Login Registrieren

Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 25: Q 25.7 (page 686)

What is the purpose of the electrodeposition step in stripping analysis?

Short Answer

Expert verified

The accumulation of the analyte on the surface of the working electrode is the main purpose of the electro deposition step in stripping analysis.

Step by step solution

01

Given information

The purpose of electrodeposition step in stripping analysis is to be stated.

02

Explanation

All the analyte those have been deposited may be reduced or oxidized and hence responsible for the production of the large current. The analyte on the surface of the working electrode is accumulated by the electro deposition step in stripping analysis. This accumulation of the analyte on the surface of the working electrode is the main purpose of the electro deposition step in stripping analysis.

The stripping analysis required an electrode known as working electrode. This working electrode is made up of various materials such as mercury, platinum, carbon, silver and gold. During the electro decomposition step only a small amount of the analyte gets concentrated. The measured time period is allowed for the decomposition due to which it is applicable only for the small fraction of the analyte.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Distinguish between (a) voltammetry and amperometry, (b) linear-scan voltammetry and pulse voltammetry, (c) differential-pulse voltammetry and square-wave voltammetry, (d) an RDE and a ringdisk electrode, (e) faradaic impedance and double-layer capacitance, (f) a limiting current and a diffusion current, (g) laminar flow and turbulent flow, (h) the standard electrode potential and the half-wave potential for a reversible reaction at a working electrode, (i) normal stripping methods and adsorptive stripping methods.

Why is it necessary to buffer solutions in organic voltammetry?

Sulfate ion can be determined by an amperometric titration procedure using ${Pb}^{2 +}$ as the titrant. If the potential of a rotating mercury film electrode is adjusted to 21.00 V versus SCE, the current can be used to monitor the ${Pb}^{2 +}$ concentration during the titration. In a calibration experiment, the limiting current, after correction for background and residual currents, was found to be related to the ${Pb}^{2 +}$ concentration by $i_{1} = 10 c_{{Pb}^{2 +}}$ , where $i_{1}$ is the limiting current in mA and $c_{{Pb}^{2 +}}$ is the ${Pb}^{2 +}$ concentration in mM.

The titration reaction is

role="math" localid="1653990390811" ${SO}_{4}^{2 -} + {Pb}^{2 +} ⇌ {PbSO}_{4} (s) K_{sp} = 1.6 \times 10^{- 8}$

If $25.00 mL of 0.020 M {Na}_{2} {SO}_{4}$ is titrated with $0.030 M Pb {({NO}_{3})}_{2}$ , develop the titration curve in spreadsheet format and plot the limiting current versus the volume of titrant.

A new method for determining ultrasmall ( $n L$ ) volumes by anodic stripping voltammetry has been proposed (W. R. Vandaveer and I. Fritsch, Anal. Chem. $2002, 74, 3575, D O I : 10.1021 / a c 011036 s$ ). 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 $V s$ is related to the total charge $Q$ required to strip the metal by

$V_{s} = \frac{Q}{n F C}$

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 $V s$ .

( $b$ ) In one experiment, the metal deposited was $A g (s)$ from a solution that was $8.00 m M$ in $A g N O 3$ . The solution was electrolyzed for $30$ min at a potential of $- 0.700 V$ 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 $0.10 V / s$ . 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 $A g (s)$ 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.

Quinone undergoes a reversible reduction at a voltammetric working electrode. The reaction is

(a) Assume that the diffusion coefficients for quinone and hydroquinone are approximately the same and calculate the approximate half-wave potential (versus SCE) for the reduction of hydroquinone at an RDE from a solution buffered to a $p H$ of $8.0$ .

(b) Repeat the calculation in (a) for a solution buffered to a $p H$ of $5.0$ .

See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free