Chapter 20.1, Problem 13SSC

(a)

Interpretation Introduction

Interpretation:

Standard cell potential for the overall reaction below should be calculated and whether the reaction is spontaneous or non spontaneous should be identified.

${\text{2Al}}^{\text{3+}}\left(\text{aq}\right)+\text{3Cu}\left(\text{s}\right)\to {\text{3Cu}}^{\text{2+}}\left(\text{aq}\right)+2\text{Al}\left(\text{s}\right)$

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 13SSC

Standard cell potential for ${\text{2Al}}^{\text{3+}}\left(\text{aq}\right)+\text{3Cu}\left(\text{s}\right)\to {\text{3Cu}}^{\text{2+}}\left(\text{aq}\right)+2\text{Al}\left(\text{s}\right)$ is $2.0039\text{V}$.

Since for redox reaction to be spontaneous, $E_{\text{cell}}^{\text{0}}$ must be greater than zero that is positive therefore, ${\text{2Al}}^{\text{3+}}\left(\text{aq}\right)+\text{3Cu}\left(\text{s}\right)\to {\text{3Cu}}^{\text{2+}}\left(\text{aq}\right)+2\text{Al}\left(\text{s}\right)$ is spontaneous.

Explanation of Solution

For the overall reaction,

${\text{2Al}}^{\text{3+}}\left(\text{aq}\right)+\text{3Cu}\left(\text{s}\right)\to {\text{3Cu}}^{\text{2+}}\left(\text{aq}\right)+2\text{Al}\left(\text{s}\right)$

The half cell reactions are as follows:

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

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

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.

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

$\begin{array}{l}{\text{Al}}^{\text{3+}}\left(\text{aq}\right)+{\text{3e}}^{-}\to \text{Al}\left(\text{s}\right);-1.662\text{V}\\ {\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cu}\left(\text{s}\right);0.3419\text{V}\end{array}$

Since copper has positive electrode potential then aluminum so reduction occurs at copper electrode and oxidation occurs at aluminum electrode.

The formula for cell potential is as follows:

$E_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}$

Where,

$E_{\text{cell}}^{\text{0}}$ is standard cell potential.

$E_{\text{Reduction}}^{\text{0}}$ is half cell potential for reduction.

$E_{\text{oxidation}}^{\text{0}}$ is half cell potential for oxidation.

Substitute $-1.662\text{V}$ for $E_{\text{oxidation}}^{\text{0}}$ and $0.3419\text{V}$ for $E_{\text{Reduction}}^{\text{0}}$ in above formula.

$\begin{array}{c}{E}_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}\\ =0.3419\text{V}-\left(-1.662\text{V}\right)\\ =2.0039\text{V}\end{array}$

Standard cell potential for ${\text{2Al}}^{\text{3+}}\left(\text{aq}\right)+\text{3Cu}\left(\text{s}\right)\to {\text{3Cu}}^{\text{2+}}\left(\text{aq}\right)+2\text{Al}\left(\text{s}\right)$ is $2.0039\text{V}$.

Since for redox reaction to be spontaneous, $E_{\text{cell}}^{\text{0}}$ must be greater than zero that is positive therefore, ${\text{2Al}}^{\text{3+}}\left(\text{aq}\right)+\text{3Cu}\left(\text{s}\right)\to {\text{3Cu}}^{\text{2+}}\left(\text{aq}\right)+2\text{Al}\left(\text{s}\right)$ is spontaneous.

(b)

Interpretation Introduction

Interpretation:

Standard cell potential for the overall reaction below should be calculated and whether the reaction is spontaneous or non spontaneous should be identified.

${\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{Cu}}^{+}\left(\text{aq}\right)\to 2{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+\text{Hg}\left(\text{l}\right)$

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 13SSC

Standard cell potential for ${\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{Cu}}^{+}\left(\text{aq}\right)\to 2{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+\text{Hg}\left(\text{l}\right)$ is $0.698\text{V}$.

Since for redox reaction to be spontaneous, $E_{\text{cell}}^{\text{0}}$ must be greater than zero that is positive therefore, ${\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{Cu}}^{+}\left(\text{aq}\right)\to 2{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+\text{Hg}\left(\text{l}\right)$ is spontaneous.

Explanation of Solution

For the overall reaction,

${\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{Cu}}^{+}\left(\text{aq}\right)\to 2{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+\text{Hg}\left(\text{l}\right)$

The half cell reactions are as follows:

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

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

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.

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

$\begin{array}{l}{\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Hg}\left(\text{l}\right);0.851\text{V}\\ {\text{Cu}}^{2+}\left(\text{aq}\right)+{\text{e}}^{-}\to 2{\text{Cu}}^{\text{+}}\left(\text{aq}\right);0.153\text{V}\end{array}$

Since mercury has positive electrode potential then copper so reduction occurs at mercury electrode and oxidation occurs at copper electrode.

The formula for cell potential is as follows:

$E_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}$

Where,

$E_{\text{cell}}^{\text{0}}$ is standard cell potential.

$E_{\text{Reduction}}^{\text{0}}$ is half cell potential for reduction.

$E_{\text{oxidation}}^{\text{0}}$ is half cell potential for oxidation.

Substitute $0.153\text{V}$ for $E_{\text{oxidation}}^{\text{0}}$ and $0.851\text{V}$ for $E_{\text{Reduction}}^{\text{0}}$ in above formula.

$\begin{array}{c}{E}_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}\\ =0.851\text{V}-\left(0.153\text{V}\right)\\ =0.698\text{V}\end{array}$

Standard cell potential for ${\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{Cu}}^{+}\left(\text{aq}\right)\to 2{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+\text{Hg}\left(\text{l}\right)$ is $0.698\text{V}$.

Since for redox reaction to be spontaneous, $E_{\text{cell}}^{\text{0}}$ must be greater than zero that is positive therefore, ${\text{Hg}}^{\text{2+}}\left(\text{aq}\right)+2{\text{Cu}}^{+}\left(\text{aq}\right)\to 2{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+\text{Hg}\left(\text{l}\right)$ is spontaneous.

(c)

Interpretation Introduction

Interpretation:

Standard cell potential for the overall reaction below should be calculated and whether the reaction is spontaneous or non spontaneous should be identified.

$\text{Cd}\left(\text{s}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+{\text{2NO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)$

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 13SSC

Standard cell potential for

$\text{Cd}\left(\text{s}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+{\text{2NO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)$ is $1.178\text{V}$.

Since for redox reaction to be spontaneous, $E_{\text{cell}}^{\text{0}}$ must be greater than zero that is positive therefore, $\text{Cd}\left(\text{s}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+{\text{2NO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)$ is

spontaneous.

Explanation of Solution

For the overall reaction,

$\text{Cd}\left(\text{s}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+{\text{2NO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)$

The half cell reactions are as follows:

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

${\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{2H}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to {\text{NO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)$

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.

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

$\begin{array}{l}{\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cd}\left(\text{s}\right);-0.4030\text{V}\\ {\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{2H}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to {\text{NO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{l}\right);0.775\text{V}\end{array}$

Since ${\text{NO}}_{\text{3}}{}^{-}$ has positive electrode potential then cadmium so reduction occurs at ${\text{NO}}_{\text{3}}{}^{-}$ copper electrode and oxidation occurs at cadmium electrode.

The formula for cell potential is as follows:

$E_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}$

Where,

$E_{\text{cell}}^{\text{0}}$ is standard cell potential.

$E_{\text{Reduction}}^{\text{0}}$ is half cell potential for reduction.

$E_{\text{oxidation}}^{\text{0}}$ is half cell potential for oxidation.

Substitute $-0.4030\text{V}$ for $E_{\text{oxidation}}^{\text{0}}$ and $0.775\text{V}$ for $E_{\text{Reduction}}^{\text{0}}$ in above formula.

$\begin{array}{c}{E}_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}\\ =0.775\text{V}-\left(-0.4030\text{V}\right)\\ =1.178\text{V}\end{array}$

Standard cell potential for

$\text{Cd}\left(\text{s}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+{\text{2NO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)$ is $1.178\text{V}$.

Since for redox reaction to be spontaneous, $E_{\text{cell}}^{\text{0}}$ must be greater than zero that is positive therefore, $\text{Cd}\left(\text{s}\right)+2{\text{NO}}_3{}^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to {\text{Cd}}^{\text{2+}}\left(\text{aq}\right)+{\text{2NO}}_2\left(\text{g}\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)$

is spontaneous.