Chapter 24, Problem 24.2QAP

Calculate the minimum difference in standard electrode potentials needed to lower the concentration of the metal M₁ to 2.00 × 10^-4 M ¡n a solution that is 1.00 × 10^-1 M in the less-reducible metal M₂ where (a) M₂ is univalent and M₁ is divalent. (b) M₂ and M₁ are both divalent, (c) M₂ is trivalent and M₁ is univalent, (d) M₂is divalent and M₁ is univalent, (e) M₂ is divalent and M₁ ¡s trivalent.

Interpretation Introduction

(a)

Interpretation:

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is to be stated when M₂ is univalent and M₁ is divalent.

Concept introduction:

The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs. The Nernst equation is used to determine the electromotive force and the reduction potential of the half life cell.

Answer to Problem 24.2QAP

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.050}\text{V}$.

Explanation of Solution

Given information:

The concentration of the metal ${\text{M}}_{\text{1}}$ is $2.00\times {10}^{-4}\text{M}$, the concentration of the solution is $1.00\times {10}^{-1}\text{M}$, the ${\text{M}}_{\text{2}}$ is univalent and ${\text{M}}_{\text{1}}$ is the divalent.

Write the expression for the Nernst at room temperature.

$E=E^{°}-\frac{0.05916\text{V}}{n}{\log }_{10}\frac{a_{\text{red}}}{a_{\text{Ox}}}$   ...... (I)

Here, the half-life potential is $E$, the standard half-cell potential is $E^{°}$, the number of moles of electrons is $n$, the reduction concentration of metal ${\text{M}}_{\text{1}}$

$a_{\text{Ox}}$ and the reduction concentration of solution is $a_{\text{red}}$.

Write the expression for the minimum difference in the standard electrode potential.

$\Delta E=E^{°}{\text{M}}_2-E^{°}{\text{M}}_1$   ...... (II)

Substitute $E_1$ for $E$, $E^{°}{\text{M}}_1$ for $E^{°}$, $2.00\times {10}^{-4}\text{M}$ for $a_{\text{Ox}}$, $2$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_1=E^{°}{\text{M}}_1-\frac{0.05916\text{V}}{2}{\log }_{10}\frac{1}{\left(2.00\times {10}^{-4}\text{M}\right)}\\ E_1=E^{°}{\text{M}}_1-0.109\text{V}\end{array}$

Here, the initial energy is $E_1$.

Substitute $E_2$ for $E$, $E^{°}{\text{M}}_2$ for $E^{°}$, $1.00\times {10}^{-1}\text{M}$ for $a_{\text{Ox}}$, $1$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_2=E^{°}{\text{M}}_2-\frac{0.05916\text{V}}{1}{\log }_{10}\frac{1}{\left(1.00\times {10}^{-1}\text{M}\right)}\\ E_2=E^{°}{\text{M}}_2-0.0592\text{V}\end{array}$

Here, the final energy is $E_2$.

Write the expression for the relation between the initial energy and final energy.

$E_1=E_2$   ...... (III)

Substitute $E^{°}{\text{M}}_2-0.0592\text{V}$ for $E_2$ and $E^{°}{\text{M}}_1-0.109\text{V}$ for $E_1$ in Equation (III).

$\begin{array}{l}{E}^{°}{\text{M}}_2-0.0592\text{V}=E^{°}{\text{M}}_1-0.109\text{V}\\ E^{°}{\text{M}}_2-E^{°}{\text{M}}_1=0.0592\text{V}-0.109\text{V}\end{array}$   ...... (IV)

Substitute $\Delta E^{°}$ for $E^{°}{\text{M}}_2-E^{°}{\text{M}}_1$ in Equation (IV).

$\begin{array}{c}\Delta E^{°}=0.0592\text{V}-0.109\text{V}\\ =-\text{0}\text{.050}\text{V}\end{array}$

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.050}\text{V}$.

Interpretation Introduction

(b)

Interpretation:

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is to be statedwhen M₁ and M₂both are divalent.

Concept introduction:

The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs. The Nernst equation is used to determine the electromotive force and the reduction potential of the half life cell.

Answer to Problem 24.2QAP

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.079}\text{V}$.

Explanation of Solution

The concentration of the metal ${\text{M}}_{\text{1}}$ is $2.00\times {10}^{-4}\text{M}$, the concentration of the solution is $1.00\times {10}^{-1}\text{M}$, the ${\text{M}}_{\text{2}}$ is divalent and ${\text{M}}_{\text{1}}$ is the divalent.

Substitute $E_1$ for $E$, $E^{°}{\text{M}}_1$ for $E^{°}$, $2.00\times {10}^{-4}\text{M}$ for $a_{\text{Ox}}$, $2$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_1=E^{°}{\text{M}}_1-\frac{0.05916\text{V}}{2}{\log }_{10}\frac{1}{\left(2.00\times {10}^{-4}\text{M}\right)}\\ E_1=E^{°}{\text{M}}_1-0.109\text{V}\end{array}$

Here, the initial energy is $E_1$.

Substitute $E_2$ for $E$, $E^{°}{\text{M}}_2$ for $E^{°}$, $1.00\times {10}^{-1}\text{M}$ for $a_{\text{Ox}}$, $2$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_2=E^{°}{\text{M}}_2-\frac{0.05916\text{V}}{2}{\log }_{10}\frac{1}{\left(1.00\times {10}^{-1}\text{M}\right)}\\ E_2=E^{°}{\text{M}}_2-0.0296\text{V}\end{array}$

Here, the final energy is $E_2$.

Substitute $E^{°}{\text{M}}_2-0.0296\text{V}$ for $E_2$ and $E^{°}{\text{M}}_1-0.109\text{V}$ for $E_1$ in Equation (III).

$\begin{array}{l}{E}^{°}{\text{M}}_2-0.0296\text{V}=E^{°}{\text{M}}_1-0.109\text{V}\\ E^{°}{\text{M}}_2-E^{°}{\text{M}}_1=0.0296\text{V}-0.109\text{V}\end{array}$   ...... (IV)

Substitute $\Delta E^{°}$ for $E^{°}{\text{M}}_2-E^{°}{\text{M}}_1$ in Equation (IV).

$\begin{array}{c}\Delta E^{°}=0.0296\text{V}-0.109\text{V}\\ =-\text{0}\text{.079}\text{V}\end{array}$

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.079}\text{V}$.

Interpretation Introduction

(c)

Interpretation:

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is to be statedwhen M₂ is trivalent and M₁ is univalent.

Concept introduction:

The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs. The Nernst equation is used to determine the electromotive force and the reduction potential of the half life cell.

Answer to Problem 24.2QAP

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.199}\text{V}$.

Explanation of Solution

The concentration of the metal ${\text{M}}_{\text{1}}$ is $2.00\times {10}^{-4}\text{M}$, the concentration of the solution is $1.00\times {10}^{-1}\text{M}$, the ${\text{M}}_{\text{2}}$ is trivalent and ${\text{M}}_{\text{1}}$ is the univalent.

Substitute $E_1$ for $E$, $E^{°}{\text{M}}_1$ for $E^{°}$, $2.00\times {10}^{-4}\text{M}$ for $a_{\text{Ox}}$, $1$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_1=E^{°}{\text{M}}_1-\frac{0.05916\text{V}}{1}{\log }_{10}\frac{1}{\left(2.00\times {10}^{-4}\text{M}\right)}\\ E_1=E^{°}{\text{M}}_1-0.219\text{V}\end{array}$

Here, the initial energy is $E_1$.

Substitute $E_2$ for $E$, $E^{°}{\text{M}}_2$ for $E^{°}$, $1.00\times {10}^{-1}\text{M}$ for $a_{\text{Ox}}$, $3$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_2=E^{°}{\text{M}}_2-\frac{0.05916\text{V}}{3}{\log }_{10}\frac{1}{\left(1.00\times {10}^{-1}\text{M}\right)}\\ E_2=E^{°}{\text{M}}_2-0.020\text{V}\end{array}$

Here, the final energy is $E_2$.

Substitute $E^{°}{\text{M}}_2-0.020\text{V}$ for $E_2$ and $E^{°}{\text{M}}_1-0.219\text{V}$ for $E_1$ in Equation (III).

$\begin{array}{l}{E}^{°}{\text{M}}_2-0.020\text{V}=E^{°}{\text{M}}_1-0.219\text{V}\\ E^{°}{\text{M}}_2-E^{°}{\text{M}}_1=0.020\text{V}-0.219\text{V}\end{array}$   ...... (IV)

Substitute $\Delta E^{°}$ for $E^{°}{\text{M}}_2-E^{°}{\text{M}}_1$ in Equation (IV).

$\begin{array}{c}\Delta E^{°}=0.020\text{V}-0.219\text{V}\\ =-\text{0}\text{.199}\text{V}\end{array}$

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.199}\text{V}$.

Interpretation Introduction

(d)

Interpretation:

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is to be statedwhen M₂ is divalent and M₁ is univalent.

Concept introduction:

The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs. The Nernst equation is used to determine the electromotive force and the reduction potential of the half life cell.

Answer to Problem 24.2QAP

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.189}\text{V}$.

Explanation of Solution

The concentration of the metal ${\text{M}}_{\text{1}}$ is $2.00\times {10}^{-4}\text{M}$, the concentration of the solution is $1.00\times {10}^{-1}\text{M}$, the ${\text{M}}_{\text{2}}$ is divalent and ${\text{M}}_{\text{1}}$ is the univalent.

Substitute $E_1$ for $E$, $E^{°}{\text{M}}_1$ for $E^{°}$, $2.00\times {10}^{-4}\text{M}$ for $a_{\text{Ox}}$, $1$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_1=E^{°}{\text{M}}_1-\frac{0.05916\text{V}}{1}{\log }_{10}\frac{1}{\left(2.00\times {10}^{-4}\text{M}\right)}\\ E_1=E^{°}{\text{M}}_1-0.219\text{V}\end{array}$

Here, the initial energy is $E_1$.

Substitute $E_2$ for $E$, $E^{°}{\text{M}}_2$ for $E^{°}$, $1.00\times {10}^{-1}\text{M}$ for $a_{\text{Ox}}$, $2$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_2=E^{°}{\text{M}}_2-\frac{0.05916\text{V}}{}{\log }_{10}\frac{1}{\left(1.00\times {10}^{-1}\text{M}\right)}\\ E_2=E^{°}{\text{M}}_2-0.0296\text{V}\end{array}$

Here, the final energy is $E_2$.

Substitute $E^{°}{\text{M}}_2-0.0296\text{V}$ for $E_2$ and $E^{°}{\text{M}}_1-0.219\text{V}$ for $E_1$ in Equation (III).

$\begin{array}{l}{E}^{°}{\text{M}}_2-0.0296\text{V}=E^{°}{\text{M}}_1-0.219\text{V}\\ E^{°}{\text{M}}_2-E^{°}{\text{M}}_1=0.0296\text{V}-0.219\text{V}\end{array}$   ...... (IV)

Substitute $\Delta E^{°}$ for $E^{°}{\text{M}}_2-E^{°}{\text{M}}_1$ in Equation (IV).

$\begin{array}{c}\Delta E^{°}=0.0296\text{V}-0.219\text{V}\\ =-\text{0}\text{.189}\text{V}\end{array}$

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.189}\text{V}$.

Interpretation Introduction

(e)

Interpretation:

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is to be statedwhen M₂ is divalent and M₁ is trivalent.

Concept introduction:

The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs. The Nernst equation is used to determine the electromotive force and the reduction potential of the half life cell.

Answer to Problem 24.2QAP

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.043}\text{V}$.

Explanation of Solution

The concentration of the metal ${\text{M}}_{\text{1}}$ is $2.00\times {10}^{-4}\text{M}$, the concentration of the solution is $1.00\times {10}^{-1}\text{M}$, the ${\text{M}}_{\text{2}}$ is divalent and ${\text{M}}_{\text{1}}$ is the trivalent.

Substitute $E_1$ for $E$, $E^{°}{\text{M}}_1$ for $E^{°}$, $2.00\times {10}^{-4}\text{M}$ for $a_{\text{Ox}}$, $3$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_1=E^{°}{\text{M}}_1-\frac{0.05916\text{V}}{3}{\log }_{10}\frac{1}{\left(2.00\times {10}^{-4}\text{M}\right)}\\ E_1=E^{°}{\text{M}}_1-0.073\text{V}\end{array}$

Here, the initial energy is $E_1$.

Substitute $E_2$ for $E$, $E^{°}{\text{M}}_2$ for $E^{°}$, $1.00\times {10}^{-1}\text{M}$ for $a_{\text{Ox}}$, $2$ for $n$ and $1$ for $a_{\text{red}}$ in Equation (I).

$\begin{array}{l}{E}_2=E^{°}{\text{M}}_2-\frac{0.05916\text{V}}{}{\log }_{10}\frac{1}{\left(1.00\times {10}^{-1}\text{M}\right)}\\ E_2=E^{°}{\text{M}}_2-0.0296\text{V}\end{array}$

Here, the final energy is $E_2$.

Substitute $E^{°}{\text{M}}_2-0.0296\text{V}$ for $E_2$ and $E^{°}{\text{M}}_1-0.073\text{V}$ for $E_1$ in Equation (III).

$\begin{array}{l}{E}^{°}{\text{M}}_2-0.0296\text{V}=E^{°}{\text{M}}_1-0.073\text{V}\\ E^{°}{\text{M}}_2-E^{°}{\text{M}}_1=0.0296\text{V}-0.073\text{V}\end{array}$   ...... (IV)

Substitute $\Delta E^{°}$ for $E^{°}{\text{M}}_2-E^{°}{\text{M}}_1$ in Equation (IV).

$\begin{array}{c}\Delta E^{°}=0.0296\text{V}-0.073\text{V}\\ =-\text{0}\text{.043}\text{V}\end{array}$

The minimum difference in the standard electrode potential needed to lower the concentration of metal ${\text{M}}_{\text{1}}$ to $2.00\times {10}^{-4}\text{M}$ in a solution of $1.00\times {10}^{-1}\text{M}$ is $-\text{0}\text{.043}\text{V}$.