Chapter 21.2, Problem 17LC

Interpretation Introduction

Interpretation: Whether the given redox reaction will occur spontaneously or not needs to be determined.

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

Concept Introduction: Whether a metal can reduce other metals from its solution or not can be determined from its reduction electrode potential. The metal with a high reduction electrode potential cannot reduce the metal with a low reduction electrode potential.

Explanation of Solution

The given redox reaction is as follows:

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

The two half-reactions can be represented as follows:

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

Here, the reduction of Al and oxidation of Mg take place.

Now, the reduction electrode potential values of Al and Mg are as follows:

  $\begin{array}{l}{\text{Al}}^{3+}\left(aq\right)+3e\to \text{Al}\left(s\right)-\text{1}\text{.68 V}\\ \text{Mg}\left(s\right)\to {\text{Mg}}^{2+}\left(aq\right)+2e-2.\text{37 V}\end{array}$

Here, the reduction electrode potential for Al is more than Mg; thus, Al undergoes reduction and Mg undergoes oxidation.

Thus, the given reaction is spontaneous.