Question

Calculate the theoretical potential of each of the following cells. Is the cell reaction spontaneous as written or spontaneous in the opposite direction?$\left(a\right)Bi\mid {BiO}^{+}(0.0300M),H^{+}(0.100M)\Vert I^{-}(0.100M),AgI(sat\text{'}d)\mid Ag$

$$\begin{gathered} \left(b\right)\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left(5.75\times {10}^{-4}\mathrm{M}\right)\Vert \mathrm{Fe}(\mathrm{CN}{)}_6^{4-}\left(5.50\times {10}^{-2}\mathrm{M}\right),\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\left(6.75\times {10}^{-2}\mathrm{M}\right)\right|\mathrm{Pt} \\ \left(c\right)\mathrm{Pt},{\mathrm{H}}_2(0.200\mathrm{atm})\mid \mathrm{HCl}\left(8.25\times {10}^{-4}\mathrm{M}\right),\mathrm{AgCl}(sat\text{'}d)\mid \mathrm{Ag} \end{gathered}$$

Short answer

1. The given cell reaction is non spontaneous in nature.
2. The theoretical potential is positive, hence the given cell reaction is spontaneous in nature.
3. The theoretical potential is positive, hence the given cell reaction is spontaneous in nature.

Step-by-step solution

part (a)

Let the sign convention be positive for a spontaneous reaction and negative for a non spontaneous reaction.

The expression for the cell potential is:

$E=E_R-E_L\dots \dots \text{(I)}$

Here, the electrode potential for the right of the cell is $E_R$and the electrode potential for the left of the cell is $E_L$.

The reaction for the left cell is:

${\mathrm{BiO}}^{+}+2{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\rightleftarrows \mathrm{Bi}\left(s\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)$

The expression for the electrode potential for the left side is:

$E_L=E_L^0-\frac{0.0592}{\mathrm{n}}\log \left(\frac{1}{{\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{BiO}}^{+}\right]}\right)\dots \dots \text{(II)}$

Here, the electrons involved in the reaction is $\mathrm{n}$, the standard electrode potential for the reaction is $E_L^0$, the concentrations of ${\mathrm{H}}^{+}$is $\left[{\mathrm{H}}^{+}\right]$, and the concentrations of ${\mathrm{BiO}}^{+}$is $\left[{\mathrm{BiO}}^{+}\right]$.

The value of standard electrode potential for the reaction is $0.320\mathrm{V}$.

$E_L^0=0.320\mathrm{V}$

The reaction for the right cell is:

$\mathrm{AgI}\left(s\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(s\right)+{\mathrm{I}}^{-}$

The expression for the electrode potential for the right side is:

$E_R=E_R^0-\frac{0.0592}{\mathrm{n}}\log \left(\left[{\mathrm{I}}^{-}\right]\right)\dots \dots \text{(III)}$

Here, the electrons involved in the reaction is $\mathrm{n}$, the standard electrode potential for the reaction is$E_R^0$, and the concentrations of ${\mathrm{I}}^{-}$is $\left[{\mathrm{I}}^{-}\right]$.

The value of standard electrode potential for the reaction is $-0.151\mathrm{V}$

$E_R^0=-0.151\mathrm{V}$

Substitute 3 for$\mathrm{n},0.320\mathrm{V}$for$E_L^0,0.030\mathrm{M}$for $\left[{\mathrm{BiO}}^{+}\right]$and $0.100\mathrm{M}$ for $\left[{\mathrm{H}}^{+}\right]$in Equation (III).

$$\begin{gathered} E_L=(0.320\mathrm{V})-\frac{0.0592}{3}\log \left(\frac{1}{(0.100\mathrm{M}{)}^2(0.030\mathrm{M})}\right) \\ =(0.320\mathrm{V})-(0.019)(3.522)\mathrm{V} \\ =0.320\mathrm{V}-0.066\mathrm{V} \\ =0.254\mathrm{V} \end{gathered}$$

Substitute 1 for $\mathrm{n},-0.151\mathrm{V}$ for $E_R^0$, and $0.100\mathrm{M}$ for $\left[{\mathrm{I}}^{-}\right]$in Equation (II).

$$\begin{gathered} E_R=(-0.151\mathrm{V})-\frac{0.0592}{1}\log (0.100\mathrm{M}) \\ =(-0.151\mathrm{V})-(0.0592)(-1)\mathrm{V} \\ =-0.151\mathrm{V}+0.0592\mathrm{V} \\ =-0.0918\mathrm{V} \end{gathered}$$

Substitute $-0.0918\mathrm{V}$ for $E_R$ and $0.254\mathrm{V}$ for $E_L$ in equation (I).

$$\begin{gathered} E=(-0.0918\mathrm{V})-(0.254\mathrm{V}) \\ =-0.345\mathrm{V} \end{gathered}$$

Since the theoretical potential is negative, hence the given cell reaction is non spontaneous in nature.

part (b)

Let the sign convention be positive for a spontaneous reaction and negative for a non spontaneous reaction.

The expression for the cell potential is:

$E=E_R-E_L\dots \dots \text{(IV)}$

Here, the electrode potential for the right of the cell is $E_R$and the electrode potential for the left of the cell is $E_L$.

The reaction for the left cell is:

${\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\rightleftarrows \mathrm{Zn}\left(s\right)$

The expression for the electrode potential for the left side is:

$E_L=E_L^0-\frac{0.0592}{\mathrm{n}}\log \left(\frac{1}{\left[{\mathrm{Zn}}^{2+}\right]}\right)\dots \dots \text{(V)}$

Here, the electrons involved in the reaction is $\mathrm{n}$, the standard electrode potential for the reaction is $E_L^0$, the concentrations of ${\mathrm{Zn}}^{2+}$is $\left[{\mathrm{Zn}}^{2+}\right]$.

The value of standard electrode potential for the reaction is $-0.763\mathrm{V}.$

$E_L^0=-0.763\mathrm{V}$

The reaction for the right cell is:

$\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Fe}(\mathrm{CN}{)}_6^4$

The expression for the electrode potential for the right side is:

$E_R=E_R^0-\frac{0.0592}{\mathrm{n}}\log \left(\frac{\left[\mathrm{Fe}(\mathrm{CN}{)}_6^4\right]}{\left[\mathrm{Fe}(\mathrm{CN}{)}_6^3\right]}\right)\dots \dots \left(\mathrm{VI}\right)$

Here, the electrons involved in the reaction is $\mathrm{n}$, the standard electrode potential for the reaction is $E_R^0$, concentrations of $\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}is\left[\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\right]$and the concentrations of $\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}$is $\left[\mathrm{Fe}(\mathrm{CN}{)}_6^4\right]$.

The value of standard electrode potential for the reaction is $0.360\mathrm{V}$

$E_R^0=0.360\mathrm{V}$

Substitute 2 for $\mathrm{n},-0.763\mathrm{V}$for $E_L^0,5.75\times {10}^{-4}\mathrm{M}$for $\left[{\mathrm{Zn}}^{2+}\right]$in Equation (V).

$$\begin{gathered} E_L=(-0.763\mathrm{V})-\frac{0.0592}{2}\log \left(\frac{1}{\left(5.75\times {10}^{-4}\mathrm{M}\right)}\right) \\ =(-0.763\mathrm{V})-(0.029)(3.240)\mathrm{V} \\ =-0.763\mathrm{V}-0.093\mathrm{V} \\ =-0.856\mathrm{V} \end{gathered}$$

Substitute 1 for $\mathrm{n},0.360\mathrm{V}$for $E_R^0,5.50\times {10}^{-2}\mathrm{M}$for $\left[\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}\right]$, and $6.75\times {10}^{-2}\mathrm{M}$for $\left[\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\right]$in Equation (VI).

$$\begin{gathered} E_R=(0.360\mathrm{V})-\frac{0.0592}{1}\log \left(\frac{\left(5.50\times {10}^{-2}\mathrm{M}\right)}{\left(6.75\times {10}^{-2}\mathrm{M}\right)}\right) \\ =(0.360\mathrm{V})-(0.0592)(-0.088)\mathrm{V} \\ =0.360\mathrm{V}+\left(5.20\times {10}^{-3}\right)\mathrm{V} \\ =0.365\mathrm{V} \end{gathered}$$

Substitute $0.365\mathrm{V}$ for $E_R$ and $-0.856\mathrm{V}$ for $E_L$ in Equation (I).

$$\begin{gathered} E=(0.365\mathrm{V})-(-0.856\mathrm{V}) \\ =1.221\mathrm{V} \end{gathered}$$

Since the theoretical potential is positive, hence the given cell reaction is spontaneous in nature.

part (c)

Let the sign convention be positive for a spontaneous reaction and negative for a non spontaneous reaction.

The expression for the cell potential is:

$E=E_R-E_L\dots \dots \text{(VII)}$

Part (c) Calculations

Here, the electrode potential for the right of the cell is $E_R$and the electrode potential for the left of the cell is $E_L$.

The reaction for the left cell is:

$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\rightleftarrows {\mathrm{H}}_2\left(g\right)$

The expression for the electrode potential for the left side is:

$E_L=E_L^0-\frac{0.0592}{\mathrm{n}}\log \left(\frac{P_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}\right)\dots \dots \text{(VIII)}$

Here, the electrons involved in the reaction is $\mathrm{n}$, the standard electrode potential for the reaction is $E_L^0$, the concentrations of ${\mathrm{H}}^{+}is\left[{\mathrm{H}}^{+}\right],$the pressure of the ${\mathrm{H}}_2$is ${\mathrm{PH}}_2$.

The value of standard electrode potential for the reaction is $0\mathrm{V}$

$E_L^0=0\mathrm{V}$

The reaction for the right cell is:

$\mathrm{AgCl}\left(s\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(s\right)+{\mathrm{Cl}}^{-}$

The expression for the electrode potential for the right side is:

$E_R=E_R^0-\frac{0.0592}{\mathrm{n}}\log \left(\left[{\mathrm{Cl}}^{-}\right]\right)\dots \dots \text{(IX)}$

Here, the electrons involved in the reaction is $\mathrm{n}$, the standard electrode potential for the reaction is $E_R^0$and concentrations of ${\mathrm{Cl}}^{-}$is $\left[{\mathrm{Cl}}^{-}\right].$

The value of standard electrode potential for the reaction is $0.222\mathrm{V}.$

$E_R^0=0.222\mathrm{V}$

Substitute 2 for $\mathrm{n},0\mathrm{V}$for $E_L^0,8.25\times {10}^{-4}\mathrm{M}$for $\left[{\mathrm{H}}^{+}\right]$and $0.20$for $P_{{\mathrm{H}}_2}$in Equation (VIII).

$$\begin{gathered} E_L=(0\mathrm{V})-\frac{0.0592}{2}\log \left(\frac{0.20}{{\left(8.25\times {10}^{-4}\mathrm{M}\right)}^2}\right) \\ =-(0.0296)(5.468)\mathrm{V} \\ =-0.161\mathrm{V} \end{gathered}$$

Substitute 1 for$\mathrm{n},0.222\mathrm{V}$for $E_R^0$, and $8.25\times {10}^{-4}\mathrm{M}$for $\left[{\mathrm{Cl}}^{-}\right]$in Equation (IX).

$$\begin{gathered} E_R=(0.222\mathrm{V})-\frac{0.0592}{1}\log \left(8.25\times {10}^{-4}\mathrm{M}\right) \\ =(0.222\mathrm{V})-(0.0592)(-3.083)\mathrm{V} \\ =0.222\mathrm{V}+0.182\mathrm{V} \\ =0.404\mathrm{V} \end{gathered}$$

Substitute $0.404\mathrm{V}$ for $E_R$ and $-0.161\mathrm{V}$ for $E_L$ in Equation (VII).

$$\begin{gathered} E=(0.404\mathrm{V})-(-0.161\mathrm{V}) \\ =0.404\mathrm{V}+0.161\mathrm{V} \\ =0.565\mathrm{V} \end{gathered}$$

Since the theoretical potential is positive, hence the given cell reaction is spontaneous in nature.