Chapter 21, Problem 55A

Interpretation Introduction

Interpretation: The result of placing an iron nail into a solution of copper (II) sulfate needs to be explained. The oxidation and reduction half-reactions (if possible) and the overall balanced equation for the reaction need to be determined.

Concept Introduction: If metals lose electron/s, oxidation takes place. If the activity series of metals is examined taking hydrogen as a reference, the metals above hydrogen can easily get oxidized and metals below hydrogen can easily get reduced (as they have more reduction potential value).

Explanation of Solution

On moving top to bottom in the activity series, the reduction electrode potential increases thus, metals can easily get reduced.

In comparing iron and copper, copper is placed below iron thus, it has more reduction electrode potential and it more easily gets reduced and oxidation of iron takes place.

Thus, if the iron nail is dipped into a solution of copper (II) sulfate, it can displace copper from its solution to form iron (II) sulfate, and a copper solid is formed.

Here, oxidation of iron and reduction of copper takes place thus, oxidation and reduction half-reactions can be written as follows:

Oxidation half-reaction:

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

Reduction half-reaction:

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

The overall reaction can be represented as follows:

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

Therefore, the overall reaction is as follows:

  $\text{Fe}\left(s\right)+{\text{Cu}}^{2+}\left(aq\right)\to {\text{Fe}}^{2+}\left(aq\right)+\text{Cu}\left(s\right)$