Chapter 18, Problem 32A

Interpretation Introduction

Interpretation: The galvanic cell for the given reaction with the details of metal ions oxidized and reduced, half-cell reaction on each metal electrode needs to be determined.

  ${\text{Zn}}_{\text{(s)}}{\text{+ Pb}}_{\text{(aq)}}^{\text{2+}}\stackrel{}{\to }{\text{Zn}}_{\text{(aq)}}^{\text{2+}}{\text{+ Pb}}_{\text{(s)}}$

Concept Introduction:In the electrochemical cell, the reactions at cathode and anode occur due to the difference in their reduction electrode potential value. The EMF of the cell can be calculated with help of electrode reduction potential values.

The reaction at each electrode is called as half-reaction and combination of both half-reaction gives the cell reaction of given electrochemical cell.

Answer to Problem 32A

  

Half-Reaction:

  $\begin{array}{l}{\text{Zn}}_{\text{(s)}}{\text{+ Pb}}_{\text{(aq)}}^{\text{2+}}\stackrel{}{\to }{\text{Zn}}_{\text{(aq)}}^{\text{2+}}{\text{+ Pb}}_{\text{(s)}}\\ {\text{Pb}}_{\text{(aq)}}^{\text{2+}}\stackrel{}{\to }{\text{ Pb}}_{\text{(s)}}\text{(reduction) at cathode}\\ {\text{Zn}}_{\text{(s)}}\stackrel{}{\to }{\text{ Zn}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation) at anode }\end{array}$

Explanation of Solution

Given information: ${\text{Zn}}_{\text{(s)}}{\text{+ Pb}}_{\text{(aq)}}^{\text{2+}}\stackrel{}{\to }{\text{Zn}}_{\text{(aq)}}^{\text{2+}}{\text{+ Pb}}_{\text{(s)}}$

In the given reaction, solid zinc involves in oxidation at anode whereas lead ions reduce at cathode.

  

The positive electrode is anode and negative electrode is cathode. Thus, half-reactions must be:

  $\begin{array}{l}{\text{Zn}}_{\text{(s)}}{\text{+ Pb}}_{\text{(aq)}}^{\text{2+}}\stackrel{}{\to }{\text{Zn}}_{\text{(aq)}}^{\text{2+}}{\text{+ Pb}}_{\text{(s)}}\\ {\text{Pb}}_{\text{(aq)}}^{\text{2+}}\stackrel{}{\to }{\text{ Pb}}_{\text{(s)}}\text{(reduction) at cathode}\\ {\text{Zn}}_{\text{(s)}}\stackrel{}{\to }{\text{ Zn}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation) at anode }\end{array}$

Conclusion

• Oxidation = at anode
• Reduction = at cathode