Chapter 21, Problem 42A

Interpretation Introduction

Interpretation: The method to determine the standard reduction potential for the aluminum half-cell needs to be explained.

Concept Introduction: In electrochemistry, the relationship between electrical potential differences and potential differences due to chemical change is explained. The two types of cells that can be studied in electrochemistry are electrolytic cells and voltaic cells. In an electrolytic cell, redox reactions are produced with the help of current. Here, electrical energy is converted into chemical energy.

In an electrolyte or voltaic cell, a half-cell is half of the cell where either oxidation or reduction takes place. Here, two half-cells are the anode and cathode. At the anode, oxidation takes place and at the cathode, reduction takes place.

Explanation of Solution

The standard electrode potential of a half-cell can be calculated using a reference electrode or standard hydrogen electrode. To determine the standard reduction potential of the aluminum half-cell, the aluminum electrode is connected to the negative terminal and the hydrogen electrode is connected to the positive terminal. Here, the oxidation of aluminum takes place at the anode, and hydrogen ions are reduced to hydrogen gas at the cathode.

The two half-cell reactions can be written as follows:

Anode-Oxidation reaction:

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

Cathode-Reduction reaction:

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

Now, the overall cell reaction will be:

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

The overall reaction is as follows:

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

Now, cell potential can be measured from the aluminum-hydrogen cell. The reduction potential of the aluminum half-cell can be calculated from the electrode potential of the cell and standard hydrogen electrode potential as follows:

  $E_{cell}^o=E_{{\text{H}}^{+}}^{\text{o}}-E_{{\text{Al}}^{3+}}^{\text{o}}$

Or,

  $E_{cell}^o=0-E_{{\text{Al}}^{3+}}^{\text{o}}$

Thus,

  $E_{{\text{Al}}^{3+}}^{\text{o}}=-E_{cell}^o$

Therefore, the reduction electrode potential of aluminum is equal to the negative electrode potential of the cell.