Chapter 21, Problem 65A

Interpretation Introduction

Interpretation: For the given spontaneous reactions, an overall balanced equation needs to be given.

Concept Introduction: The overall cell reaction is the sum of two half-reactions. Here, two half-reactions are added in such a way that there are no electrons in the overall cell reaction.

Explanation of Solution

The given spontaneous reactions are as follows:

1. $\text{Zn}+{\text{Pb}}^{2+}\to ?$
2. $\text{Fe}+{\text{Pb}}^{2+}\to ?$
3. $\text{Na}+{\text{Cl}}_2\to ?$

The two half-reactions for the Zn and ${\text{Pb}}^{2+}$ reaction are as follows:

Anode: Oxidation reaction

  $\text{Zn}\to {\text{Zn}}^{2+}+2e$

Cathode: Reduction reaction

  ${\text{Pb}}^{2+}+2e\to \text{Pb}$

Thus, the overall cell reaction will be:

  $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{2+}+2e\\ {\text{Pb}}^{2+}+2e\to \text{Pb}\\ \overline{\underset{¯}{\text{Zn}+{\text{Pb}}^{2+}\to {\text{Zn}}^{2+}+\text{Pb}}}\end{array}$

Thus, the overall cell reaction is as follows:

  $\text{Zn}+{\text{Pb}}^{2+}\to {\text{Zn}}^{2+}+\text{Pb}$

The two half-reactions for the $\text{Fe}$ and ${\text{Pb}}^{2+}$ reaction are as follows:

Anode: Oxidation reaction

  $\text{Fe}\to {\text{Fe}}^{2+}+2e$

Cathode: Reduction reaction

  ${\text{Pb}}^{2+}+2e\to \text{Pb}$

Thus, the overall cell reaction will be:

  $\begin{array}{l}\text{Fe}\to {\text{Fe}}^{2+}+2e\\ {\text{Pb}}^{2+}+2e\to \text{Pb}\\ \overline{\underset{¯}{\text{Fe}+{\text{Pb}}^{2+}\to {\text{Fe}}^{2+}+\text{Pb}}}\end{array}$

Therefore, the overall cell reaction is as follows:

  $\text{Fe}+{\text{Pb}}^{2+}\to {\text{Fe}}^{2+}+\text{Pb}$

Now, two half-reactions for the $\text{Na}$ and ${\text{Cl}}_2$ reaction are as follows:

Anode: Oxidation reaction

  $\text{Na}\to {\text{Na}}^{+}+e$

Cathode: Reduction reaction

  ${\text{Cl}}_2+2e\to {\text{2Cl}}^{-}$

Thus, the overall cell reaction will be:

  $\begin{array}{l}2\left(\text{Na}\to {\text{Na}}^{+}+e\right)\\ {\text{Cl}}_2+2e\to {\text{2Cl}}^{-}\\ \overline{\underset{¯}{\text{2Na}+{\text{Cl}}_2\to 2{\text{Na}}^{+}+{\text{2Cl}}^{-}}}\end{array}$

Therefore, the overall cell reaction is as follows:

  $\text{2Na}+{\text{Cl}}_2\to 2{\text{Na}}^{+}+{\text{2Cl}}^{-}$