Chapter 13, Problem 13.22PAE

Using values from the table of standard reduction potentials, calculate the cell potentials of the following cells.

(a) $\text{Fe}\left(\text{s}\right)|{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\Vert {\text{Hg}}^{\text{2+}}\left(\text{aq}\right)|\text{Hg}\left(\mathcal{l}\right)$

(b) $\text{Pt}\left(\text{s}\right)|{\text{Fe}}^{\text{2+}}\left(\text{aq}\right),{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\Vert {\text{MnO}}_4{}^{-}\left(\text{aq}\right),{\text{H}}^{\text{+}}\left(\text{aq}\right),{\text{Mn}}^{2+}\left(\text{aq}\right)|\text{Pt}\left(\text{s}\right)$

(c) $\text{Pt}\left(\text{s}\right)|{\text{Cl}}_2\left(\text{g}\right)|{\text{Cl}}^{\text{-}}\left(\text{aq}\right)\Vert {\text{Au}}^{\text{+}}\left(\text{aq}\right)|\text{Au}\left(\text{s}\right)$

Interpretation Introduction

To determine:

The cell potential of the cell.

Explanation of Solution

The electrode potentials of the cell are given as:

$\begin{array}{l}{E}^o\left(F{e}^{2+}/Fe\right)=-0.44V\\ E^o\left(H{g}^{2+}/Hg\right)=0.85V\end{array}$

Since $E^0{}_{cell} =E^0{}_{cathode} -E^0{}_{anode}$

Hence,E⁰_cell = $0.85-\left(-0.44\right)=1.29V$

Since,the concentration of the solutions is not given, it is taken as 1 molar

So, in Nernst equation

$\text{log}\frac{\left[\text{product}\right]}{\left[\text{reactant}\right]}\text{=log }\frac{1}{1}\text{=0}$

i.e. ${\text{E}}_{\text{cell }}{\text{ = E}}^{\text{0}}{}_{\text{cell}}\text{= 1}\text{.29 V}$

Interpretation Introduction

To determine:

The cell potential of the cell.

Explanation of Solution

The electrode potentials of the cell are given as:

$\begin{array}{l}{E}^o\left(Mn{O}_4{}^{-}/M{n}^{2+}\right)=1.51V\hfill \\ E^o\left(F{e}^{3+}/F{e}^{2+}\right)=0.77V\hfill \end{array}$

Since, ${\text{E}}^{\text{0}}{}_{\text{cell}}{\text{ = E}}^{\text{0}}{}_{\text{cathode}} -{\text{ E}}^{\text{0}}{}_{\text{anode}}$

Hence, ${\text{E}}^{\text{0}}{}_{\text{cell}} =1.51–0.77=0.74\text{ V}$

${\text{E}}_{\text{cell }}{\text{ = E}}^{\text{0}}{}_{\text{cell}}\text{= 0}\text{.77 V}$

Interpretation Introduction

To determine:

The cell potential of the cell.

Explanation of Solution

The electrode potentials of the cell are given as:

$\begin{array}{l}{E}^o\left(A{u}^{+}/Au\right)=1.69V\hfill \\ E^o\left(C{l}_2/C{l}^{-}\right)=1.36V\hfill \end{array}$

Since ${\text{E}}^{\text{0}}{}_{\text{cell}}{\text{ = E}}^{\text{0}}{}_{\text{cathode}} -{\text{ E}}^{\text{0}}{}_{\text{anode}}$

Hence, ${\mathrm{E}}^0{}_{cell} =1.69-1.36=\text{ 0}\text{.33 V}$

${\text{E}}_{\text{cell }}{\text{ = E}}^{\text{0}}{}_{\text{cell}}\text{=0}\text{.33 V}$

Conclusion

${\text{E}}_{\text{cell }}\text{= 1}\text{.29 V}$

${\text{E}}_{\text{cell }}\text{= 0}\text{.77 V}$

${\text{E}}_{\text{cell }}\text{= 0}\text{.33 V}$