/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Explain why each of the followin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 16

Explain why each of the following decreases the analysis time in controlled- potential coulometry: a larger surface area for the working electrode; a smaller volume of solution; and a faster stirring rate.

Short Answer

Expert verified
A larger electrode surface speeds reaction kinetics, smaller volume enhances reactant concentration, and faster stirring quickens mass transport.

Step by step solution

01

Understanding Controlled-Potential Coulometry

Controlled-potential coulometry involves maintaining a constant potential at the working electrode to drive the electrochemical reaction. The time taken for complete analysis can be influenced by various factors, including electrode surface area, solution volume, and stirring rate.
02

Role of Electrode Surface Area

A larger surface area for the working electrode increases the number of sites available for the electrochemical reaction to occur. This enhances the rate of electron transfer and accelerates the reaction kinetics, thereby reducing the total analysis time.
03

Impact of Solution Volume

A smaller solution volume means that there are fewer molecules to react within the solution. This leads to a higher concentration of the reactant at the electrode surface, improving mass transport and allowing the reaction to proceed more quickly, thus decreasing the analysis time.
04

Effect of Stirring Rate

Increasing the stirring rate helps in homogenizing the solution, reducing the thickness of the diffusion layer at the electrode surface. This improvement in mass transport ensures that reactants are rapidly brought to the electrode surface, facilitating faster reaction completion and thus shorter analysis time.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrode Surface Area
In controlled-potential coulometry, the electrode surface area plays a crucial role in determining how fast the electrochemical reaction can proceed. A larger surface area for the working electrode means more sites are available for the reaction to occur. Imagine trying to finish a painting: the bigger the canvas, the more brushes can be used at once.

This increases the number of electron transfers happening simultaneously, speeding up the reaction kinetics.
  • More active sites mean quicker completion of the analysis.
  • Larger surfaces allow for increased accessibility for reactants.
Consequently, having a larger electrode surface area significantly decreases the time required for the complete analysis, as reactions can occur more rapidly across the broader expanse of the electrode.
Solution Volume
The volume of the solution is another factor that can affect the time required for an analysis in controlled-potential coulometry. When the solution volume is smaller, there are fewer molecules needing to react, making it easier for the reaction to reach completion quickly.

Picture a small pool versus an ocean; the small pool will heat up faster than the ocean under the sun.
  • Smaller volume results in higher reactant concentration near the electrode.
  • This improves mass transport as molecules move more easily to the electrode.
Thus, reducing the solution volume can lead to shorter analysis times as the reactants have a shorter distance to travel and can reach the electrode surface more swiftly.
Stirring Rate
The rate at which the solution is stirred is a key factor in controlled-potential coulometry because it directly affects how well-mixed the solution is. Increased stirring rates help in breaking down the diffusion layer, a boundary where molecule movement is slower.

Think of stirring cream into coffee; the faster you stir, the quicker it blends uniformly.
  • Enhanced mixing transports reactants swiftly to the electrode surface.
  • Decreases the build-up of reactant concentration in areas away from the electrode.
As a result, faster stirring speeds minimize the time it takes for the reaction to conclude by ensuring reactants are always replenished at the electrode surface, leading to quicker completion of the analysis.

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

Mifflin and associates described a membrane electrode for the quantitative analysis of penicillin in which the enzyme penicillinase is immobilized in a polyacrylamide gel coated on the glass membrane of a \(\mathrm{pH}\) electrode. \({ }^{22}\) The following data were collected using a set of penicillin standards. \begin{tabular}{cc} [penicillin] (M) & potential (mV) \\ \hline \(1.0 \times 10^{-2}\) & 220 \\ \(2.0 \times 10^{-3}\) & 204 \\ \(1.0 \times 10^{-3}\) & 190 \\ \(2.0 \times 10^{-4}\) & 153 \\ \(1.0 \times 10^{-4}\) & 135 \\ \(1.0 \times 10^{-5}\) & 96 \\ \(1.0 \times 10^{-6}\) & 80 \end{tabular} (a) Over what range of concentrations is there a linear response? (b) What is the calibration curve's equation for this concentration range? (c) What is the concentration of penicillin in a sample that yields a potential of \(142 \mathrm{mV?}\)

Sittampalam and Wilson described the preparation and use of an amperometric sensor for glucose. \(^{27}\) The sensor is calibrated by measuring the steady- state current when it is immersed in standard solutions of glucose. A typical set of calibration data is shown here. \begin{tabular}{cc} [glucose] \((\mathrm{mg} / 100 \mathrm{~mL})\) & current (arb. units) \\ \hline 2.0 & 17.2 \\ 4.0 & 32.9 \\ 6.0 & 52.1 \\ 8.0 & 68.0 \\ 10.0 & 85.8 \end{tabular} A \(2.00-\mathrm{mL}\) sample is diluted to \(10 \mathrm{~mL}\) in a volumetric flask and a steady-state current of 23.6 (arbitrary units) is measured. What is the concentration of glucose in the sample in \(\mathrm{mg} / 100 \mathrm{~mL}\) ?

The standard-state reduction potential for \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) is \(+0.342 \mathrm{~V}\) versus the SHE. Given that \(\mathrm{Cu}^{2+}\) forms a very stable complex with the ligand EDTA, do you expect that the standard- state reduction potential for \(\mathrm{Cu}(\mathrm{EDTA})^{2-}\) is greater than \(+0.342 \mathrm{~V}\), less than \(+0.342 \mathrm{~V}\), or equal to \(+0.342 \mathrm{~V}\) ? Explain your reasoning.

An ion-selective electrode can be placed in a flow cell into which we inject samples or standards. As the analyte passes through the cell, a potential spike is recorded instead of a steady-state potential. The concentration of \(\mathrm{K}^{+}\) in serum has been determined in this fashion using standards prepared in a matrix of \(0.014 \mathrm{M} \mathrm{NaCl}^{23}\) \begin{tabular}{cccc} {\(\left[\mathrm{K}^{+}\right](\mathrm{mM})\)} & \(E\) (arb. units) & {\(\left[\mathrm{K}^{+}\right](\mathrm{mM})\)} & \(E(\) arb. units \()\) \\ \hline 0.10 & 25.5 & 0.60 & 58.7 \\ 0.20 & 37.2 & 0.80 & 64.0 \\ 0.40 & 50.8 & 1.00 & 66.8 \end{tabular} A \(1.00-\mathrm{mL}\) sample of serum is diluted to volume in a \(10-\mathrm{mL}\) volumetric flask and analyzed, giving a potential of 51.1 (arbitrary units). Report the concentration of \(\mathrm{K}^{+}\) in the sample of serum.

The concentration of \(\mathrm{H}_{2} \mathrm{~S}\) in the drainage from an abandoned mine is determined by a coulometric titration using KI as a mediator and \(I_{3}^{-}\) as the titrant. $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{I}_{5}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)=2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+3 \mathrm{I}^{-}(a q)+\mathrm{S}_{(s)} $$ A \(50.00-\mathrm{mL}\) sample of water is placed in a coulometric cell, along with an excess of KI and a small amount of starch as an indicator. Electrolysis is carried out at a constant current of \(84.6 \mathrm{~mA}\), requiring \(386 \mathrm{~s}\) to reach the starch end point. Report the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) in the sample in \(\mu \mathrm{g} / \mathrm{mL}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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.