Chapter 21, Problem 21.118P

(a)

Interpretation Introduction

Interpretation:

The number of days required to produce the given current by silver button battery 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.

Galvanic cell consists of two half-cells. The redox reaction occurs in these half-cells. The half-cell in which the reduction reaction occurs is known as the reduction half-cell, whereas the half-cell in which the oxidation reaction occurs is known as the oxidation half-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.

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

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

In order to know the number of electrons involved the reaction for silver button battery is determined first, which then the given zinc mass is converted into moles.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Zn}\left(\text{s}\right){\text{+2OH}}^{\text{-}}\left(\text{aq}\right)\to \text{ZnO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\\ {\text{Ag}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\to 2\text{Ag}\left(\text{s}\right)+{\text{2OH}}^{\text{-}}\left(\text{aq}\right)\end{array}}\\ \text{Zn}\left(\text{s}\right)+{\text{Ag}}_{\text{2}}\text{O}\left(\text{s}\right)\to \text{ZnO}\left(\text{s}\right)+2\text{Ag}\left(\text{s}\right)\\ \\ Znmoles=0.75gZn\left(\frac{80\%}{100\%}\right)\left(\frac{1molZn}{65.41gZn}\right)\\ =0.00917291mol\end{array}$

$\begin{array}{c}\text{No}\text{.}\text{of}\text{days}\text{=}\text{0}\text{.00917291}\text{mol}\text{Zn}\frac{\text{2}\text{mol}{\text{e}}^{\text{-}}}{\text{1}\text{mol}\text{Zn}}\left(\frac{\text{96485}\text{C}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\right)\left(\frac{\text{A}}{\text{C/s}}\right)\left(\frac{1\mu A}{{10}^{-6}A}\right)\left(\frac{1}{0.85\mu A}\right)\left(\frac{1h}{3600s}\right)\left(\frac{1day}{24h}\right)\\ =2.4\times {10}^{4}days\end{array}$

(b)

Interpretation Introduction

Interpretation:

The silver grams used to make the given batter 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.

Galvanic cell consists of two half-cells. The redox reaction occurs in these half-cells. The half-cell in which the reduction reaction occurs is known as the reduction half-cell, whereas the half-cell in which the oxidation reaction occurs is known as the oxidation half-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.

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

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

In order to know the number of electrons involved the reaction for silver button battery is determined first, which then the given zinc mass is converted into moles.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Zn}\left(\text{s}\right){\text{+2OH}}^{\text{-}}\left(\text{aq}\right)\to \text{ZnO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\\ {\text{Ag}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\to 2\text{Ag}\left(\text{s}\right)+{\text{2OH}}^{\text{-}}\left(\text{aq}\right)\end{array}}\\ \text{Zn}\left(\text{s}\right)+{\text{Ag}}_{\text{2}}\text{O}\left(\text{s}\right)\to \text{ZnO}\left(\text{s}\right)+2\text{Ag}\left(\text{s}\right)\\ \\ Znmoles=0.75gZn\left(\frac{80\%}{100\%}\right)\left(\frac{1molZn}{65.41gZn}\right)\\ =0.00917291mol\end{array}$

$\begin{array}{c}\text{Ag}\text{mass}\text{=}\text{0}\text{.00917291}\text{mol}\text{Zn}\frac{\text{1}\text{mol}{\text{Ag}}_{\text{2}}\text{O}}{\text{1}\text{mol}\text{Zn}}\left(\frac{\text{100}\text{\%}}{95\%}\right)\left(\frac{2\text{mol}Ag}{\text{1}\text{mol}{\text{Ag}}_{\text{2}}\text{O}}\right)\left(\frac{107.9gAg}{\text{1}\text{mol}Ag}\right)\\ =2.1g\end{array}$

(c)

Interpretation Introduction

Interpretation:

The cost of silver consumed for each day has to be identified.

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.

Galvanic cell consists of two half-cells. The redox reaction occurs in these half-cells. The half-cell in which the reduction reaction occurs is known as the reduction half-cell, whereas the half-cell in which the oxidation reaction occurs is known as the oxidation half-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.

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

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

In order to know the number of electrons involved the reaction for silver button battery is determined first, which then the given zinc mass is converted into moles.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Zn}\left(\text{s}\right){\text{+2OH}}^{\text{-}}\left(\text{aq}\right)\to \text{ZnO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\\ {\text{Ag}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\to 2\text{Ag}\left(\text{s}\right)+{\text{2OH}}^{\text{-}}\left(\text{aq}\right)\end{array}}\\ \text{Zn}\left(\text{s}\right)+{\text{Ag}}_{\text{2}}\text{O}\left(\text{s}\right)\to \text{ZnO}\left(\text{s}\right)+2\text{Ag}\left(\text{s}\right)\\ \\ Znmoles=0.75gZn\left(\frac{80\%}{100\%}\right)\left(\frac{1molZn}{65.41gZn}\right)\\ =0.00917291mol\end{array}$

$\begin{array}{c}\text{Cost}\text{=}2.1g\text{Ag}\left(\frac{95\%}{\text{100}\text{\%}}\right)\left(\frac{1troyoz}{\text{31}\text{.10}\text{g}\text{Ag}}\right)\left(\frac{\$23}{\text{1}troyoz}\right)\frac{1}{2.410262\times {10}^{4}days}\\ =\$6.1\times {10}^{-5}/day\end{array}$