Chapter 18.2, Problem 6RQ

Interpretation Introduction

Interpretation: The balanced redox equation for the reaction between ${\text{Cu}}_{\text{(s)}}$ and ${\text{Ag}}_{\text{(aq)}}^{\text{+}}$ needs to be determined.

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved in loss or gain of electron. Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.

Answer to Problem 6RQ

  ${\text{2 Ag}}_{\text{(aq)}}^{\text{+}}{\text{+Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Ag}}_{\text{(s)}}{\text{ + Cu}}_{\text{(aq)}}^{\text{2+}}$

Explanation of Solution

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

Assign oxidation number to each atom:

  $\begin{array}{l}{\text{Cu}}_{\text{(s)}}{\text{+ Ag}}_{\text{(aq)}}^{\text{+}}\stackrel{}{\to }{\text{Cu}}_{\text{(aq)}}^{\text{2+}}{\text{+ Ag}}_{\text{(s)}}\\ \text{0 +1 +2 0}\end{array}$

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of Ag decreases from +1 to 0. Thus, the reduction half-reaction can be written as;

  ${\text{ Ag}}_{\text{(aq)}}^{\text{+}}\stackrel{}{\to }{\text{ Ag}}_{\text{(s)}}$

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of Cu increases from 0 to +2. Thus, the oxidation half-reaction can be written as;

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

Balance charges:

  $\begin{array}{l}{\text{ Ag}}_{\text{(aq)}}^{\text{+}}{\text{+e}}^{\text{-}}\stackrel{}{\to }{\text{ Ag}}_{\text{(s)}}\text{ (Reduction)}\\ {\text{Cu}}_{\text{(s)}}\text{ + 2 e-}\stackrel{}{\to }{\text{ Cu}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation)}\end{array}$

Step-5:

Add both equation:

  $\begin{array}{l}{\text{ Ag}}_{\text{(aq)}}^{\text{+}}{\text{+e}}^{\text{-}}\stackrel{}{\to }{\text{ Ag}}_{\text{(s)}}\text{ (Reduction)}\\ {\text{Cu}}_{\text{(s)}}\text{ + 2 e-}\stackrel{}{\to }{\text{ Cu}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{ 2 Ag}}_{\text{(aq)}}^{\text{+}}{\text{+Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Ag}}_{\text{(s)}}{\text{ + Cu}}_{\text{(aq)}}^{\text{2+}}\end{array}$

Thus, the balance equation is:

  ${\text{2 Ag}}_{\text{(aq)}}^{\text{+}}{\text{+Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Ag}}_{\text{(s)}}{\text{ + Cu}}_{\text{(aq)}}^{\text{2+}}$

Conclusion

  ${\text{2 Ag}}_{\text{(aq)}}^{\text{+}}{\text{+Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Ag}}_{\text{(s)}}{\text{ + Cu}}_{\text{(aq)}}^{\text{2+}}$