=2.686\times {10}^5{n}^{3/2}A{c}_2{D}_2^{1/2}{v}_2^{1/2}$

Put the values in above equation

${\left(i_D\right)}_2=2.686\times {10}^5{n}^{3/2}A(4.26\mathrm{mM}){\left(0.72\times {10}^{-5}{\mathrm{cm}}^2{\mathrm{s}}^{-1}\right)}^{1/2}{\left(v_2\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}..............\left(2\right)$

By dividing equation 2 by equation 1, one gets

$\begin{array}{c}\frac{{\left(i_{\mathrm{p}}\right)}_2}{{\left(i_{\mathrm{p}}\right)}_1}=\frac{2.686\times {10}^5{n}^{3/2}A(4.26\mathrm{mM}){\left(0.72\times {10}^{-5}{\mathrm{cm}}^2{\mathrm{s}}^{-1}\right)}^{1/2}{\left(v_2\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}}{2.686\times {10}^5{n}^{3/2}A(0.159\mathrm{mM}){\left(0.98\mathrm{cm}{\mathrm{s}}^{-1}\right)}^{1/2}{\left(2.9\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}}\\ =19.684\left(\frac{{\left(v_2\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}}{{\left(2.9\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}}\right)\end{array}$

Step 4. Scan rate

But it is known that the peak current of Experiment 1 is equal to the peak current of Experiment 2.

$\frac{{\left(i_p\right)}_2}{{\left(i_p\right)}_1}=1$

Now put the values

$1=19.684\left(\frac{{\left(v_2\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}}{{\left(2.9\mathrm{V}{\mathrm{s}}^{-1}\right)}^{1/2}}\right)$

Take squares on both sides:

$\mathrm{I}=(19.684{)}^2\frac{\left(v_2\mathrm{V}{\mathrm{s}}^{-1}\right)}{\left(2.9\mathrm{V}{\mathrm{s}}^{-1}\right)}$

Simplify the above equation

$\begin{array}{c}{v}_2=\frac{\left(2.9\mathrm{V}{\mathrm{s}}^{-1}\right)}{19.684\times 19.684}\\ =2.581\times {10}^{-3}\times 2.9\mathrm{V}{\mathrm{s}}^{-1}\\ =7.5\times {10}^{-3}\mathrm{V}{\mathrm{s}}^{-1}\\ =7.5{\mathrm{mVs}}^{-1}\end{array}$