Chapter 21, Problem 27A

Interpretation Introduction

Interpretation: The half-reactions that occur when an aluminum strip is dipped into copper (II) sulfate need to be determined.

Concept Introduction: In an overall cell reaction, there are two half-reactions. In one reaction, oxidation takes place and in another reaction, reduction takes place. Here, species with higher reduction electrode potential undergo reduction, and species with lower reduction electrode potential undergo oxidation.

Explanation of Solution

The reduction electrode potential for ${\text{Al}}^{3+}$ and ${\text{Cu}}^{2+}$ is -1.66 V and 0.34 V respectively. Since, the reduction electrode potential for ${\text{Cu}}^{2+}$ is higher than that of ${\text{Al}}^{3+}$ , the aluminium strip dipped into copper (II) sulfate can replace Cu from its solution and aluminum sulfate is formed.

The overall reaction can be written as follows:

  $\text{2Al}\left(s\right)+3{\text{CuSO}}_{\text{4}}\left(aq\right)\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+3\text{Cu}\left(s\right)$

Here, aluminium gets oxidized and copper gets reduced. The two half-reactions can be represented as follows:

Oxidation:

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

Reduction:

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