Chapter 20.1, Problem 12SSC

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the spontaneous cell reaction that occurs in a cell with the reduction half reaction given as follows:

${\text{Ag}}^{\text{+}}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Ag}\left(\text{s}\right)$ and ${\text{Ni}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Ni}\left(\text{s}\right)$

should be written.

Concept introduction:

Electrode potential is capacity of electrode to gain or lose electron when it is dipped in solution of its own ions. The absolute magnitude of cell potential of an electrode cannot be determined as oxidation half reaction or reduction half reaction cannot occur alone. It can be measured by taking a reference electrode. The reference electrode used is standard hydrogen electrode.

Oxidation potential is specific term used for cell potential if oxidation occurs at electrode and reduction potential is the term used if reduction occurs at electrode, with respect to standard hydrogen electrode.

An electrochemical cell is formed of two electrodes that is two half cells. One of these electrodes has higher electrode potential than the other due to which potential difference is created and current flows.

Answer to Problem 12SSC

Therefore, balanced equation for spontaneous cell reaction is ${\text{2Ag}}^{\text{+}}\left(\text{aq}\right)+\text{Ni}\left(\text{s}\right)\to 2\text{Ag}\left(\text{s}\right)+{\text{Ni}}^{\text{2+}}\left(\text{aq}\right)$.

Explanation of Solution

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reactions can be determined as follows:

Step1: The two half reactions are identified as follows:

${\text{Ag}}^{\text{+}}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Ag}\left(\text{s}\right)$

${\text{Ni}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Ni}\left(\text{s}\right)$

As per the latest convention of sign, the electrode at which reduction occurs with respect to standard hydrogen electrode is assigned positive sign or has higher reduction potential and the electrode at which oxidation occurs with respect to standard hydrogen electrode is assigned negative sign or has lower reduction potential.

Step 2: The electrode potential for two half reactions are compared.

As per table 20.1, standard potential for the half cell reactions are as follows:

$\begin{array}{l}{\text{Ag}}^{\text{+}}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Ag}\left(\text{s}\right);0.7996\text{V}\\ {\text{Ni}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Ni}\left(\text{s}\right);-0.257\text{V}\end{array}$

Since silver has positive electrode potential then nickel so reduction occurs at silver electrode and oxidation occurs at nickel electrode.

Step 3: Write oxidation half reaction in reverse manner and retain reduction half reaction as follows:

$\begin{array}{l}{\text{Ag}}^{\text{+}}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Ag}\left(\text{s}\right)\\ \text{Ni}\left(\text{s}\right)\to {\text{Ni}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\end{array}$

Step 4: Balance electrons in two half reaction by multiplying each by a factor, if required and then adding them.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{multiply}\text{by}{\text{2: 2Ag}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to 2\text{Ag}\left(\text{s}\right)\\ \text{ Ni}\left(\text{s}\right)\to {\text{Ni}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\end{array}}\\ \text{Add}\text{the}\text{equation}:{\text{2Ag}}^{\text{+}}\left(\text{aq}\right)+\text{Ni}\left(\text{s}\right)\to 2\text{Ag}\left(\text{s}\right)+{\text{Ni}}^{\text{2+}}\left(\text{aq}\right)\end{array}$

Therefore, balanced equation for spontaneous cell reaction is ${\text{2Ag}}^{\text{+}}\left(\text{aq}\right)+\text{Ni}\left(\text{s}\right)\to 2\text{Ag}\left(\text{s}\right)+{\text{Ni}}^{\text{2+}}\left(\text{aq}\right)$.

(b)

Interpretation Introduction

Interpretation:

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reaction given as follows:

${\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Mg}\left(\text{s}\right)$ and ${\text{2H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right)$

should be written.

Concept introduction:

Electrode potential is capacity of electrode to gain or lose electron when it is dipped in solution of its own ions. The absolute magnitude of cell potential of an electrode cannot be determined as oxidation half reaction or reduction half reaction cannot occur alone. It can be measured by taking a reference electrode. The reference electrode used is standard hydrogen electrode.

Oxidation potential is specific term used for cell potential if oxidation occurs at electrode and reduction potential is the term used if reduction occurs at electrode, with respect to standard hydrogen electrode.

An electrochemical cell is formed of two electrodes that is two half cells. One of these electrodes has higher electrode potential than the other due to which potential difference is created and current flows.

Answer to Problem 12SSC

Therefore, balanced equation for spontaneous cell reaction is $\text{Mg}\left(\text{s}\right)+{\text{2H}}^{\text{+}}\left(\text{aq}\right)\to {\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+{\text{H}}_2\left(\text{g}\right)$.

Explanation of Solution

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reactions can be determined as follows:

Step1: The two half reactions are identified as follows:

${\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Mg}\left(\text{s}\right)$

${\text{2H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right)$

As per the latest convention of sign, the electrode at which reduction occurs with respect to standard hydrogen electrode is assigned positive sign or has higher reduction potential and the electrode at which oxidation occurs with respect to standard hydrogen electrode is assigned negative sign or has lower reduction potential.

Step 2: The electrode potential for two half reactions are compared.

As per table 20.1, standard potential for the half cell reactions are as follows:

$\begin{array}{l}{\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Mg}\left(\text{s}\right);-2.372\text{V}\\ {\text{2H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right);-0.000\text{V}\end{array}$

Since hydrogen has higher electrode potential then magnesium so reduction occurs at hydrogen electrode and oxidation occurs at magnesium electrode.

Step 3: Write oxidation half reaction in reverse manner and retain reduction half reaction as follows:

$\begin{array}{l}\text{Mg}\left(\text{s}\right)\to {\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-};\\ {\text{2H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right);\end{array}$

Step 4: Balance electrons in two half reaction by multiplying each by a factor, if required and then adding them.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Mg}\left(\text{s}\right)\to {\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{2H}}^{\text{+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right)\end{array}}\\ \text{Add}\text{the}\text{equation}:\text{Mg}\left(\text{s}\right)+{\text{2H}}^{\text{+}}\left(\text{aq}\right)\to {\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+{\text{H}}_2\left(\text{g}\right)\end{array}$

Therefore, balanced equation for spontaneous cell reaction is $\text{Mg}\left(\text{s}\right)+{\text{2H}}^{\text{+}}\left(\text{aq}\right)\to {\text{Mg}}^{\text{2+}}\left(\text{aq}\right)+{\text{H}}_2\left(\text{g}\right)$.

(c)

Interpretation Introduction

Interpretation:

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reaction given as follows:

${\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Sn}\left(\text{s}\right)$ and ${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)$

should be written.

Concept introduction:

Electrode potential is capacity of electrode to gain or lose electron when it is dipped in solution of its own ions. The absolute magnitude of cell potential of an electrode cannot be determined as oxidation half reaction or reduction half reaction cannot occur alone. It can be measured by taking a reference electrode. The reference electrode used is standard hydrogen electrode.

Oxidation potential is specific term used for cell potential if oxidation occurs at electrode and reduction potential is the term used if reduction occurs at electrode, with respect to standard hydrogen electrode.

An electrochemical cell is formed of two electrodes that is two half cells. One of these electrodes has higher electrode potential than the other due to which potential difference is created and current flows.

Answer to Problem 12SSC

Therefore, balanced equation for spontaneous cell reaction is $\text{3Sn}\left(\text{s}\right)+{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\to 3{\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2\text{Fe}\left(\text{s}\right)$.

Explanation of Solution

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reactions can be determined as follows:

Step1: The two half reactions are identified as follows:

${\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Sn}\left(\text{s}\right)$

${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)$

As per the latest convention of sign, the electrode at which reduction occurs with respect to standard hydrogen electrode is assigned positive sign or has higher reduction potential and the electrode at which oxidation occurs with respect to standard hydrogen electrode is assigned negative sign or has lower reduction potential.

Step 2: The electrode potential for two half reactions are compared.

As per table 20.1, standard potential for the half cell reactions are as follows:

$\begin{array}{l}{\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Sn}\left(\text{s}\right);-0.1375\text{V}\\ {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right);-0.037\text{V}\end{array}$

Since iron has higher electrode potential then tin so reduction occurs at iron electrode and oxidation occurs at tin electrode.

Step 3: Write oxidation half reaction in reverse manner and retain reduction half reaction as follows:

$\begin{array}{l}\text{Sn}\left(\text{s}\right)\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)\end{array}$

Step 4: Balance electrons in two half reaction by multiplying each by a factor and then adding them.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{multiply}\text{by}3\text{: 3Sn}\left(\text{s}\right)\to 3{\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+6{\text{e}}^{-}\\ \text{multiply}\text{by}2{\text{: 2Fe}}^{\text{3+}}\left(\text{aq}\right)+6{\text{e}}^{-}\to 2\text{Fe}\left(\text{s}\right)\end{array}}\\ \text{Add}\text{the}\text{equation}:\text{3Sn}\left(\text{s}\right)+{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\to 3{\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2\text{Fe}\left(\text{s}\right)\end{array}$

Therefore, balanced equation for spontaneous cell reaction is $\text{3Sn}\left(\text{s}\right)+{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\to 3{\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2\text{Fe}\left(\text{s}\right)$.

(d)

Interpretation Introduction

Interpretation:

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reaction given as follows:

${\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{e}}^{-}\to \text{Pb}\left(\text{s}\right)$ and ${\text{Pt}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Pt}\left(\text{s}\right)$

should be written.

Concept introduction:

Electrode potential is capacity of electrode to gain or lose electron when it is dipped in solution of its own ions. The absolute magnitude of cell potential of an electrode cannot be determined as oxidation half reaction or reduction half reaction cannot occur alone. It can be measured by taking a reference electrode. The reference electrode used is standard hydrogen electrode.

Oxidation potential is specific term used for cell potential if oxidation occurs at electrode and reduction potential is the term used if reduction occurs at electrode, with respect to standard hydrogen electrode.

An electrochemical cell is formed of two electrodes that is two half cells. One of these electrodes has higher electrode potential than the other due to which potential difference is created and current flows.

Answer to Problem 12SSC

Therefore, balanced equation for spontaneous cell reaction is $\text{2Pb}\left(\text{s}\right)+{\text{Pt}}^{\text{2+}}\left(\text{aq}\right)\to 2{\text{PbI}}_{\text{2}}\left(\text{s}\right)+\text{Pt}\left(\text{s}\right)$.

Explanation of Solution

Balanced equation for spontaneous cell reaction that occurs in a cell with the reduction half reactions can be determined as follows:

Step1: The two half reactions are identified as follows:

${\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{e}}^{-}\to \text{Pb}\left(\text{s}\right)$

${\text{Pt}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Pt}\left(\text{s}\right)$

As per the latest convention of sign, the electrode at which reduction occurs with respect to standard hydrogen electrode is assigned positive sign or has higher reduction potential and the electrode at which oxidation occurs with respect to standard hydrogen electrode is assigned negative sign or has lower reduction potential.

Step 2: The electrode potential for two half reactions are compared.

As per table 20.1, standard potential for the half cell reactions are as follows:

$\begin{array}{l}{\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{e}}^{-}\to \text{Pb}\left(\text{s}\right);-0.365\text{V}\\ {\text{Pt}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Pt}\left(\text{s}\right);1.18\text{V}\end{array}$

Since platinum has positive electrode potential then lead so reduction occurs at platinum electrode and oxidation occurs at lead electrode.

Step 3: Write oxidation half reaction in reverse manner and retain reduction half reaction as follows:

$\begin{array}{l}\text{Pb}\left(\text{s}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{e}}^{-};\\ {\text{Pt}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Pt}\left(\text{s}\right);\end{array}$

Step 4: Balance electrons in two half reaction by multiplying each by a factor and then adding them.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{multiply}\text{by}\text{2: 2Pb}\left(\text{s}\right)\to 2{\text{PbI}}_{\text{2}}\left(\text{s}\right)+2{\text{e}}^{-}\\ {\text{ Pt}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Pt}\left(\text{s}\right)\end{array}}\\ \text{Add}\text{the}\text{equation}:\text{ 2Pb}\left(\text{s}\right)+{\text{Pt}}^{\text{2+}}\left(\text{aq}\right)\to 2{\text{PbI}}_{\text{2}}\left(\text{s}\right)+\text{Pt}\left(\text{s}\right)\end{array}$

Therefore, balanced equation for spontaneous cell reaction is $\text{2Pb}\left(\text{s}\right)+{\text{Pt}}^{\text{2+}}\left(\text{aq}\right)\to 2{\text{PbI}}_{\text{2}}\left(\text{s}\right)+\text{Pt}\left(\text{s}\right)$.