Question

A new method for determining ultrasmall ($nL$) volumes by anodic stripping voltammetry has been proposed (W. R. Vandaveer and I. Fritsch, Anal. Chem.$2002,74,3575,DOI:10.1021/ac011036s$). In this method, a metal is exhaustively deposited from the small volume to be measured onto an electrode, from which it is later stripped. The solution volume $Vs$is related to the total charge $Q$required to strip the metal by

$V_{\mathrm{s}}=\frac{Q}{nFC}$

where $n$is the number of moles of electrons per mole of analyte, $F$is the faraday, and $C$is the molar concentration of the metal ion before electrolysis.

($a$) Beginning with Faraday’s law (see Equation $22-8$), derive the above equation for $Vs$.

($b$) In one experiment, the metal deposited was $Ag\left(s\right)$from a solution that was $8.00mM$in $AgNO3$. The solution was electrolyzed for $30$min at a potential of $-0.700V$versus a gold top layer as a pseudo reference. A tubular nano band electrode was used. The silver was then anodically stripped off the electrode using a linear-sweep rate of $0.10V/s$. The following table represents idealized anodic stripping results. By integration, determine the total charge required to strip the silver from the tubular electrode. You can do a manual Simpson’s rule integration or do the integration with Excel From the charge, determine the volume of the solution from which the silver was deposited.

($c$) Suggest experiments to show whether all the $A{g}^{+}$was reduced to $Ag\left(s\right)$in the deposition step.

($d$) Would it matter if the droplet were not a hemisphere? Why or why not?

($e$) Describe an alternative method against which you might test the proposed method.

Short answer

($a$) The derived equation for $V_s$is equal to $V_s=\frac{n_A}{C}$.

($b$) The volume of the solution is $0.326nL$.

($c$) Use cyclic, pulse, or linear scan voltammetry is another way to reduce from $A{g}^{+}$to $Ag\left(s\right)$.

($d$) Reason for the given is not a hemisphere because the only method to visually measure the size of a droplet was previously, and the hemispherical or spherical shape makes that measurement easier and more accurate.

($e$) Another way to figure out the volume of a solution is to look at the shape of the droplet.

Step-by-step solution

Derive the equation Vs (part a).

Consider the equation below:

${\mathrm{v}}_{\mathrm{s}}=\frac{Q}{\mathrm{nFC}}$

where $Q$denotes the charge, $n$denotes the number of moles of electrons, $F$denotes the faraday, and $C$is the metal ion's molar concentration.

Using the following equation, derive the equation for$V_s$.

${\mathrm{n}}_{\mathrm{A}}=\frac{Q}{\mathrm{nF}}$

Rearrange both equations so that they equal$Q$.

$Q=V_s\mathrm{nFC}$

$Q={\mathrm{n}}_A\mathrm{nF}$

Rearrange the equations to equal$V_s$and make them equal to each other.

${\mathrm{n}}_{\mathrm{A}}\mathrm{nF}=V_s\mathrm{nFC}$

$$\begin{gathered} V_s=\frac{{\mathrm{n}}_{\mathrm{A}}\mathrm{nF}}{\mathrm{nFC}} \\ {V}_s=\frac{{\mathrm{n}}_{\mathrm{A}}}{C} \end{gathered}$$

As a result, the $V_s=\frac{n_A}{C}$equation is derived.

Determine the Volume of the solution (part b)

$-0.140nA$Use Excel to do a Simpson's rule integration, as shown in the equation below.

localid="1650442636266" $I_t=\Delta x\left[\frac{y_0+4y_1+2y_2+4y_3+\dots +4y_{m-1}+y_n}{3}\right]$

where $I$denotes total current, $x$denotes interval width, and $n$denotes the total number of values

Some data points were excluded since $x$must be the same for every group of data points.

The charge can then be computed using the equation below:

$Q=It$

where $I$is the current and $t$is the time.

Use to convert the current from $nA$to$A$.

$I=(-0.140\mathrm{nA})\times \left(\frac{{10}^{-4}\mathrm{A}}{1\mathrm{nA}}\right)$

$=-1.40\times {10}^{-10}\mathrm{A}$

To convert minutes to seconds, use the calculator below.$30$minutes is recommended.

$t=(30\min )\times \left(\frac{60\sec }{1\min }\right)$

$=1800s$

Make a charge calculation. For $I$, use $-1.40\times {10}^{-10}nA$, and for $t$, use the$1800s$.

$Q=\left(-1.40\times {10}^{-10}\mathrm{A}\right)\times (1800\mathrm{s})$

$=-2.52\times {10}^{-7}C$

Now, determine the volume of solution. first convert concentration from $mm$to$M$.

$M=(8.00\mathrm{mM})\times \left(\frac{1\mathrm{M}}{1000\mathrm{mM}}\right)$

$=0.00800M$

Determine the volume of the solution.

$V_s=\frac{\mid -2.52\times {10}^{-7}\mathrm{C}|}{(1\mathrm{mole}{\mathrm{e}}^{-}/\mathrm{mol})\times (96485\mathrm{C}/\mathrm{mole}{e}^{-})\times (0.00800\mathrm{M})}$

$=\frac{2.52\times {10}^{-7}\mathrm{C}}{771.88\mathrm{C}\times \mathrm{M}}$

$=3.26\times {10}^{-10}L$

Convert a volume in $L$to a volume in$nL$.

$V=\left(3.26\times {10}^{-10}\mathrm{L}\right)\times \left(\frac{1\mathrm{nL}}{{10}^{-9}\mathrm{L}}\right)$

$=0.326nL$

Find the experiment to show Ag+reduced to Ag(s) (part c).

Comparing data from a cyclic, pulse, or linear scan voltammetry with data from a cyclic, pulse, or linear scan voltammetry is another way to see if $A{g}^{+}$ totally decreased $Ag\left(s\right)$. Any of these voltammetry techniques can be employed, depending on what is available at the moment.

Reason for it is not a hemisphere (part d).

It wouldn't matter if the droplet was hemispherical if the necessary equipment to measure the size of the droplet was available. The only method to visually measure the size of a droplet was previously, and the hemispherical or spherical shape makes that measurement easier and more accurate.

Alternation method to test the proportion method (part e).

Another way to figure out the volume of a solution is to look at the shape of the droplet. This would include measuring the diameter of the droplet and then calculating the volume of a sphere or a hemisphere using the equations shown below.

$V_{yp}=\frac{4}{3}{\pi }^3$

$V_{hpp}=\frac{V_{yp}}{2}$