Chapter 21, Problem 45A

(a)

Interpretation Introduction

Interpretation: The standard cell potentials for a given voltaic cell need to be calculated.

  $\text{Zn}\left|{\text{Zn}}^{2+}\right|\left|{\text{Cu}}^{2+}\right|\text{Cu}$

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. In oxidation, electrons are released and in reduction, electrons are added to the reactant. The oxidation state of the reactant increases during oxidation and decreases during reduction.

Explanation of Solution

The given voltaic cell is as follows:

  $\text{Zn}\left|{\text{Zn}}^{2+}\right|\left|{\text{Cu}}^{2+}\right|\text{Cu}$

From the above cell notation, it can be seen that oxidation of Zn takes place and reduction of Cu takes place. The two half-reactions can be represented as follows:

  $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{2+}+2e\\ {\text{Cu}}^{2+}+2e\to \text{Cu}\end{array}$

The standard reduction electrode potential for the above two reactions is as follows:

  $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{2+}+2e{E}_{}^{\text{o}}=-0.76\text{ V}\\ {\text{Cu}}^{2+}+2e\to \text{Cu }{E}_{}^{\text{o}}=+0.34\text{ V}\end{array}$

The standard cell potential can be calculated as follows:

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

Substitute the values,

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=+0.34\text{ V}-\left(-0.76\text{ V}\right)\\ =\left(0.34+0.76\right)\text{ V}\\ =1.1\text{ V}\end{array}$

Therefore, the standard cell potential is $1.1\text{ V}$ .

(b)

Interpretation Introduction

Interpretation: The standard cell potentials for a given voltaic cell need to be calculated.

  $\text{Ni}\left|{\text{Ni}}^{2+}\right|\left|{\text{Cl}}_2\right|{\text{Cl}}^{-}$

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. In oxidation, electrons are released and in reduction, electrons are added to the reactant. The oxidation state of the reactant increases during oxidation and decreases during reduction.

Explanation of Solution

The given voltaic cell is as follows:

  $\text{Ni}\left|{\text{Ni}}^{2+}\right|\left|{\text{Cl}}_2\right|{\text{Cl}}^{-}$

From the above cell notation, it can be seen that oxidation of Ni takes place and reduction of Cl takes place. The two half-reactions can be represented as follows:

  $\begin{array}{l}\text{Ni}\to {\text{Ni}}^{2+}+2e\\ {\text{Cl}}_2+2e\to 2{\text{Cl}}^{-}\end{array}$

The standard reduction electrode potential for the above two reactions is as follows:

  $\begin{array}{l}\text{Ni}\to {\text{Ni}}^{2+}+2e{E}_{}^{\text{o}}=-0.257\text{ V}\\ {\text{Cl}}_2+2e\to 2{\text{Cl}}^{-}{E}_{}^{\text{o}}=+1.396\text{ V}\end{array}$

The standard cell potential can be calculated as follows:

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

Substitute the values,

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=+1.396\text{ V}-\left(-0.257\text{ V}\right)\\ =\left(1.396+0.257\right)\text{ V}\\ =1.653\text{ V}\end{array}$

Therefore, the standard cell potential is $1.653\text{ V}$ .

(c)

Interpretation Introduction

Interpretation: The standard cell potentials for a given voltaic cell need to be calculated.

  $\text{Sn}\left|{\text{Sn}}^{2+}\right|\left|{\text{Ag}}^{+}\right|\text{Ag}$

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. In oxidation, electrons are released and in reduction, electrons are added to the reactant. The oxidation state of the reactant increases during oxidation and decreases during reduction.

Explanation of Solution

The given voltaic cell is as follows:

  $\text{Sn}\left|{\text{Sn}}^{2+}\right|\left|{\text{Ag}}^{+}\right|\text{Ag}$

From the above cell notation, it can be seen that oxidation of Sn takes place and reduction of Ag takes place. The two half-reactions can be represented as follows:

  $\begin{array}{l}\text{Sn}\to {\text{Sn}}^{2+}+2e\\ {\text{Ag}}^{+}+e\to \text{Ag}\end{array}$

The standard reduction electrode potential for the above two reactions is as follows:

  $\begin{array}{l}\text{Sn}\to {\text{Sn}}^{2+}+2e{E}_{}^{\text{o}}=-0.136\text{ V}\\ {\text{Ag}}^{+}+e\to \text{Ag }{E}_{}^{\text{o}}=+0.7996\text{ V}\end{array}$

The standard cell potential can be calculated as follows:

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

Substitute the values,

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=+0.7996\text{ V}-\left(-0.136\text{ V}\right)\\ =\left(0.7996+0.136\right)\text{ V}\\ =0.9356\text{ V}\end{array}$

Therefore, the standard cell potential is $0.9356\text{ V}$ .