Chapter 21, Problem 63A

(a)

Interpretation Introduction

Interpretation: Whether the given cell reaction will proceed spontaneously or not needs to be determined.

  $\text{Zn}+{\text{Pb}}^{2+}\to ?$

Concept Introduction: The metal with high reduction electrode potential can easily get reduced and metal with low reduction electrode potential will undergo oxidation.

Explanation of Solution

The given cell reaction is as follows:

  $\text{Zn}+{\text{Pb}}^{2+}\to ?$

The above reaction shows the oxidation of Zn and reduction of ${\text{Pb}}^{2+}$ . Whether this reaction proceeds spontaneously or not depends on the reduction electrode potential of Zn and Pb. The metal with a high reduction electrode potential can easily get reduced and metal with a low reduction electrode potential will undergo oxidation.

The reduction electrode potential of ${\text{Zn}}^{2+}$ and ${\text{Pb}}^{2+}$ is -0.76 V and -0.13 V respectively. Thus, the reduction potential of ${\text{Pb}}^{2+}$ is higher than ${\text{Zn}}^{2+}$ . Here, Zn gets oxidized to form ${\text{Zn}}^{2+}$ and ${\text{Pb}}^{2+}$ gets reduced to Pb.

Therefore, the given reaction proceeds spontaneously.

(b)

Interpretation Introduction

Interpretation: Whether the given cell reaction will proceed spontaneously or not needs to be determined.

  $\text{Cu}+{\text{Fe}}^{2+}\to ?$

Concept Introduction: The metal with high reduction electrode potential can easily get reduced and metal with low reduction electrode potential will undergo oxidation.

Explanation of Solution

The given cell reaction is as follows:

  $\text{Cu}+{\text{Fe}}^{2+}\to ?$

The above reaction shows the oxidation of Cu and reduction of ${\text{Fe}}^{2+}$ . Whether this reaction proceeds spontaneously or not depends on the reduction electrode potential of Cu and Fe. The metal with a high reduction electrode potential can easily get reduced and metal with a low reduction electrode potential will undergo oxidation.

The reduction electrode potential of ${\text{Cu}}^{2+}$ and ${\text{Fe}}^{2+}$ is 0.34 V and -0.44 V respectively. Thus, the reduction potential of ${\text{Cu}}^{2+}$ is higher than ${\text{Fe}}^{2+}$ thus, reduction of ${\text{Cu}}^{2+}$ takes place and oxidation of Fe takes place. Therefore, the above reaction will not proceed spontaneously.

(c)

Interpretation Introduction

Interpretation: Whether the given cell reaction will proceed spontaneously or not needs to be determined.

  $\text{Ag}+{\text{Cu}}^{2+}\to ?$

Concept Introduction: The metal with high reduction electrode potential can easily get reduced and metal with low reduction electrode potential will undergo oxidation.

Explanation of Solution

The given cell reaction is as follows:

  $\text{Ag}+{\text{Cu}}^{2+}\to ?$

The above reaction shows the oxidation of Ag and reduction of ${\text{Cu}}^{2+}$ . Whether this reaction proceeds spontaneously or not depends on the reduction electrode potential of ${\text{Ag}}^{+}$ and ${\text{Cu}}^{2+}$ . The metal with a high reduction electrode potential can easily get reduced and metal with a low reduction electrode potential will undergo oxidation.

The reduction electrode potential of ${\text{Ag}}^{+}$ and ${\text{Cu}}^{2+}$ is 0.80 V and 0.34 V respectively. Since, the reduction electrode potential for ${\text{Ag}}^{+}$ is higher than ${\text{Cu}}^{2+}$ , ${\text{Ag}}^{+}$ gets reduced to Ag and Cu gets oxidized. The given reaction will not occur spontaneously.

(d)

Interpretation Introduction

Interpretation: Whether the given cell reaction will proceed spontaneously or not needs to be determined.

  ${\text{H}}_2+\text{Cu}\to ?$

Concept Introduction: The metal with high reduction electrode potential can easily get reduced and metal with low reduction electrode potential will undergo oxidation.

Explanation of Solution

The given cell reaction is as follows:

  ${\text{H}}_2+\text{Cu}\to ?$

The above reaction shows the oxidation of Cu and reduction of ${\text{H}}_{\text{2}}$ . Whether this reaction proceeds spontaneously or not depends on the reduction electrode potential of ${\text{Cu}}^{2+}$ and ${\text{H}}_{\text{2}}$ . The metal with a high reduction electrode potential can easily get reduced and metal with a low reduction electrode potential will undergo oxidation.

The reduction electrode potential of ${\text{Cu}}^{2+}$ and ${\text{H}}_{\text{2}}$ is 0.34 V and 0 V respectively. Since the reduction electrode potential of ${\text{Cu}}^{2+}$ is higher than ${\text{H}}_{\text{2}}$ thus, reduction of ${\text{Cu}}^{2+}$ takes place and ${\text{H}}_{\text{2}}$ gets oxidized.

The given reaction will not occur spontaneously.

(e)

Interpretation Introduction

Interpretation: Whether the given cell reaction will proceed spontaneously or not needs to be determined.

  $\text{Fe}+{\text{Pb}}^{2+}\to ?$

Concept Introduction: The metal with high reduction electrode potential can easily get reduced and metal with low reduction electrode potential will undergo oxidation.

Explanation of Solution

The given cell reaction is as follows:

  $\text{Fe}+{\text{Pb}}^{2+}\to ?$

The above reaction shows the oxidation of Fe and reduction of ${\text{Pb}}^{2+}$ . Whether this reaction proceeds spontaneously or not depends on the reduction electrode potential of ${\text{Fe}}^{2+}$ and ${\text{Pb}}^{2+}$ . The metal with a high reduction electrode potential can easily get reduced and metal with a low reduction electrode potential will undergo oxidation.

The reduction electrode potential of ${\text{Fe}}^{2+}$ and ${\text{Pb}}^{2+}$ is -0.44 V and -0.13 V respectively. Since the reduction electrode potential of ${\text{Pb}}^{2+}$ is higher than ${\text{Fe}}^{2+}$ , ${\text{Pb}}^{2+}$ will undergo reduction and oxidation of Fe takes place.

Therefore, the given cell reaction proceeds spontaneously.

(f)

Interpretation Introduction

Interpretation: Whether the given cell reaction will proceed spontaneously or not needs to be determined.

  $\text{Na}+{\text{Cl}}_2\to ?$

Concept Introduction: The metal with high reduction electrode potential can easily get reduced and metal with low reduction electrode potential will undergo oxidation.

Explanation of Solution

The given cell reaction is as follows:

  $\text{Na}+{\text{Cl}}_2\to ?$

The above reaction shows the oxidation of Na and reduction of ${\text{Cl}}_{\text{2}}$ . Whether this reaction proceeds spontaneously or not depends on the reduction electrode potential of Na and ${\text{Cl}}_{\text{2}}$ . The metal with a high reduction electrode potential can easily get reduced and metal with a low reduction electrode potential will undergo oxidation.

The reduction electrode potential of ${\text{Na}}^{+}$ and ${\text{Cl}}_{\text{2}}$ is -2.71 V and 1.36 V respectively. Since the reduction electrode potential of ${\text{Cl}}_{\text{2}}$ is higher than ${\text{Na}}^{+}$ thus, it will undergo reduction, and oxidation of $\text{Na}$ takes place.

Therefore, the given cell reaction proceeds spontaneously.