Question

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}} & {\mathscr{E}^{\circ}=1.50 \mathrm{V}} \\ {\mathrm{T} 1^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl}} & {\mathscr{E}^{\circ}=-0.34 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text { cell }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text { cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{T} 1^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Short answer

The overall cell reaction is: $$\mathrm{Au}^{3+} + 3\,\mathrm{Tl}^{+} \longrightarrow \mathrm{Au} + 3\,\mathrm{Tl}$$ The standard cell potential is: \(\mathscr{E}_\text{cell} = 1.84\,\text{V}\). The Gibbs free energy change under standard conditions is: \(\Delta G^\circ = -532,869\,\mathrm{J/mol}\). The equilibrium constant is: \(K = 2.37 \times 10^{86}\). Finally, the cell potential at given concentrations is: \(\mathscr{E}_\text{cell} = 1.82\,\text{V}\).

Step-by-step solution

Determine the overall cell reaction

To determine the overall cell reaction, combine the two half-reactions: $$\mathrm{Au}^{3+} + 3\,\mathrm{e}^- \longrightarrow \mathrm{Au}$$ $$\mathrm{Tl}^{+} + \mathrm{e}^- \longrightarrow \mathrm{Tl}$$ To balance the electrons, we multiply the second reaction by 3: $$3 \times (\mathrm{Tl}^{+} + \mathrm{e}^- \longrightarrow \mathrm{Tl})$$ Now, we can combine the two reactions: $$\mathrm{Au}^{3+} + 3\,\mathrm{Tl}^{+} \longrightarrow \mathrm{Au} + 3\,\mathrm{Tl}$$

Calculate the standard cell potential

The standard cell potential, \(\mathscr{E}_\text{cell}\), is the difference between the reduction potentials of the two half-cells: \(\mathscr{E}_\text{cell} = \mathscr{E}_\text{Au}^\circ - \mathscr{E}_\text{Tl}^\circ = 1.50\,\text{V} - (-0.34\,\text{V}) = 1.84\,\text{V}\)

Calculate the Gibbs free energy change

Then we can calculate the Gibbs free energy change under standard conditions using the formula: \(\Delta G^\circ = -nFE^\circ\) where \(n\) denotes the number of moles of electrons transferred, \(F\) is the Faraday constant (\(F \approx 96,485\,\mathrm{C/mol}\)), and \(E^\circ\) refers to the standard cell potential. In this case, \(n = 3\), as there are 3 moles of electrons transferred in the overall cell reaction. \(\Delta G^\circ = -3 \times 96,485\,\mathrm{C/mol} \times 1.84\,\text{V} = -532,869\,\mathrm{J/mol}\)

Calculate the equilibrium constant

We can find the equilibrium constant, \(K\), by using the relationship between Gibbs free energy change and equilibrium constant: \(\Delta G^\circ = -RT \ln K\) where \(R\) denotes the ideal gas constant (\(8.314\,\mathrm{J/(mol \cdot K)}\)) and \(T\) is the temperature in Kelvin, which is \(25^\circ\mathrm{C} + 273.15 = 298.15\,\text{K}\). Rearranging for \(K\): \(K = e^{-\Delta G^\circ / (RT)} = e^{532,869\,\mathrm{J/mol} / (8.314\,\mathrm{J/(mol \cdot K)} \times 298.15\,\text{K})} = 2.37 \times 10^{86}\)

Calculate the cell potential at given concentrations

Finally, we calculate the cell potential at non-standard conditions using the Nernst equation: \(\mathscr{E}_\text{cell} = \mathscr{E}^\circ - \frac{RT}{nF} \ln Q\) where \(Q\) refers to the reaction quotient. For our cell reaction, the reaction quotient is: \(Q = \frac{[\mathrm{Au}^{3+}]}{[\mathrm{Tl}^{+}]^3} = \frac{1.0 \times 10^{-2}\,\mathrm{M}}{(1.0 \times 10^{-4}\,\mathrm{M})^3} = 1.0 \times 10^9\) Now, we can find the cell potential: \(\mathscr{E}_\text{cell} = 1.84\,\text{V} - \frac{8.314\,\mathrm{J/(mol \cdot K)} \times 298.15\,\text{K}}{3 \times 96,485\,\mathrm{C/mol}} \ln (1.0 \times 10^9) = 1.82\,\text{V}\)

Key concepts

Standard Cell Potential

The standard cell potential, represented as \( \mathscr{E}^\circ \), is a measure of the voltage difference between two half-cells in a galvanic cell under standard conditions (all concentrations at 1 M, gasses at 1 atm, and the temperature at 25°C or 298.15 K)..
This potential difference is an indicator of the driving force of the electrochemical reaction, with a more positive value indicating a stronger tendency for the reaction to proceed.

In a galvanic cell, the standard cell potential is calculated by subtracting the standard reduction potential of the anode half-cell from that of the cathode.

• For the given cell reaction, the cathode (reduction) reaction is \( \mathrm{Au}^{3+} + 3 \mathrm{e}^- \rightarrow \mathrm{Au} \), with a standard reduction potential \( \mathscr{E}^\circ = 1.50\, \text{V} \).
• The anode (oxidation) reaction is \( \mathrm{Tl}^{+} + \mathrm{e}^- \rightarrow \mathrm{Tl} \), with a standard reduction potential \( \mathscr{E}^\circ = -0.34\, \text{V} \).

Thus, the standard cell potential \( \mathscr{E}_{\text{cell}} = 1.50\, \text{V} - (-0.34\, \text{V}) = 1.84\, \text{V}. \)

Gibbs Free Energy

Gibbs Free Energy Change \( \Delta G^\circ \) is a thermodynamic property that provides information about the spontaneity of a reaction.
A negative \( \Delta G^\circ \) indicates that a reaction is spontaneous under standard conditions, while a positive value suggests non-spontaneity.

The relation between Gibbs Free Energy and the standard cell potential in electrochemical cells is given by:
\[ \Delta G^\circ = -nF \mathscr{E}^\circ \]
where:

• \( n \) is the number of moles of electrons exchanged in the cell reaction.
• \( F \) is the Faraday constant, approximately \( 96,485\, \mathrm{C/mol} \).
• \( \mathscr{E}^\circ \) is the standard cell potential.

For our reaction:

• \( n = 3 \) as three electrons are exchanged.

Calculation:
\( \Delta G^\circ = -3 \times 96,485 \, \mathrm{C/mol} \times 1.84\, \text{V} \approx -532,869 \, \text{J/mol} \).
This large negative value tells us that the reaction is highly spontaneous.

Equilibrium Constant

The equilibrium constant, \( K \), of a reaction provides an idea of the position of equilibrium and determines the extent of a reaction.

For galvanic cells, \( K \) can be linked with \( \Delta G^\circ \) using the formula:
\[ \Delta G^\circ = -RT \ln K \]
where:

• \( R \) is the ideal gas constant, \( 8.314 \mathrm{J/(mol \cdot K)} \).
• \( T \) is the temperature in Kelvin.

By rearranging and solving for \( K \):
\[ K = e^{\frac{-\Delta G^\circ}{RT}} \]
For our example:

• \( T = 298.15 \text{K} \) (which is standard temperature of 25°C).
• \( \Delta G^\circ \approx -532,869 \, \text{J/mol} \).

Plugging in these values:
\[ K = e^{\frac{532,869}{8.314 \times 298.15}} \approx 2.37 \times 10^{86} \].
This extremely large \( K \) value indicates that the reaction proceeds nearly completely to the products, {brace generating \( \text{Au} \) and reducing \( \text{Tl}^+ \).

Nernst Equation

The Nernst Equation allows the calculation of the cell potential under non-standard conditions by incorporating the effects of concentration on the cell potential.
The equation is:
\[ \mathscr{E}_{\text{cell}} = \mathscr{E}^\circ - \frac{RT}{nF} \ln Q \]
where:

• \( R \) is 8.314 \mathrm{J/(mol \cdot K)}.
• \( T \) is the absolute temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred.
• \( F \) is the Faraday constant.
• \( Q \) is the reaction quotient.

For our reaction, given concentrations:

• \( [\mathrm{Au}^{3+}] = 1.0 \times 10^{-2}\, \mathrm{M} \).
• \( [\mathrm{Tl}^+] = 1.0 \times 10^{-4}\, \mathrm{M} \).

The reaction quotient, \( Q \), is calculated as:
\[ Q = \frac{[\mathrm{Au}^{3+}]}{[\mathrm{Tl}^+]^3} = \frac{1.0 \times 10^{-2}}{(1.0 \times 10^{-4})^3} = 1.0 \times 10^9 \]
Now, calculate \( \mathscr{E}_{\text{cell}} \):
\[ \mathscr{E}_{\text{cell}} = 1.84 - \frac{8.314 \times 298.15}{3 \times 96,485} \ln(1.0 \times 10^9) \approx 1.82\, \text{V} \].
This result shows the adjustment needed on the cell potential due to the deviation from standard conditions.