Q21.154CP - Chemistry: Molecular Nature Of Matter And Change

Question

Two voltaic cells are to be joined so that one will run the other as an electrolytic cell. In the first cell, one half-cell has $\mathbf{Au}$ foil in data-custom-editor="chemistry" $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{MAu}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}$ , and the other half-cell has a Cr bar in $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{MCr}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}$ . In the second cell, one half-cell has a Co bar in $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{MCo}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}$ , and the other half-cell has a  $\mathbf{Zn}$ bar in   - $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{MZn}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}$-

(a) Calculate data-custom-editor="chemistry" ${\mathbf{E}}_{\mathbf{cell}}^{\mathbf{o}}$  for each cell. 

(b) Calculate the total potential if the two cells are connected as voltaic cells in series. 

(c) When the electrode wires are switched in one of the cells, which cell will run as the voltaic cell and which as the electrolytic cell? 

(d) Which metal ion is being reduced in each cell? 

(e) If  $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{g}$ of metal plates out in the voltaic cell, how much metal ion plates out in the electrolytic cell?

Step-by-Step Solution

Answer

(a) The value of  $E_{cell}^o$ for each cell is $2.24\mathrm{V} \mathrm{and} 0.48\mathrm{V}.$

(b) The potential if the two cells are connected as voltaic cells in series is $2.72\mathrm{V}$.

(c) The value is $1.76V$.

(d) ${\mathrm{Au}}^{3+}$ and ${\mathrm{Co}}^{2+}$ metal ion is being reduced in each cell.

(e) The metal ion plates out in the electrolytic cell is $0.996\mathrm{gZn}$.

Step 1: Definition of Concept

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Step 2: Subpart (a)

Consider the reaction given below,

According to the information provided, two voltaic cells are combined. The half reactions are illustrated as follows: 

$$\begin{gathered} {\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Cr}\left(\mathrm{s}\right) {\mathrm{E}}^{\mathrm{o}}=-0.74 \mathrm{V} \\ {\mathrm{Co}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right) {\mathrm{E}}^{\mathrm{o}}=-0.28 \mathrm{V} \\ {\mathrm{Z}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.76 \mathrm{V} \end{gathered}$$

Using the formula ${\mathrm{E}}_{\mathrm{cell}}^o{=\mathrm{E}}_{\mathrm{cathode}}^o{-\mathrm{E}}_{\mathrm{anode}}^o$, cell potential is determined as follows,

$\begin{array}{c}\mathrm{Au}/\mathrm{Cr}\Rightarrow E_{\mathrm{cell}}^o=1.50 \mathrm{V}-(-0.74 \mathrm{V})\\ =2.24 V\end{array}$

for cell:

$\begin{array}{c}\mathrm{Co}/\mathrm{Zn}\Rightarrow {E^o}_{\mathrm{cell}}=-0.28 \mathrm{V}-(-0.76 \mathrm{V})\\ =0.48V\end{array}$

Therefore, the required value is  $2.24\mathrm{V} \mathrm{and} 0.48\mathrm{V}.$

Step 3: Subpart (b)

Consider the reaction given below,

According to the information provided, two voltaic cells are combined. The half reactions are illustrated as follows: 

${\mathrm{Au}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Au}\left(\mathrm{s}\right){\mathrm{E}}^o=1.50\mathrm{V}$

$$\begin{gathered} {\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Cr}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.74\mathrm{V} \\ {\mathrm{Co}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.28 \mathrm{V} \\ {\mathrm{Z}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.76 \mathrm{V} \end{gathered}$$

Using the formula ${\mathrm{E}}_{\mathrm{cell}}^o{=\mathrm{E}}_{\mathrm{cathode}}^o{-\mathrm{E}}_{\mathrm{anode}}^o$ , cell potential is determined as follows,

$$\begin{gathered} \mathrm{Au}/\mathrm{Cr}\Rightarrow E_{\mathrm{cell}}^o=1.50 \mathrm{V}-(-0.74 \mathrm{V}) \\ =2.24 \end{gathered}$$

$$\begin{gathered} \mathrm{Co}/\mathrm{Zn}\Rightarrow {E^o}_{\mathrm{cell}}=-0.28 \mathrm{V}-(-0.76 \mathrm{V}) \\ =0.48V \end{gathered}$$
When two cells are linked in series, the total cell potential is calculated by adding the potentials of the individual cells.

$\begin{array}{c}{\mathrm{E}}_{\mathrm{series}}^o{=\mathrm{E}}_{\mathrm{Au}/\mathrm{Cr}}^o+{\mathrm{E}}^o\mathrm{Cor}/\mathrm{n}\\ =2.24 \mathrm{V}+0.48 \mathrm{V}\\ =2.72 \mathrm{V}\end{array}$

Therefore, the required value is $2.72\mathrm{V}$.

Step 4: Subpart (c)

Consider the reaction given below,

When the given cells are connected in series, the voltages differ because (Au/Cr) is a voltaic cell and (Co/Zn) is an electrolytic cell. The voltage difference is as follows:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{series}}^o{=\mathrm{E}}_{\mathrm{Au}/\mathrm{Cr}}^o-{\mathrm{E}}^o\mathrm{Co}/\mathrm{Zn}\\ =2.24 \mathrm{V}-0.48 \mathrm{V}\\ =1.76 \mathrm{V}\end{array}$

Therefore, the required value is  $1.76\mathrm{V}$.

Step 5: Subpart (d)

Consider the reaction given below,

According to the information provided, two voltaic cells are combined. The half reactions are illustrated as follows: 

${\mathrm{Au}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Au}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=1.50\mathrm{V}$

$$\begin{gathered} {\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Cr}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.74\mathrm{V} \\ {\mathrm{Co}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.28 \mathrm{V} \\ {\mathrm{Z}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right){\mathrm{E}}^{\mathrm{o}}=-0.76 \mathrm{V} \end{gathered}$$

Analyzing the reduction potentials of  ${\mathrm{Au}}^{3+}$ and ${\mathrm{Cr}}^{3+}$   it is clear that   gets easily reduced since its reduction potential is higher than ${\mathrm{Cr}}^{3+}$.

Similarly, analyzing the reduction potentials of ${\mathrm{Co}}^{2+}$ and  ${\mathrm{Zn}}^{2+}$. it is clear that   gets easily reduced since its reduction potential is higher than ${\mathrm{Zn}}^{2+}$.

Step 6: Subpart (e)

Consider the reaction given below,

The voltage difference occurs when the given cells are connected in series, which connects (Au/Cr) as voltaic and  $(\mathrm{Co}/\mathrm{Zn})$  as electrolytic cell. The voltage difference is as follows:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{series}}^o{=\mathrm{E}}_{\mathrm{Au}/\mathrm{Cr}}^o-{\mathrm{E}}^o\mathrm{Co}/\mathrm{Zn}\\ =2.24 \mathrm{V}-0.48 \mathrm{V}\\ =1.76 \mathrm{V}\end{array}$

As a result of the above connection, $\mathrm{Co}/\mathrm{Zn}$ is forced to operate in the opposite direction, resulting in a reduction of ${\mathrm{Zn}}^{2+}$ .

As a result, the amount of zinc produced is calculated as follows:

$\begin{array}{c}{\mathrm{Zn}}_{\mathrm{Mass}}=2\mathrm{g} \mathrm{Au}\left(\frac{1\mathrm{molAu}}{197 \mathrm{gAu}}\right)\left(\frac{3 {\mathrm{mole}}^{-}}{1\mathrm{molAu}}\right)\left(\frac{1\mathrm{molZn}}{2 {\mathrm{mole}}^{-}}\right)\left(\frac{65.41 \mathrm{gZn}}{1\mathrm{molZn}}\right)\\ =0.996 \mathrm{g} \mathrm{Zn}\end{array}$