Chapter 21.2, Problem 18LC

Interpretation Introduction

Interpretation: The cell reaction and standard cell potential 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.

The standard cell potential is calculated as follows:

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

Here, $E_{\text{cathode}}$ is reduction potential of cathode and $E_{\text{anode}}$ is reduction potential of anode.

Explanation of Solution

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

  ${\text{Li}}^{+}\left(aq\right)+e^{-}\to \text{Li}\left(s\right){\text{ E}}_{{\text{Li}}^{\text{+}}}^{\text{o}}=-3.05\text{ V}$

  ${\text{Mg}}^{2+}\left(aq\right)+2e^{-}\to \text{Mg}\left(s\right){\text{ E}}_{{\text{Mg}}^{\text{2+}}}^{\text{o}}=-2.37\text{ V}$

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

The two half reactions can be written as follows:

Anode-Oxidation reaction:

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

Cathode-Reduction reaction:

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

Overall cell reaction can be obtained as:

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

Therefore, the overall cell reaction is as follows:

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

Now, standard cell potential can be calculated as follow:

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

Or,

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=-2.37\text{ V}-\left(-3.05\text{ V}\right)\\ =-2.37\text{ V}+\text{3}\text{.05 V}\\ =\text{0}\text{.68 V}\end{array}$

Thus, the standard cell potential is $\text{0}\text{.68 V}$ .