Chapter 21, Problem 80A

(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.

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

Concept Introduction: In a redox reaction, reduction and oxidation take place simultaneously. Here, oxidation takes place at the anode and reduction takes place at the cathode.

Explanation of Solution

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

  $\begin{array}{l}\text{AgCl}\left(s\right)+e\to \text{Ag}\left(s\right)+{\text{Cl}}^{-}\left(aq\right)\text{ 0}\text{.22 V}\\ {\text{Ni}}^{2+}\left(aq\right)+2e\to \text{Ni}\left(s\right)\text{ -0}\text{.23 V}\end{array}$

Since the reduction electrode potential for the first reaction is higher than the second reduction of $\text{AgCl}\left(s\right)$ and oxidation of Ni takes place.

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

Cathode-Reduction reaction:

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

Anode-Oxidation reaction:

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

The overall cell reaction will be as follow:

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

Thus, the overall cell reaction will be:

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

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

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

Substitute the values,

  $\begin{array}{c}{E}^o=E_{\text{cathode}}-E_{\text{anode}}\\ =0.22\text{ V}-\left(-0.23\text{ V}\right)\\ =\left(0.22+0.23\right)\text{ V}\\ \text{=0}\text{.45 V}\end{array}$

Thus, the standard electrode potential for the voltaic cell is $\text{0}\text{.45 V}$ .

(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.

  $\begin{array}{l}{\text{Al}}^{3+}\left(aq\right)+3e\to \text{Al}\left(s\right)\\ {\text{Cl}}_2\left(g\right)+2e\to {\text{2Cl}}^{-}\left(aq\right)\end{array}$

Concept Introduction: In a redox reaction, reduction and oxidation take place simultaneously. Here, oxidation takes place at the anode and reduction takes place at the cathode.

Explanation of Solution

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

  $\begin{array}{l}{\text{Al}}^{3+}\left(aq\right)+3e\to \text{Al}\left(s\right)\text{ -1}\text{.66 V}\\ {\text{Cl}}_2\left(g\right)+2e\to {\text{2Cl}}^{-}\left(aq\right)\text{ 1}\text{.36 V}\end{array}$

Since, the reduction electrode potential for the second reaction is higher than the first, reduction of ${\text{Cl}}_2\left(g\right)$ and oxidation of $\text{Al}\left(s\right)$ takes place.

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

Cathode-Reduction reaction:

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

Anode-Oxidation reaction:

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

The overall cell reaction will be as follow:

  $\begin{array}{l}3\left({\text{Cl}}_2\left(g\right)+2e\to {\text{2Cl}}^{-}\left(aq\right)\right)\\ 2\left(\text{Al}\left(s\right)\to {\text{Al}}^{3+}\left(aq\right)+3e\right)\\ \overline{\underset{¯}{{\text{3Cl}}_2\left(g\right)+2\text{Al}\left(s\right)\to 6{\text{Cl}}^{-}\left(aq\right)+2{\text{Al}}^{3+}\left(aq\right)}}\end{array}$

Thus, the overall cell reaction will be:

  ${\text{3Cl}}_2\left(g\right)+2\text{Al}\left(s\right)\to 6{\text{Cl}}^{-}\left(aq\right)+2{\text{Al}}^{3+}\left(aq\right)$

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

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

Substitute the values,

  $\begin{array}{c}{E}^o=E_{\text{cathode}}-E_{\text{anode}}\\ =1.36\text{ V}-\left(-1.66\text{ V}\right)\\ =\left(1.36+1.66\right)\text{ V}\\ =\text{3}\text{.02 V}\end{array}$

Thus, the standard electrode potential for the voltaic cell is $\text{3}\text{.02 V}$ .