Chapter 13, Problem 13.38PAE

Interpretation Introduction

To determine: The standard free energy change for the cell:

$Ga(s)|G{a}^{3+}(aq)\Vert A{g}^{+}(aq)|Ag(s)$

Explanation of Solution

In this reaction,

Half-cell reaction at the anode

$Ga(s)\to G{a}^{3+}(aq)+3e^{-}$

Half-cell reaction at the cathode

$A{g}^{+}(aq)+e^{-}\to Ag(s)$

Complete balanced cell reaction is:

$Ga(s)+3A{g}^{+}(aq)\to G{a}^{3+}(aq)+3Ag(s)$

Number of electrons taking part in this reaction is n=3

Now,

$\begin{array}{l}{\text{E}}_{\text{right}}{\text{=E}}_{{\text{Ag}}^{\text{+}}\text{/Ag}}\text{=0}\text{.80 V}\\ {\text{E}}_{\text{left}}{\text{=E}}_{{\text{Ga}}^{\text{3+}}\text{/Ga}}\text{=-0}\text{.53 V}\\ \text{So,}\\ {\text{E}}_{\text{cell}}\text{=0}\text{.8-(-0}\text{.53)=1}\text{.33 V}\\ \text{Now,}\\ {\text{ΔG}}^{\text{0}}{\text{=-nFE}}_{\text{cell}}\text{=-3×96500×1}\text{.33=-385 kJ/mol}\end{array}$

(b)

Interpretation Introduction

To determine:

The standard free energy change for the cell:

$Zn(s)|Z{n}^{2+}(aq)\Vert C{r}^{3+}(aq)|Cr(s)$

Explanation of Solution

In this reaction,

Half-cell reaction at the anode

$Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$

Half-cell reaction at the cathode

$C{r}^{3+}(aq)+3e^{-}\to Cr(s)$

The complete balanced cell reaction

$3Zn(s)+2C{r}^{3+}(aq)\to 3Z{n}^{2+}(aq)+2Cr(s)$

Number of electrons taking part in this reaction is 6

Now,

$\begin{array}{l}{\text{E}}_{\text{right}}{\text{=E}}_{{\text{Cr}}^{\text{3+}}\text{/Cr}}\text{=-0}\text{.74 V}\\ {\text{E}}_{\text{left}}{\text{=E}}_{{\text{Zn}}^{\text{2+}}\text{/Zn}}\text{=-0}\text{.763 V}\\ \text{So,}\\ {\text{E}}_{\text{cell}}\text{=-0}\text{.74-(-0}\text{.763)=0}\text{.023 V}\\ \text{Now,}\\ {\text{ΔG}}^{\text{0}}{\text{=-nFE}}_{\text{cell}}\text{=-6×96500×0}\text{.023=-13}\text{.2 kJ/mol}\end{array}$

(c)

Interpretation Introduction

To determine:

The standard free energy change for the cell:

$Fe(s),FeS(s)|S^{2-}(aq)\Vert S{n}^{2+}(aq)|Sn(s)$

Explanation of Solution

In this reaction,

Half-cell reaction at the anode

$FeS(s)\to F{e}^{2+}(aq)+S^{2-}(aq)$

Half-cell reaction at the cathode

$S{n}^{2+}(aq)+2e^{-}\to Sn(s)$

Complete balanced cell reaction:

$FeS(s)+S{n}^{2+}(aq)\to F{e}^{2+}(aq)+S^{2-}(aq)+Sn(s)$

Number of electrons taking part in this reaction is 2

Now,

$\begin{array}{l}{\text{E}}_{\text{right}}{\text{=E}}_{{\text{Sn}}^{\text{2+}}\text{/Sn}}\text{=-0}\text{.14 V}\\ {\text{E}}_{\text{left}}{\text{=E}}_{{\text{Fe}}^{\text{2+}}\text{/Fe}}\text{=-0}\text{.57 V}\\ \text{So,}\\ {\text{E}}_{\text{cell}}\text{=-0}\text{.14-(-0}\text{.57)=0}\text{.43 V}\\ \text{Now,}\\ {\text{ΔG}}^{\text{0}}{\text{=-nFE}}_{\text{cell}}\text{=-2×96500×0}\text{.43= -83 kJ/mol}\end{array}$

Conclusion

Therefore, standard free energy changes are:

a. -385 kJ/mol,

b. -13.2 kJ/mol and

c. -83 kJ/mol