Chapter 21, Problem 72A

(a)

Interpretation Introduction

Interpretation: The overall cell reaction and standard electrode potential for the voltaic cell with the following half-reactions need to be determined.

  $\begin{array}{l}{\text{Ag}}^{+}\left(aq\right)+e\to \text{Ag}\left(s\right)\\ {\text{Cr}}^{3+}\left(aq\right)+3e\to \text{Cr}\left(s\right)\end{array}$

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. The overall reaction is the sum of two half-reactions. The standard electrode potential of the cell is calculated from the standard electrode potential of the cathode and anode as follows:

  $E^o=E_{\text{cathode}}-E_{\text{anode}}$

Explanation of Solution

The reduction electrode potentials for the given two half-reactions are as follows:

  $\begin{array}{l}{\text{Ag}}^{+}\left(aq\right)+e\to \text{Ag}\left(s\right)\text{ 0}\text{.80 V}\\ {\text{Cr}}^{3+}\left(aq\right)+3e\to \text{Cr}\left(s\right)\text{ -0}\text{.74 V}\end{array}$

Since the reduction electrode potential for the first reaction is higher than a second reduction of ${\text{Ag}}^{+}$ and oxidation of Cr takes place.

The reactions at the cathode and anode will be as follow:

Cathode-Reduction reaction:

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

Anode-Oxidation reaction:

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

The overall cell reaction will be as follow:

  $\begin{array}{l}3\left({\text{Ag}}^{+}\left(aq\right)+e\to \text{Ag}\left(s\right)\right)\\ \text{Cr}\left(s\right)\to {\text{Cr}}^{3+}\left(aq\right)+3e\\ \overline{\underset{¯}{{\text{3Ag}}^{+}\left(aq\right)+\text{Cr}\left(s\right)\to 3\text{Ag}\left(s\right)+{\text{Cr}}^{3+}\left(aq\right)}}\end{array}$

Thus, the overall cell reaction will be:

  ${\text{3Ag}}^{+}\left(aq\right)+\text{Cr}\left(s\right)\to 3\text{Ag}\left(s\right)+{\text{Cr}}^{3+}\left(aq\right)$

The standard electrode potential of the cell is calculated as follows:

  $E^o=E_{\text{cathode}}-E_{\text{anode}}$

Substitute the values,

  $\begin{array}{c}{E}^o=E_{\text{cathode}}-E_{\text{anode}}\\ =0.80\text{ V}-\left(-0.74\text{ V}\right)\\ =\left(0.80+0.74\right)\text{ V}\\ \text{=1}\text{.54 V}\end{array}$

Thus, the standard electrode potential for the voltaic cell is $\text{1}\text{.54 V}$ .

(b)

Interpretation Introduction

Interpretation: The overall cell reaction and standard electrode potential for the voltaic cell with the following half-reactions need to be determined.

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

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. The overall reaction is the sum of two half-reactions. The standard electrode potential of the cell is calculated from the standard electrode potential of the cathode and anode as follows:

  $E^o=E_{\text{cathode}}-E_{\text{anode}}$

Explanation of Solution

The reduction electrode potentials for the given two half-reactions are as follows:

  $\begin{array}{l}{\text{Al}}^{3+}\left(aq\right)+3e\to \text{Al}\left(s\right)\text{ -1}\text{.66 V}\\ {\text{Cd}}^{2+}\left(aq\right)+2e\to \text{Cd}\left(s\right)\text{ 0}\text{.40 V}\end{array}$

Since the reduction electrode potential for the second reaction is higher than the first, reduction of ${\text{Cd}}^{2+}$ and oxidation of Al takes place.

The reactions at the cathode and anode will be as follow:

Cathode-Reduction reaction:

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

Anode-Oxidation reaction:

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

The overall cell reaction will be as follow:

  $\begin{array}{l}3\left({\text{Cd}}^{2+}\left(aq\right)+2e\to \text{Cd}\left(s\right)\right)\\ 2\left(\text{Al}\left(s\right)\to {\text{Al}}^{3+}\left(aq\right)+3e\right)\\ \overline{\underset{¯}{{\text{3Cd}}^{2+}\left(aq\right)+2\text{Al}\left(s\right)\to 3\text{Cd}\left(s\right)+2{\text{Al}}^{3+}\left(aq\right)}}\end{array}$

Thus, the overall cell reaction will be:

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

The standard electrode potential of the cell is calculated as follows:

  $E^o=E_{\text{cathode}}-E_{\text{anode}}$

Substitute the values,

  $\begin{array}{c}{E}^o=E_{\text{cathode}}-E_{\text{anode}}\\ =0.40\text{ V}-\left(-1.66\text{ V}\right)\\ =\left(0.80+1.66\right)\text{ V}\\ \text{=2}\text{.46 V}\end{array}$

Thus, the standard electrode potential for the voltaic cell is $\text{2}\text{.46 V}$ .