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 \(\left[\mathrm{T} 1^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Step by step solution

01

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: $$\mathrm{Au}^{3+} + 3\,\mathrm{Tl}^{+} \longrightarrow \mathrm{Au} + 3\,\mathrm{Tl}$$

02

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: \(\mathscr{E}_\text{cell} = \mathscr{E}_\text{Au}^\circ - \mathscr{E}_\text{Tl}^\circ = 1.50\,\text{V} - (-0.34\,\text{V}) = 1.84\,\text{V}\)

03

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. \(\Delta G^\circ = -3 \times 96,485\,\mathrm{C/mol} \times 1.84\,\text{V} = -532,869\,\mathrm{J/mol}\)

04

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 \(K\): \(K = e^{-\Delta G^\circ / (RT)} = e^{532,869\,\mathrm{J/mol} / (8.314\,\mathrm{J/(mol \cdot K)} \times 298.15\,\text{K})} = 2.37 \times 10^{86}\)

05

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: \(\mathscr{E}_\text{cell} = 1.84\,\text{V} - \frac{8.314\,\mathrm{J/(mol \cdot K)} \times 298.15\,\text{K}}{3 \times 96,485\,\mathrm{C/mol}} \ln (1.0 \times 10^9) = 1.82\,\text{V}\)

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