Question

Calculate the theoretical potential of each of the following cells. Is the cell reaction spontaneous as written or spontaneous in the opposite direction?

(a). $\mathrm{Bi}/{\mathrm{BiO}}^{+}(0.0300\mathrm{M}),{\mathrm{H}}^{+}(0.100\mathrm{M})//{\mathrm{I}}^{-}(0.100\mathrm{M}),\mathrm{AgI}(\mathrm{sat}\text{'}\mathrm{d})/\mathrm{Ag}$

(b). $\mathrm{Zn}/{\mathrm{Zn}}^{2+}(5.75\mathrm{x}{10}^{-4}\mathrm{M})//\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{4-}(5.50\mathrm{x}{10}^{-2}\mathrm{M}),\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{3-}(6.75\mathrm{x}{10}^{-2}\mathrm{M})/\mathrm{Pt}$

(c). $\mathrm{Pt},{\mathrm{H}}_2(0.200\mathrm{atm})/\mathrm{HCI}(8.25\mathrm{x}{10}^{-4}\mathrm{M}),\mathrm{AgCI}(\mathrm{sat}\text{'}\mathrm{d})/\mathrm{Ag}$

Short answer

(a). The theoretical potential is $-0.342\mathrm{V}$.

(b). The theoretical potential is $1.22\mathrm{V}$.

(c). The theoretical potential is $0.566\mathrm{V}$.

Step-by-step solution

Part (a) Step 1: Given information 

The electromotive force of a cell made up of two electrodes, or the potential difference formed between the metal electrode and the solution, is known as the electrode potential. The nernst equation is used to compute the electrode potential of a cell, which is written as E.

The nernst equation at ${25}^{°}C$

$E=E^{°}-\frac{0.0592}{n}\log \frac{\left(a_p\right)}{\left(a_R\right)}..........\left(1\right)$

The cell potential for the complete cell is calculated as:

$E_{cell}=E_{right}-E_{left}................\left(2\right)$

The sign of cell potential indicates whether the reaction of cell as written is quick.

Part (a) Step 2: Explanation

The cell contains two half-cells whose reaction and standard electrode reduction potential is as follow:

At left of cell:

${\mathrm{BiO}}^{+}+2{\mathrm{H}}^{+}+3^{\mathrm{e}-}\mathrm{Bi}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right){{\mathrm{E}}^{°}}_{\mathrm{left}}=0.320\mathrm{V}$

At right of cell:

$\mathrm{AgI}\left(\mathrm{s}\right)+{\mathrm{e}}^{-}\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{I}}^{-}{{\mathrm{E}}^{°}}_{\mathrm{right}}=-0.151\mathrm{V}$

The number of electrons involved in the half-cell reaction is$3.$

Part (a) Step 3: The Nernst equation for the half cell reaction at left

$E_{left}={E^{°}}_{left}-\frac{0.0592}{3}\log \frac{1}{{\left[H^{+}\right]}^2\left[Bi{O}^{+}\right]}$

localid="1646118276817" $$\begin{gathered} E_{left}=0.320\mathrm{V}-\frac{0.0592}{3}\log \frac{1}{{(0.100)}^2(0.0300)} \\ {E}_{left}=0.250\mathrm{V} \end{gathered}$$

Part (a) Step 4: The Nernst equation for the half cell reaction at the right

The number of electrons involved in half cell reaction at the right $1.$

localid="1646118340422" $$\begin{gathered} E_{right}={E^o}_{right}-\frac{0.0592}{1}\log \frac{\left[I^{-}\right]}{1} \\ {E}_{right}=-0.151\mathrm{V}-\frac{0.0592\mathrm{V}}{1}\log (0.100) \\ {E}_{right}=-0.092\mathrm{V} \end{gathered}$$

Part (a) Step 5: The theoretical potential for the cell

$$\begin{gathered} E_{cell}=E_{right}-E_{left} \\ {E}_{cell}=-0.092\mathrm{V}-0.250\mathrm{V} \\ {E}_{cell}=-0.342\mathrm{V} \end{gathered}$$

Part (b) Step 1: Given information 

The cell contains two half cells whose reactions and standard electrode reduction potential is:

At left of the cell:

${\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\mathrm{Zn}\left(\mathrm{s}\right){{\mathrm{E}}_{\mathrm{left}}}^{°}=-0.763\mathrm{V}$

At right of the cell:

$\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{3-}+{\mathrm{e}}^{-}\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{4-}{{\mathrm{E}}_{\mathrm{right}}}^{°}=0.360\mathrm{V}$

Considering the activity of solid compound as $1.$

Part (b) Step 2: The Nernst equation for the half cell reaction at left

The number of electrons in the half cell reaction is $2.$The Nernst equation for the half cell reaction at the left of the cell can be written as:

localid="1646118643709" $$\begin{gathered} E_{left}={E^{°}}_{left}-\frac{0.0592}{2}\log \frac{1}{Z{n}^{2+}} \\ {E}_{left}=-0.763\mathrm{V}-\frac{0.0592\mathrm{V}}{2}\log \frac{1}{(5.75\times {10}^{-4})} \\ {E}_{left}=-0.859\mathrm{V} \end{gathered}$$

Part (b) Step 3: The Nernst equation for the half cell reaction at the right

The number of electrons involved in a half cell at the right is $1.$

${\mathrm{E}}_{\mathrm{right}}={{\mathrm{E}}_{\mathrm{right}}}^{°}-\frac{0.0592}{1}\log \frac{\left[\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{4-}\right]}{\left[\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{3-}\right]}$

localid="1646118553416" $$\begin{gathered} E_{right}=0.360\mathrm{V}-\frac{0.0592}{1}\log \left[\frac{(5.50\times {10}^{-2})}{(6.75\times {10}^{-2})}\right] \\ {E}_{right}=0.365\mathrm{V} \end{gathered}$$

Part (b) Step 4: The theoretical potential for the cell

The theoretical potential for the cell:

$$\begin{gathered} E_{cell}=E_{right}-E_{left} \\ {E}_{cell}=0.365\mathrm{V}-\hspace{0.17em}(-0.859\mathrm{V}) \\ {E}_{cell}=1.22\mathrm{V} \end{gathered}$$

Part (c) Step 1: Given information  

The cell contains two half cells whose reaction and standard electrode reduction potential are as follow:

At left of the cell:

$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}{\mathrm{H}}_2\left(\mathrm{g}\right){{\mathrm{E}}_{\mathrm{left}}}^{°}=0.000\mathrm{V}$

At right of the cell:

$\mathrm{AgCI}\left(\mathrm{s}\right)+{\mathrm{e}}^{-}\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Cl}}^{-}{{\mathrm{E}}_{\mathrm{right}}}^{°}=0.222\mathrm{V}$

Considering the activity of gaseous hydrogen as equal to pressure of${\mathrm{H}}_2$ in atmospheric unit.

Part (c) Step 2: The Nernst equation for the half cell reaction at left

The number of electrons involved in the half cell reaction at the left is $2.$

localid="1646118758435" $E_{left}={E_{left}}^{°}-\frac{0.0592}{2}\log \frac{{\mathrm{p}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}$

localid="1646118789238" $$\begin{gathered} E_{left}=0.000\mathrm{V}-\frac{0.592\mathrm{V}}{2}\log \left[\frac{(0.200)}{{(8.25\times {10}^{-4})}^2}\right] \\ {E}_{left}=-0.162\mathrm{V} \end{gathered}$$

Part (c) Step 3:  The Nernst equation for the half cell reaction at the right

The number of electrons involved in the half cell reaction at the right is $1.$

${\mathrm{E}}_{\mathrm{right}}={{\mathrm{E}}_{\mathrm{right}}}^{°}-\frac{0.0592}{1}\log \frac{{\mathrm{Cl}}^{-}}{1}$

localid="1646118876025" $$\begin{gathered} E_{right}=0.222\mathrm{V}-\frac{0.0592\mathrm{V}}{1}\log (8.25\times {10}^{-4}) \\ {E}_{right}=0.404\mathrm{V} \end{gathered}$$

Part (c) Step 4: The theoretical potential for the cell

The theoretical potential for the cell is:

$$\begin{gathered} E_{cell}=E_{right}-E_{left} \\ {E}_{cell}=0.404\mathrm{V}-(-0.162\mathrm{V}) \\ {E}_{cell}=0.566\mathrm{V} \end{gathered}$$