Question

For \(\mathrm{Na}^{+}\), the value of symbol \(\lambda_{\mathrm{m}}^{\circ}\) is \(50.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} .\) The speed of \(\mathrm{Na}^{+}\) ion in the solution, if in the cell, electrodes are \(5 \mathrm{~cm}\) apart and to which a potential of \(19.3\) volt is applied is (a) \(2 \times 10^{-3} \mathrm{~cm} / \mathrm{s}\) (b) \(1 \times 10^{-3} \mathrm{~cm} / \mathrm{s}\) (c) \(2 \times 10^{-4} \mathrm{~cm} / \mathrm{s}\) (d) \(2 \times 10^{-2} \mathrm{~cm} / \mathrm{s}\)

Short answer

(b) \(1 \times 10^{-3} \mathrm{~cm} / \mathrm{s}\)

Step-by-step solution

Understand the Given Data and Formula

We are given the molar conductivity \(\lambda_{\mathrm{m}}^{\circ} = 50.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). The distance between electrodes is \(5 cm\), and the potential difference applied is \(19.3 V\). We will be using the formula relating conductivity \(\kappa\), molar conductivity \(\lambda_{\mathrm{m}}^{\circ}\), and concentration \(c\): \[\kappa = \lambda_{\mathrm{m}}^{\circ} \cdot c\]. We need to find the speed of \(\mathrm{Na}^{+}\) ions, which can be calculated using the relation between electric field \(E\), mobility \(u\), and potential difference \(V\): \[u = \frac{E}{V}\] where \(E = \frac{V}{d}\).

Calculate the Electric Field

The electric field \(E\) is defined as the potential difference \(V\) divided by the distance \(d\) between the electrodes. Using the given values: \[E = \frac{V}{d} = \frac{19.3 \text{ V}}{5 \text{ cm}} = 3.86 \text{ V/cm}\].

Calculate Mobility of Na+ Ions

The mobility \(u\) of \(\mathrm{Na}^{+}\) ions can be linked to \(\lambda_{\mathrm{m}}^{\circ}\) because molar conductivity is equal to mobility times the Faraday constant \(F\), \(\lambda_{\mathrm{m}}^{\circ} = u \cdot F\). Since \(F\) is approximately \(96500 \text{ C/mol}\), we can solve for \(u\): \[u = \frac{\lambda_{\mathrm{m}}^{\circ}}{F} = \frac{50.0 \Omega^{-1} \mathrm{cm}^2 \mathrm{mol}^{-1}}{96500 \text{ C/mol}}\].

Calculate Speed of Na+ Ions

With the mobility \(u\), we can find the speed \(v\) of \(\mathrm{Na}^{+}\) ions by multiplying the mobility by the electric field determined in Step 2: \[v = u \cdot E = \frac{\lambda_{\mathrm{m}}^{\circ} \cdot E}{F} = \frac{50.0 \times 3.86}{96500}\].

Evaluate and Choose the Correct Option

Evaluate the calculation from the previous step, convert the units of speed from \(\text{V/cm}\) to \(\text{cm/s}\) by taking note of the units that cancel out and choosing the correct answer from the provided options.

Key concepts

Electrolyte Mobility

Electrolyte mobility refers to the speed at which an ion moves through a solution when subjected to an electric field. It is intrinsic to the nature of ions and is affected by factors such as the size and charge of the ion, the viscosity of the solvent, and the temperature of the solution.

In physical chemistry, understanding ion mobility is crucial as it helps in calculating current flow in electrochemical cells. In the context of our exercise, the mobility was found by using the molar conductivity and Faraday's constant. The mobility equation, expressed as \( u = \frac{\lambda_{\mathrm{m}}^{\circ}}{F} \), where \( u \) is the mobility, \( \lambda_{\mathrm{m}}^{\circ} \) is the molar conductivity, and \( F \) is Faraday's constant, shows the direct relationship between mobility and molar conductivity.

Electric Field Calculation

The electric field in a region is a measure of the potential energy per unit charge at any point within that field. It's a vector quantity, with both magnitude and direction, and is defined as the force per unit charge.

In the given problem, the electric field \( E \) was calculated using the formula \( E = \frac{V}{d} \), where \( V \) represents the voltage applied across the electrodes and \( d \) is the distance between them. Practically, the electric field provides the ‘push’ that drives the charged ions through the solution, and it's fundamental in determining the rate of ionic movement, or mobility.

Faraday's Constant

Faraday's constant is a cornerstone of electrochemistry, representing the total electric charge carried by one mole of electrons. It is approximately \( 96500 \) coulombs per mole (C/mol).

This constant links the amount of electric charge to the substance amount in moles, connecting the macroscopic world of chemical reactions with the microscopic world of electric currents. In the solution for the exercise, Faraday's constant was used to relate the molar conductivity of sodium ions (\( \lambda_{\mathrm{m}}^{\circ} \)) with their mobility (\( u \)), highlighting its importance in converting between different units and concepts in electrochemical processes.

Physical Chemistry

Physical chemistry is a branch of chemistry focused on the understanding of the physical properties of molecules, the forces that act upon them, and the energy changes associated with chemical reactions. Concepts like molar conductivity, mobility, and the role of electric fields are all fundamental topics studied within this field.

As illustrated through our exercise, physical chemistry applies principles of physics to chemically significant problems, using mathematical equations to express and predict chemical behavior. It bridges the gap between the small-scale interactions of particles and the observable phenomena in the lab.

Conductivity and Concentration Relation

The relationship between conductivity and concentration is an essential aspect of understanding how ions contribute to the overall conductivity of a solution. Conductivity, denoted by \( \kappa \), is the measure of a solution's ability to conduct electricity and is directly proportional to the concentration of ions in the solution.

The formula \( \kappa = \lambda_{\mathrm{m}}^{\circ} \cdot c \) shows how the molar conductivity (\( \lambda_{\mathrm{m}}^{\circ} \)) varies with concentration (\( c \)). In dilute solutions, as concentration decreases, the molar conductivity tends to increase due to the reduced inter-ionic interactions, thereby enhancing the mobility of ions and in turn the conductivity of the solution.