Chapter 21, Problem 44A

(a)

Interpretation Introduction

Interpretation: Whether the given redox reaction will occur spontaneously or not needs to be determined and the standard cell potential needs to be calculated.

  $\text{Cu}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

Concept Introduction: The overall cell reaction can be obtained by adding the two half-reactions. Here, in one half-reaction, oxidation takes place and in other half-reaction, reduction takes place.

Oxidation takes place at anode, here oxidation state of the species increases and removal of electrons takes place. Similarly, reduction takes place at cathode and oxidation state of species decreases as electrons are added to the reactant.

Explanation of Solution

The given cell reaction is as follows:

  $\text{Cu}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

From the above reaction, two half-cell reactions can be obtained as follows:

  $\begin{array}{l}\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)+2e\\ 2{\text{H}}^{+}\left(aq\right)+2e\to {\text{H}}_{\text{2}}\left(g\right)\\ \text{Or,}\\ {\text{H}}^{+}\left(aq\right)+e\to \frac{1}{2}{\text{H}}_{\text{2}}\left(g\right)\end{array}$

This indicates oxidation of Cu and reduction of hydrogen.

Now, reduction potentials for both the reactions are as follows:

  $\begin{array}{l}\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)+2e{\text{ E}}^{\text{o}}=0.34\text{ V}\\ {\text{H}}^{+}\left(aq\right)+e\to \frac{1}{2}{\text{H}}_{\text{2}}\left(g\right){\text{ E}}^{\text{o}}=0\text{ V}\end{array}$

Since reduction potential of copper is more than that of hydrogen, copper will show reduction. Thus, two possible half-cell reactions are as follows:

  $\begin{array}{l}{\text{Cu}}^{2+}\left(aq\right)+2e\to \text{Cu}\left(s\right){\text{ E}}^{\text{o}}=0.34\text{ V}\\ \frac{1}{2}{\text{H}}_{\text{2}}\left(g\right)\to {\text{H}}^{+}\left(aq\right)+e{\text{ E}}^{\text{o}}=0\text{ V}\end{array}$

Thus, the overall reaction will be:

  ${\text{Cu}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)\to \text{Cu}\left(s\right)+2{\text{H}}^{+}\left(aq\right)$

Therefore, the give redox reaction will not occur spontaneously.

(b)

Interpretation Introduction

Interpretation: Whether the given redox reaction will occur spontaneously or not needs to be determined and the standard cell potential needs to be calculated.

  $\text{2Ag}\left(s\right)+{\text{Fe}}^{2+}\left(aq\right)\to 2{\text{Ag}}^{+}\left(aq\right)+\text{Fe}\left(s\right)$

Concept Introduction: The overall cell reaction can be obtained by adding the two half-reactions. Here, in one half-reaction, oxidation takes place and in other half-reaction, reduction takes place.

Oxidation takes place at anode, here oxidation state of the species increases and removal of electrons takes place. Similarly, reduction takes place at cathode and oxidation state of species decreases as electrons are added to the reactant.

Explanation of Solution

The given redox reaction is as follows:

  $\text{2Ag}\left(s\right)+{\text{Fe}}^{2+}\left(aq\right)\to 2{\text{Ag}}^{+}\left(aq\right)+\text{Fe}\left(s\right)$

From the above reaction, two half-reactions take place as follows:

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

The reduction electrode potentials for the above two reactions are as follows:

  $\begin{array}{l}\text{Ag}\left(s\right)\to {\text{Ag}}^{+}\left(aq\right)+e{E}^o=0.80\text{ V}\\ {\text{Fe}}^{2+}\left(aq\right)+2e\to \text{Fe}\left(s\right)E^o=-0.45\text{ V}\end{array}$

Now, reduction electrode potential of Ag is more than Fe thus, Ag undergoes reduction and oxidation of Fe takes place.

Thus, the two half-reactions will be:

  $\begin{array}{l}{\text{Ag}}^{+}\left(aq\right)+e\to \text{Ag}\left(s\right)E^o=0.80\text{ V}\\ \text{Fe}\left(s\right)\to {\text{Fe}}^{2+}\left(aq\right)+2e{E}^o=-0.45\text{ V}\end{array}$

The overall cell reaction can be represented as follows:

  $2{\text{Ag}}^{+}\left(aq\right)+\text{Fe}\left(s\right)\to \text{2Ag}\left(s\right)+{\text{Fe}}^{2+}\left(aq\right)$

Therefore, the give redox reaction will not occur spontaneously.