Chapter 21.2, Problem 10SP

Interpretation Introduction

Interpretation: The cell reaction for the given half-reactions of a voltaic cell needs to be determined.

Concept Introduction: In a voltaic cell, oxidation takes place at anode and reduction takes place at cathode. In oxidation, electrons are released and in reduction, electrons are added to the reactant. The oxidation state of the reactant increases during oxidation and it decreases during reduction.

Explanation of Solution

The two half-reactions for given voltaic cell are as follows:

  ${\mathrm{Cu}}^{2+}\left(aq\right)+2e^{-}\to \text{Cu}\left(s\right){\text{ E}}_{{\text{Cu}}^{\text{2+}}}^{\text{o}}=+0.3\text{4 V}$

  ${\text{Al}}^{3+}\left(aq\right)+3e^{-}\to \text{Al}\left(s\right){\text{ E}}_{{\text{Al}}^{\text{3+}}}^{\text{o}}=-1.66\text{ V}$

Now, the reduction potential for two reactions is given. The species with higher reduction potential will undergo reduction. Here, $E_{{\text{Cu}}^{\text{2+}}}^{\text{o}}>E_{{\text{Al}}^{\text{3+}}}^{\text{o}}$ thus, ${\text{Cu}}^{2+}$ will undergo reduction and $\text{Al}$ gets oxidized to form ${\text{Al}}^{3+}$ .

The two half reactions can be written as follows:

Anode-Oxidation reaction:

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

Cathode-Reduction reaction:

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

Overall cell reaction can be obtained as:

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

Therefore, the overall cell reaction is as follows:

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