Chapter 21, Problem 64A

Interpretation Introduction

Interpretation: For spontaneous reactions in the previous problem, half-reaction for the anode and cathode needs to be determined.

Concept Introduction: The two half-reactions in cells are oxidation and reduction reactions. Here, oxidation takes place at the anode and reduction takes place at the cathode. In oxidation, the oxidation state of species increases, and reduction, the oxidation state of species decreases.

Explanation of Solution

The spontaneous reactions are as follows:

  $\text{Zn}+{\text{Pb}}^{2+}\to ?$

  $\text{Fe}+{\text{Pb}}^{2+}\to ?$

  $\text{Na}+{\text{Cl}}_2\to ?$

Now, at the anode oxidation takes place and reduction takes place at the cathode. 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

  ${\text{Pb}}^{2+}+2e\to \text{Pb}$

Similarly, for the reaction of $\text{Fe}$ and ${\text{Pb}}^{2+}$:

Anode: Oxidation reaction

  $\text{Fe}\to {\text{Fe}}^{2+}+2e$

Cathode: Reduction reaction

  ${\text{Pb}}^{2+}+2e\to \text{Pb}$

Now, for $\text{Na}$ and ${\text{Cl}}_2$:

Anode: Oxidation reaction

  $\text{Na}\to {\text{Na}}^{+}+e$

Cathode: Reduction reaction

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