Chapter 16, Problem 16B.6P

(a)

Interpretation Introduction

Interpretation:

The drift speed of ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, and ${\text{K}}^{+}$ in water have to be calculated.

Concept introduction:

Conductivity is a property of a material that depends upon the charge carriers in a sample.  Molar conductivity is the conductivity per unit molar concentration.  According to Kohlrausch’s law, the molar conducivities of the dilute solutions containing strong electrolytes depend upon the square root of the concentration.  The conductivity of a solution when the concentration tends to zero is known as the limiting molar conductivity.  Limiting molar conductivity is the sum of contributions from individual ions.

Answer to Problem 16B.6P

The drift speed of ${\text{Li}}^{+}$ is $\underset{\_}{8.02\times 10^{-3}cm{s}^{-1}}$.

The drift speed of ${\text{Na}}^{+}$ is $\underset{\_}{1.038\times 10^{-2}cm{s}^{-1}}$.

The drift speed of ${\text{K}}^{+}$ is $\underset{\_}{1.524\times 10^{-2}cm{s}^{-1}}$.

Explanation of Solution

The drift speed is calculated by the formula shown below.

    $s=uE$        (1)

Where,

• $u$ is the mobility of ion.
• $E$ is the electric field strength.

The electric field is given by the formula shown below.

    $E=\frac{\Delta \varphi }{l}$        (2)

Where,

• $\Delta \varphi$ is the potential difference.
• $l$ is the distance between two plane parallel plates.

The value of potential difference and distance between two plane parallel plates are $10\text{0}\text{V}$ and $5.0\text{0}\text{cm}$ respectively.

Substitute the values of potential difference and distance between two plane parallel plates in equation (2).

    $\begin{array}{c}E=\frac{\Delta \varphi }{l}\\ =\frac{10\text{0}\text{V}}{5.0\text{0}\text{cm}}\\ =20\text{V}{\text{cm}}^{-1}\end{array}$

For ${\text{Li}}^{+}$ ion, the value of mobility is $4.01\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}$.

Substitute the value of mobility of ${\text{Li}}^{+}$ ion and $E$ in equation (1).

    $\begin{array}{c}s=uE\\ =4.01\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}\times 20\text{V}{\text{cm}}^{-1}\\ =\underset{\_}{8.02\times 10^{-3}cm{s}^{-1}}\end{array}$

Therefore, the drift speed of ${\text{Li}}^{+}$ is $\underset{\_}{8.02\times 10^{-3}cm{s}^{-1}}$.

For ${\text{Na}}^{+}$ ion, the value of mobility is $5.19\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}$.

Substitute the value of mobility of ${\text{Na}}^{+}$ ion and $E$ in equation (1).

    $\begin{array}{c}s=uE\\ =5.19\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}\times 20\text{V}{\text{cm}}^{-1}\\ =\underset{\_}{1.038\times 10^{-2}cm{s}^{-1}}\end{array}$

Therefore, the drift speed of ${\text{Na}}^{+}$ is $\underset{\_}{1.038\times 10^{-2}cm{s}^{-1}}$.

For ${\text{K}}^{+}$ ion, the value of mobility is $7.62\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}$.

Substitute the value of mobility of ${\text{K}}^{+}$ ion and $E$ in equation (1).

    $\begin{array}{c}s=uE\\ =7.62\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}\times 20\text{V}{\text{cm}}^{-1}\\ =\underset{\_}{1.524\times 10^{-2}cm{s}^{-1}}\end{array}$

Therefore, the drift speed of ${\text{K}}^{+}$ is $\underset{\_}{1.524\times 10^{-2}cm{s}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The time taken by ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, and ${\text{K}}^{+}$ ions to move from one electrode to other has to be calculated.

Concept introduction:

As mentioned in concept introduction in part (a).

Answer to Problem 16B.6P

The time taken by ${\text{Li}}^{+}$ ion to move from one electrode to other is $\underset{\_}{623.44s}$.

The time taken by ${\text{Na}}^{+}$ ion to move from one electrode to other is $\underset{\_}{481.69s}$.

The time taken by ${\text{K}}^{+}$ ion to move from one electrode to other is $\underset{\_}{328.08s}$.

Explanation of Solution

The time taken by the each ion to move from one electrode to other is calculated by the formula shown below.

    $t=\frac{l}{s}$        (1)

Where,

• $l$ is the distance between two plane parallel plates.
• $s$ is the drift speed.

For ${\text{Li}}^{+}$ ion, the value of $s$ and $l$ is $8.02\times {10}^{-3}\text{cm}{\text{s}}^{-1}$ and $5.0\text{0}\text{cm}$ respectively.

Substitute the value of $s$ and $l$ in equation (1).

    $\begin{array}{c}t=\frac{l}{s}\\ =\frac{5.0\text{0}\text{cm}}{8.02\times {10}^{-3}\text{cm}{\text{s}}^{-1}}\\ =\underset{\_}{623.44s}\end{array}$

Therefore, the time taken by ${\text{Li}}^{+}$ ion to move from one electrode to other is $\underset{\_}{623.44s}$.

For ${\text{Na}}^{+}$ ion, the value of $s$ and $l$ is $1.038\times {10}^{-2}\text{cm}{\text{s}}^{-1}$ and $5.0\text{0}\text{cm}$ respectively.

Substitute the value of $s$ and $l$ in equation (1).

    $\begin{array}{c}t=\frac{l}{s}\\ =\frac{5.0\text{0}\text{cm}}{1.038\times {10}^{-2}\text{cm}{\text{s}}^{-1}}\\ =\underset{\_}{481.69s}\end{array}$

Therefore, the time taken by ${\text{Na}}^{+}$ ion to move from one electrode to other is $\underset{\_}{481.69s}$.

For ${\text{K}}^{+}$ ion, the value of $s$ and $l$ is $1.524\times {10}^{-2}\text{cm}{\text{s}}^{-1}$ and $5.0\text{0}\text{cm}$ respectively.

Substitute the value of $s$ and $l$ in equation (1).

    $\begin{array}{c}t=\frac{l}{s}\\ =\frac{5.0\text{0}\text{cm}}{1.524\times {10}^{-2}\text{cm}{\text{s}}^{-1}}\\ =\underset{\_}{328.08s}\end{array}$

Therefore, the time taken by ${\text{K}}^{+}$ ion to move from one electrode to other is $\underset{\_}{328.08s}$.

(c)

Interpretation Introduction

Interpretation:

The displacement of each ion in centimeter and in term of solvent diameter during half cycle has to be calculated.

Concept introduction:

As mentioned in concept introduction in part (a).

Answer to Problem 16B.6P

The displacement of ${\text{Li}}^{+}$ ion in centimeter during half cycle is $\underset{\_}{1.28\times 10^{-6}cm}$.

The the displacement of ${\text{Li}}^{+}$ ion in term of solvent diameters during half cycle is $\underset{\_}{42.66d_s}$.

The displacement of ${\text{Na}}^{+}$ ion in centimeter during half cycle is $\underset{\_}{1.65\times 10^{-6}cm}$.

The displacement of ${\text{Na}}^{+}$ ion in term of solvent diameters during half cycle is $\underset{\_}{55d_s}$.

The displacement of ${\text{K}}^{+}$ ion in centimeter during half cycle is $\underset{\_}{2.43\times 10^{-6}cm}$.

The displacement of ${\text{K}}^{+}$ ion in term of solvent diameters during half cycle is $\underset{\_}{81d_s}$.

Explanation of Solution

The displacement during half cycle is calculated as shown below.

    $\begin{array}{c}d={\displaystyle {\int }_0^{1/2v}s\text{d}t}\\ ={\displaystyle {\int }_0^{1/2v}uE\text{d}t}\left(\because s=uE\right)\\ ={\displaystyle {\int }_0^{1/2v}u{E}_0\text{sin}\left(2\pi vt\right)\text{d}t}\left(E=E_0\text{sin}\left(2\pi vt\right)\right)\\ =\frac{u{E}_0}{2\pi v}{\left[\cos \left(2\pi vt\right)\right]}_0^{1/2v}\end{array}$

Solve the above equation.

    $\begin{array}{c}d=\frac{u{E}_0}{2\pi v}{\left[\cos \left(2\pi vt\right)\right]}_0^{1/2v}\\ =\frac{u{E}_0}{2\pi v}\left[\cos \left(2\pi v\frac{1}{2v}\right)-\cos \left(0\right)\right]\\ =\frac{u{E}_0}{2\pi v}\left[\cos \left(\pi \right)-\cos \left(0\right)\right]\\ =\frac{u{E}_0}{2\pi v}\left[-1-1\right]\end{array}$

Solve the above equation.

    $\begin{array}{c}d=\frac{u{E}_0}{2\pi v}\left[-1-1\right]\\ =-\frac{u{E}_0}{\pi v}\end{array}$

The displacement cannot be negative.  Therefore, the above equation can be written as shown below.

    $d=\frac{u{E}_0}{\pi v}$

Where,

• $u$ is the mobility.
• $E_0$ is the potential difference.
• $v$ is the frequency.

The value of $E_0$ and $v$ is $20\text{V}{\text{cm}}^{-1}$ and $2\text{kHz}$ respectively.

For ${\text{Li}}^{+}$ ion, the value of mobility is $4.01\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}$.

Substitute the value of mobility, $E_0$ and $v$ in equation (1).

    $\begin{array}{c}d=\frac{u{E}_0}{\pi v}\\ =\frac{4.01\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}\times 20\text{V}{\text{cm}}^{\text{-1}}}{3.14\times 2\text{kHz}\left(\frac{{10}^3{\text{s}}^{-1}}{1\text{kHz}}\right)}\\ =\frac{8.02\times {10}^{-6}}{6.28}\text{cm}\\ =\underset{\_}{1.28\times 10^{-6}cm}\end{array}$

Therefore, the displacement of ${\text{Li}}^{+}$ ion in centimeter during half cycle is $\underset{\_}{1.28\times 10^{-6}cm}$.

The solvent diameter $\left(d_s\right)$ is given as $30\text{0}\text{pm}$.

     $\begin{array}{c}{d}_s=300\text{pm}\\ =300\text{pm}\frac{{10}^{-10}\text{cm}}{1\text{pm}}\\ =3\times {10}^{-8}\text{cm}\\ 1\text{cm}=\frac{d_s}{3}\times {10}^8\end{array}$

Use $1\text{cm}=\frac{d_s}{3}\times {10}^8$ to calculate the the displacement of ${\text{Li}}^{+}$ ion in term of solvent diameters during half cycle as shown below.

    $\begin{array}{c}d=1.28\times {10}^{-6}\text{cm}\\ =1.28\times {10}^{-6}\left(\frac{d_s}{3}\times {10}^8\right)\\ =\underset{\_}{42.66d_s}\end{array}$

The the displacement of ${\text{Li}}^{+}$ ion in term of solvent diameters during half cycle is $\underset{\_}{42.66d_s}$.

For ${\text{Na}}^{+}$ ion, the value of mobility is $5.19\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}$.

Substitute the value of mobility, $E_0$ and $v$ in equation (1).

    $\begin{array}{c}d=\frac{u{E}_0}{\pi v}\\ =\frac{5.19\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}\times 20\text{V}{\text{cm}}^{\text{-1}}}{3.14\times 2\text{kHz}\left(\frac{{10}^3{\text{s}}^{-1}}{1\text{kHz}}\right)}\\ =\frac{10.38\times {10}^{-6}}{6.28}\text{cm}\\ =\underset{\_}{1.65\times 10^{-6}cm}\end{array}$

Therefore, the displacement of ${\text{Na}}^{+}$ ion in centimeter during half cycle is $\underset{\_}{1.65\times 10^{-6}cm}$.

Use $1\text{cm}=\frac{d_s}{3}\times {10}^8$ to calculate the the displacement of ${\text{Na}}^{+}$ ion in term of solvent diameters during half cycle as shown below.

    $\begin{array}{c}d=1.65\times {10}^{-6}\text{cm}\\ \text{=}1.65\times {10}^{-6}\left(\frac{d_s}{3}\times {10}^8\right)\\ =\underset{\_}{55d_s}\end{array}$

The displacement of ${\text{Na}}^{+}$ ion in term of solvent diameters during half cycle is $\underset{\_}{55d_s}$

For ${\text{K}}^{+}$ ion, the value of mobility is $7.62\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}$.

Substitute the value of mobility, $E_0$ and $v$ in equation (1).

    $\begin{array}{c}d=\frac{u{E}_0}{\pi v}\\ =\frac{7.62\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{V}}^{-1}{\text{s}}^{-1}\times 20\text{V}{\text{cm}}^{\text{-1}}}{3.14\times 2\text{kHz}\left(\frac{{10}^3{\text{s}}^{-1}}{1\text{kHz}}\right)}\\ =\frac{10.38\times {10}^{-6}}{6.28}\text{cm}\\ =\underset{\_}{2.43\times 10^{-6}cm}\end{array}$

Therefore, the displacement of ${\text{K}}^{+}$ ion in centimeter during half cycle is $\underset{\_}{2.43\times 10^{-6}cm}$.

Use $1\text{cm}=\frac{d_s}{3}\times {10}^8$ to calculate the the displacement of ${\text{K}}^{+}$ ion in term of solvent diameters during half cycle as shown below.

    $\begin{array}{c}d=2.43\times {10}^{-6}\text{cm}\\ \text{=}2.43\times {10}^{-6}\left(\frac{d_s}{3}\times {10}^8\right)\\ =\underset{\_}{81d_s}\end{array}$

The displacement of ${\text{K}}^{+}$ ion in term of solvent diameters during half cycle is $\underset{\_}{81d_s}$.