Chapter 13, Problem 13.32PAE

The following half-cells are available: Ag(s); ${\text{Ag}}^{\text{+}}\left(\text{aq,}\text{1}\text{.0}\text{M}\right)|\text{Ag};{\text{Zn}}^{\text{2}+}\left(\text{aq},\text{1}.0\text{ M}\right)|\text{Zn}\left(\text{s}\right);{\text{ Cu}}^{\text{2}+}\left(\text{aq},\text{ 1}.0\text{ M}\right)|\text{Cu}\left(\text{s}\right)$ ; and ${\text{Co}}^{\text{2}+}\left(\text{aq},\text{1}.0\text{ M}\right)|\text{Co}\left(\text{s}\right)$ . Linking any two half-cells makes a voltaic cell. Given four different half-cells, six voltaic cells are possible. These are labeled, for simplicity, Ag-Zn, Ag-Cu, Ag-Co, Zn-Cu, Zn-Co, and Cu-Co.

(a) In which of the voltaic cells does the copper electrode serve as the cathode? In which of the voltaic cells does the cobalt electrode serve as the anode?

(b) Which combination of half-cells generates the highest potential? Which combination generates the lowest potential?

Interpretation Introduction

To determine:

The nature of Cu and Co in the different cells

Explanation of Solution

The electrode potentials are given as:

$\begin{array}{l}{\text{E}}^{\text{o}}\left({\text{Zn}}^{\text{2+}}\text{/Zn}\right)=-0.\text{76 V}\\ {\text{E}}^{\text{o}}\left({\text{Co}}^{\text{2+}}\text{/Co}\right)=-\text{0}\text{.28 V}\\ {\text{E}}^{\text{o}}\left({\text{Ag}}^{\text{+}}\text{/Ag}\right)=+0.80\text{ V}\\ {\text{E}}^o{\text{(Cu}}^{\text{2+}}\text{/Cu)}=0.\text{34 V}\end{array}$

Since in $\text{Zn}-\text{Cu}$ and $\text{Co}-\text{Cu}$ cell, copper has more reduction potential than $\text{Zn}$ and $\text{Co}$, hence in these cells $\text{Cu}$ act as cathode.

Similarly, in $\text{Co}-\text{Ag}$ and $\text{Co}-\text{Cu}$ cell, cobalt has less reduction potential than $\text{Ag}$ and Cu, hence in these cells $\text{Co}$ acts as anode.

Interpretation Introduction

To determine:

The highest and the lowest cell potential

Explanation of Solution

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

The reactions of the six voltaic cells are:

$\text{Zn}\left(\text{s}\right){\text{ + 2Ag}}^{\text{+}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)\text{ + 2Ag}\left(\text{s}\right), {\text{ E}}^{\text{0}}{}_{\text{cell}}\text{= 0}\text{.80}-\left(-0.76\right)=1.\text{56 V}$

$\text{Cu}\left(\text{s}\right){\text{ + 2Ag}}^{\text{+}}\left(\text{aq}\right)\to {\text{Cu}}^{\text{2+}}\left(\text{aq}\right)\text{ + 2Ag}\left(\text{s}\right), {\text{ E}}^{\text{0}}{}_{\text{cell}}\text{= }0.80-0.34=0.\text{46 V}$

$\mathrm{Co}\left(s\right) + 2A{g}^{+}\left(aq\right) \to C{o}^{2+}\left(aq\right) +2Ag\left(s\right), {\text{ E}}^{\text{0}}{}_{\text{cell}}\text{= 0}\text{.80 }–\left(-0.28\right)=\text{ 1}\text{.08 V}$

$\text{Zn}\left(\text{s}\right){\text{ + Cu}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)\text{ + Cu}\left(\text{s}\right)\text{,} {\text{ E}}^{\text{0}}{}_{\text{cell}}\text{= 0}\text{.34}-\left(-0.76\right)=\text{ 1}\text{.10 V}$

$\text{Zn}\left(\text{s}\right){\text{ + Co}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)\text{ + Co}\left(\text{s}\right), {\text{E}}^{\text{0}}{}_{\text{cell}}\text{= }-0.28–\left(-0.76\right)=\text{0}\text{.48 V}$

$\text{Co}\left(\text{s}\right){\text{ + Cu}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Co}}^{\text{2+}}\left(\text{aq}\right)\text{ + Cu}\left(\text{s}\right)\text{, } {\text{E}}^{\text{0}}{}_{\text{cell}}\text{=0}\text{.34}-\left(-0.28\text{ V}\right)=\text{ 0}\text{.62 V}$

Since, Nernst equation at 298K is given by:

${\text{E}}_{\text{cell }}={\text{ E}}^{\text{0}}{}_{\text{cell}}–\left(\text{0}\text{.0591/n}\right)\text{ log }\frac{\left[\text{product}\right]}{\left[\text{reactant}\right]}$

Hence,for the first cell

${\text{E}}_{\text{cell }}\text{ = 1}\text{.56 }–\left(0.0591/2\right)\text{ log}\left(\left[\text{1}\right]\text{/}\left[\text{1}\right]\right)$

$=\text{ 1}\text{.56 V}$

Similarly,

${\text{E}}_{\text{cell}}\text{= 0}\text{.46 V}$

${\text{E}}_{\text{cell}}\text{= 1}\text{.08 V}$

${\text{E}}_{\text{cell}}\text{= 1}\text{.10 V}$

${\text{E}}_{\text{cell}}\text{= 0}\text{.48 V}$

${\text{E}}_{\text{cell}}\text{= 0}\text{.62 V}$

From the above solution:

Highest cell potential: $\text{Zn}-\text{Ag}$

1. cell

Lowest cell potential: $\text{Cu }–\text{ Ag}$

2. cell

Conclusion

Hence in the given cells, $\text{Cu}$ act as the cathode and $\text{Co}$ acts as the anode. The $\text{Zn}-\text{Ag}$ cell has the highest cell potential while the $\text{Cu }–\text{ Ag}$ cell has the lowest cell potential