Chapter 21, Problem 83A

(a)

Interpretation Introduction

Interpretation: For the given spontaneous redox reaction of a voltaic cell, anode and cathode electrodes needs to be identified.

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode.

Explanation of Solution

The given spontaneous redox reaction is as follows:

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

According to the above reaction, the reduction of ${\text{Ni}}^{2+}$ to Ni and oxidation of Fe to ${\text{Fe}}^{2+}\left(aq\right)$ takes place.

Since oxidation takes place at the anode and reduction occurs at the cathode; thus, at the anode, Fe is converted to ${\text{Fe}}^{2+}\left(aq\right)$ and at the cathode ${\text{Ni}}^{2+}$ is getting reduced to Ni.

(b)

Interpretation Introduction

Interpretation: For the given spontaneous redox reaction of a voltaic cell, charges to the electrodes need to be assigned.

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode.

Explanation of Solution

The charge on the electrode depends on the type of reaction taking place. Here, oxidation occurs at the anode thus, it will have negative potential with respect to the solution and reduction occurs at the cathode; thus, it will have positive potential with respect to the solution.

(c)

Interpretation Introduction

Interpretation: For the given spontaneous redox reaction of a voltaic cell, half reactions need to be written.

Concept Introduction: In oxidation, the oxidation state of metal increases, and loss of electron/s takes place. In reduction, the oxidation state of metal decreases, and the gain of electron/s takes place.

Explanation of Solution

The spontaneous redox reaction is as follows:

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

Here, reduction of ${\text{Ni}}^{2+}$ and oxidation of Fe takes place thus, two half-reactions will be:

Anode-Oxidation reaction:

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

Cathode-Reduction reaction:

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

(d)

Interpretation Introduction

Interpretation: For the given spontaneous redox reaction of a voltaic cell, the standard cell potential at standard conditions for half-cells needs to be calculated.

Concept Introduction: The standard electrode potential of the cell is calculated from the standard electrode potential of the cathode and anode as follows:

  $E^o=E_{\text{cathode}}-E_{\text{anode}}$

Explanation of Solution

For the given voltaic cell, two half-reactions and their respective reduction electrode potentials are as follows:

Anode:

  $\text{Fe}\left(s\right)\to {\text{Fe}}^{2+}\left(aq\right)+2e-\text{0}\text{.41 V}$

Cathode:

  ${\text{Ni}}^{2+}\left(aq\right)\to \text{Ni}\left(s\right)+2e-0.23\text{ V}$

The standard electrode potential can be calculated as follows:

  $E^{\text{o}}=E_{\text{cathode}}-E_{\text{anode}}$

Substitute the values,

  $\begin{array}{c}{E}^{\text{o}}=-0.23\text{ V}-\left(-\text{0}\text{.41 V}\right)\\ =-0.23\text{ V}+\text{0}\text{.41 V}\\ =0.18\text{ V}\end{array}$

Thus, the standard electrode potential of a given voltaic cell is 0.18 V.