Chapter 21, Problem 21.71P

(a)

Interpretation Introduction

Interpretation:

The initial cell potential for given cell has to be calculated.

Concept Introduction:

An electrochemical cell is a device in which a redox reaction is used to convert chemical energy into electrical energy. Such device is also known as the galvanic or voltaic cell.

Anode: The electrode where the oxidation occurs is called as an anode. It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode. It is a positively charged electrode.

Redox reaction: Redox reaction is a type of chemical reaction, where both the oxidation and reduction occur at the same time. In a redox reaction, one of the reactant is oxidized and the other is reduced simultaneously.

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}$

n - Number of electrons involved per equivalent of the net redox reaction in the cell

F - Faraday’s Constant (96500 C)

$\text{E}{°}_{\text{cell}}$ - Standard cell potential.

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The voltaic cell has 2 half cells namely ${\text{Mn/Mn}}^{\text{2+}}$ and ${\text{Cd/Cd}}^{\text{2+}}$.

The initial cell potential is determined by first determining the ${\text{E}}^{\text{o}}{}_{\text{cell}}$ which then substituting it in Nernst equation, the cell potential is calculated as shown below,

$\begin{array}{c}{\text{E}}^{\text{o}}{}_{\text{cell}}\text{=}{\text{E}}^{\text{o}}{}_{\text{cathode}}{\text{-E}}^{\text{o}}{}_{\text{anode}}\\ \text{=}\text{-0}\text{.40}\text{V-}\left(\text{-1}\text{.18}\text{V}\right)\\ =0.78V\end{array}$

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ ={\text{E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log \frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]}\\ =0.78V-\frac{0.0592V}{2}\log \frac{\left[0.090\right]}{\left[0.060\right]}\\ =0.77V\end{array}$

(b)

Interpretation Introduction

Interpretation:

For given $\left[{\text{Cd}}^{\text{2+}}\right]$ concentration the ${\text{E}}_{\text{cell}}$ has to be calculated.

Concept Introduction:

An electrochemical cell is a device in which a redox reaction is used to convert chemical energy into electrical energy. Such device is also known as the galvanic or voltaic cell.

Anode: The electrode where the oxidation occurs is called as an anode. It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode. It is a positively charged electrode.

Redox reaction: Redox reaction is a type of chemical reaction, where both the oxidation and reduction occur at the same time. In a redox reaction, one of the reactant is oxidized and the other is reduced simultaneously.

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}$

n - Number of electrons involved per equivalent of the net redox reaction in the cell

F - Faraday’s Constant (96500 C)

$\text{E}{°}_{\text{cell}}$ - Standard cell potential.

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The $\left[{\text{Cd}}^{\text{2+}}\right]$ concentration decreases from $\text{0}\text{.060}\text{M}$ to $\text{0}\text{.050}\text{M}$. The change in concentration is $\text{0}\text{.010}\text{M}$, which the same concentration amount has to be increase in $\left[M{n}^{\text{2+}}\right]$ hence, $\left[M{n}^{\text{2+}}\right]$ concentration increases from $\text{0}\text{.090}\text{M}$ to $\text{0}\text{.100}\text{M}$. The $E_{cell}$ is determined by using Nernst equation as follows,

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592}{n}\log Q\\ =0.78V-\frac{0.0592}{2}\log \frac{\left[0.10\right]}{0.050}\\ =0.77V\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $\left[M{n}^{\text{2+}}\right]$ concentration has to be calculated when cell potential reaches given concentration.

Concept Introduction:

An electrochemical cell is a device in which a redox reaction is used to convert chemical energy into electrical energy. Such device is also known as the galvanic or voltaic cell.

Anode: The electrode where the oxidation occurs is called as an anode. It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode. It is a positively charged electrode.

Redox reaction: Redox reaction is a type of chemical reaction, where both the oxidation and reduction occur at the same time. In a redox reaction, one of the reactant is oxidized and the other is reduced simultaneously.

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}$

n - Number of electrons involved per equivalent of the net redox reaction in the cell

F - Faraday’s Constant (96500 C)

$\text{E}{°}_{\text{cell}}$ - Standard cell potential.

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The $\left[M{n}^{\text{2+}}\right]$ increase and decrease in $\left[{\text{Cd}}^{\text{2+}}\right]$ concentration are in equal amounts.

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592}{n}\log Q\\ 0.055=0.78-\frac{0.0592}{2}\log \frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]}\\ \frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]}=3.11\times {10}^{24}\\ \left[M{n}^{2+}\right]=3.11\times {10}^{24}\times \left[C{d}^{2+}\right]\end{array}$

$\begin{array}{c}\left[M{n}^{2+}\right]+\left[C{d}^{2+}\right]=0.150M\\ 3.11\times {10}^{24}\times \left[C{d}^{2+}\right]+\left[C{d}^{2+}\right]=0.150M\\ \left[C{d}^{2+}\right]=4.82\times {10}^{-26}M\\ \left[M{n}^{2+}\right]=0.150M-4.82\times {10}^{-26}M\\ =0.150M.\end{array}$

(d)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ions have to be given.

Concept Introduction:

An electrochemical cell is a device in which a redox reaction is used to convert chemical energy into electrical energy. Such device is also known as the galvanic or voltaic cell.

Anode: The electrode where the oxidation occurs is called as an anode. It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode. It is a positively charged electrode.

Redox reaction: Redox reaction is a type of chemical reaction, where both the oxidation and reduction occur at the same time. In a redox reaction, one of the reactant is oxidized and the other is reduced simultaneously.

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}$

n - Number of electrons involved per equivalent of the net redox reaction in the cell

F - Faraday’s Constant (96500 C)

$\text{E}{°}_{\text{cell}}$ - Standard cell potential.

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The cell potential at equilibrium will be equal to 0. Then, using Nernst equation

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{ = E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592}{n}\log Q\\ 0=0.78-\frac{0.0592}{2}\log \frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]}\\ \frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]}=2.25\times {10}^{26}\\ \left[M{n}^{2+}\right]=2.25\times {10}^{26}\times \left[C{d}^{2+}\right]\end{array}$

$\begin{array}{c}\left[M{n}^{2+}\right]+\left[C{d}^{2+}\right]=0.150M\\ 2.25\times {10}^{26}\times \left[C{d}^{2+}\right]+\left[C{d}^{2+}\right]=0.150M\\ \left[C{d}^{2+}\right]=7\times {10}^{-28}M\\ \left[M{n}^{2+}\right]=0.150M-7\times {10}^{-28}M\\ =0.150M.\end{array}$