/*! 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 75 The specific conductance of \(0.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The specific conductance of \(0.1 \mathrm{~N} \mathrm{KCl}\) solution at \(23^{\circ} \mathrm{C}\) is \(0.012 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (a) \(0.142 \mathrm{~cm}^{-1}\) (b) \(0.616 \mathrm{~cm}^{-1}\) (c) \(6.16 \mathrm{~cm}^{-1}\) (d) \(616 \mathrm{~cm}^{-1}\)

Short Answer

Expert verified
The closest correct answer based on the approximate calculation is not explicitly listed. Re-evaluate or check possible errors in given options.

Step by step solution

01

Understanding the Relationship Between Conductance and Cell Constant

The cell constant (also known as the cell’s geometric factor), denoted as \( G^* \), is connected to the specific conductance \( k \) by the formula:\[ G^* = k \times R \] where \( R \) is the resistance of the cell. This formula implies that the cell constant is the product of the specific conductance and the resistance of the cell.
02

Calculating the Cell Constant

Given the specific conductance \( k = 0.012 \ \text{ohm}^{-1} \ \text{cm}^{-1} \) and the resistance \( R = 55 \ \text{ohms} \), we plug these values into the formula to find the cell constant: \[ G^* = 0.012 \times 55 = 0.66 \ \text{cm}^{-1} \]. Hence, the calculation gives us a cell constant of \( 0.66 \ \text{cm}^{-1} \).
03

Matching the Calculated Cell Constant to Given Options

After calculating the cell constant to be \( 0.66 \ \text{cm}^{-1} \), we compare this value to the given options. None of the options exactly match \( 0.66 \ \text{cm}^{-1} \). Upon revisiting the calculations, if a rounding error is a possibility in the exercise, the closest approximation in the provided answers should be considered.

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 91Ó°ÊÓ!

Key Concepts

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

Specific Conductance
In the realm of electrochemistry, specific conductance is a pivotal concept. It refers to the efficiency with which a solution can conduct electricity. Unlike conductance, which measures the flow of electric current through a particular system, specific conductance is a measure of this ability per unit length and unit area.
Specific conductance is analytically denoted as the reciprocal of the resistivity, extending the measure of how easily ions can move in solution.
This is typically expressed with units of \( ext{ohm}^{-1} \ ext{cm}^{-1}\).
  • A higher specific conductance implies a more conductive solution.
  • This plays a crucial role in determining how well an electrolytic cell will function.
  • Affected by temperature, concentration of ions, and the type of ions present.
Electrochemistry exercises often incorporate specific conductance, as it helps to relate the physical properties of the cell to its electrical characteristics. By understanding how specific conductance impacts the overall conductance of an electrolytic solution, we can better analyze and optimize electrochemical cells for various applications.
Resistance in Electrochemistry
Resistance, a key factor in electrochemistry, is the opposition to the flow of electric current through a conductor. Within an electrolytic cell, resistance depends on several factors including the nature of the electrolyte and the dimensions of the cell.
The resistance, denoted as \(R\), can be determined by Ohm’s Law, which states: \( ext{Voltage} = I imes R\), where \( ext{Voltage}\) is the difference in electric potential and \(I\) is the current. In electrochemical contexts, the resistance is tied closely with the measuring of conductance through:
  • Changing dimensions - Resistance is influenced by the length and cross-sectional area of the cell.
  • Type of material - Different materials conduct electricity differently, which is essential when considering the electrolyte used.
  • Temperature effects - As temperature increases, resistance typically decreases in electrolytes.
Understanding resistance helps in calculating the cell constant, which is an important measure of the cell's geometry and how it impacts the ability to conduct electricity. Resistance measurements are foundational in determining the overall efficiency and functioning of electrochemical cells.
Electrolytic Cells
Electrolytic cells are fundamental devices in the field of electrochemistry, converting electrical energy into chemical energy by driving non-spontaneous chemical reactions. These cells are essential for applications such as electroplating, electrolysis of water, and battery recharging.
Key features characterize electrolytic cells:
  • An external voltage is applied to drive the chemical reactions.
  • Comprised of two electrodes (anode and cathode) immersed in an electrolyte solution.
  • Ions in the solution transfer charge between electrodes, enabling the reaction.
The geometry of an electrolytic cell impacts its efficiency, with the cell constant being one of the pivotal geometric factors. It's a measure of the cell's dimensions and affects how the specific conductance and resistance manifest as observable conductance in the lab.
It’s crucial to iteratively refine and calibrate electrolytic cells to ensure their optimal functioning in desired processes. By grasping the workings of specific conductance and resistance in these systems, we lay the groundwork for enhancing performances and efficiencies across various applications.

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

The electrical conductivity of the flowing aqueous solutions is highest for [2008] (a) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (b) \(0.1 \mathrm{M} \mathrm{CH}_{2} \mathrm{FCOOH}\) (c) \(0.1 \mathrm{M} \mathrm{CHF}_{3} \mathrm{COOH}\) (d) \(0.1 \mathrm{M} \mathrm{CH}_{2} \mathrm{ClCOOH}\)

Calculate the weight of copper that will be deposited at the cathode in the electrolysis of a \(0.2 \mathrm{M}\) solution of copper sulphate, when quantity of electricity, equal to the required to liberate \(2.24 \mathrm{~L}\) of hydrogen at STP from a \(0.1 \mathrm{M}\) aqueous sulphuric acid, is passed (Atomic mass of \(\mathrm{Cu}=63.5\) ) (a) \(6.35 \mathrm{~g}\) (b) \(3.17 \mathrm{~g}\) (c) \(12.71 \mathrm{~g}\) (d) \(63.5 \mathrm{~g}\)

\(\mathrm{K}_{\text {p }}\) of \(\mathrm{BaSO}_{4}\) is \(1 \times 10^{-10 .}\) If the ionic conductances of \(\mathrm{Ba}^{+}\)and \(\mathrm{SO}_{4}^{2-}\) ions are 64 and \(80 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}\) - respectively, then its specific conductance is (a) \(1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (b) \(144 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (c) \(1.44 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (d) \(144 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

The \(E\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)\) values for \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) are \(-0.41,+1.57,+0.77\) and \(+1.97 \mathrm{~V}\) respectively. For which one of these metals the change in oxidation state form \(+2\) to \(+3\) is easiest? \(\quad\) [2004] (a) \(\mathrm{Cr}\) (b) \(\mathrm{Mn}\) (c) \(\mathrm{Fe}\) (d) \(\mathrm{Co}\)

Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are (a) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{2+}\) and dissolved oxygen in water is reduced to \(\mathrm{OH}^{-}\) (b) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}_{2} \mathrm{O}\) is reduced to \(\mathrm{O}_{2}^{2-}\) (c) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{2+}\) and \(\mathrm{H}_{2} \mathrm{O}\) is reduced to \(\mathrm{O}_{2}\) (d) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{2+}\) and \(\mathrm{H}_{2} \mathrm{O}\) is reduced to \(\mathrm{O}_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.