Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}} & {\mathscr{E}^{\circ}=-0.76 \mathrm{V}} \\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {\mathscr{E}^{\circ}=-0.44 \mathrm{V}}\end{array}$$ 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 \(\left[\mathrm{Zn}^{2+}\right]=0.10 M\) and $\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} .$

Based on the standard electrode potentials, the reaction with a higher standard electrode potential (less negative) is more likely to proceed in the direction of reduction, and the less positive reaction is more likely to proceed in the direction of oxidation. So, Fe2+ must be reduced, and Zn2+ must be oxidized: $$\begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \ (\text {Oxidation})\\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \ (\text {Reduction}) \end{array}$$

Since both half-reactions have a transfer of two electrons, there is no need to balance the electrons in this case.

Add together the oxidation and reduction reactions to obtain the overall cell reaction: $$\mathrm{Zn} + \mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+} + \mathrm{Fe}$$

Using the given standard electrode potentials, calculate the cell potential using the equation: $$\mathscr{E}_{\text {cell}} = \mathscr{E}_{\text{cathode}} -\mathscr{E}_{\text{anode}}$$ $$\mathscr{E}_{\text {cell}} = (-0.44\,\mathrm{V}) - (-0.76\,\mathrm{V}) = 0.32\,\mathrm{V}$$ #b. Calculate ΔG° and K for the cell reaction at 25°C#

Calculate the change in Gibbs free energy (\(\Delta G^\circ\)) using the formula: $$\Delta G^\circ = -nF\mathscr{E}^\circ_{\text{cell}}$$ where \(n\) is the number of moles of electrons transferred; \(F\) is the Faraday constant (\(96485\,\text{C}\, \text{mol}^{-1}\)); and \(\mathscr{E}_{\text {cell}}^\circ\) is the standard cell potential. In this case, \(n = 2\) moles of electrons are transferred, thus: $$\Delta G^\circ = -2 \times 96485\,\mathrm{C\, mol}^{-1} \times 0.32\,\mathrm{V} = -61830\,\mathrm{J\, mol^{-1}}$$

Find the equilibrium constant \(K\) using the relationship: $$\Delta G^\circ = -RT \ln K$$ Where \(R\) is the gas constant (\(8.314\, \mathrm{J \, K^{-1} mol^{-1}}\)), \(T\) is the temperature in Kelvin (\(298\, \text{K}\)), and \(K\) is the equilibrium constant. Rearrange the equation and solve for \(K\): $$K = e^{(-\Delta G^\circ) / (RT)}$$ $$K = e^{61830\, \mathrm{J \ mol^{-1}} / (8.314\,\mathrm{J \, K^{-1} mol^{-1}} \times 298\,\text{K})} = 4.79 \times 10^{25}$$ #c. Calculate E_cell at 25°C when [Zn2+]=0.10 M and [Fe2+]=1.0 × 10⁻⁵ M.#

The Nernst equation describes the relationship between the equilibrium reduction potential of half-cell reactions and the standard cell potential under non-standard conditions. Calculate the cell potential (\(\mathscr{E}_{\text {cell}}\)) using the Nernst equation: $$\mathscr{E}_{\text {cell}} = \mathscr{E}_{\text {cell}}^{\circ} - \frac{RT}{nF} \ln\left(\frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Fe}^{2+}]}\right)$$ Plug in the values, with \(T = 298\,\text{K}\), \(R = 8.314\,\mathrm{J \, K^{-1} mol^{-1}}\), \(F = 96485\,\mathrm{C\, mol^{-1}}\), \(n = 2\), and \([\mathrm{Zn}^{2+}] = 0.10\,\text{M}\, [\mathrm{Fe}^{2+}] = 1.0 \times 10^{-5}\,\text{M}\): $$\mathscr{E}_{\text {cell}} = 0.32\,\mathrm{V} - \frac{(8.314 \,\mathrm{J \, K^{-1} mol^{-1}}) (298\,\text{K})}{ 2(96485\,\mathrm{C \, mol}^{-1})} \ln\left(\frac{0.10\,\text{M}}{1.0 \times 10^{-5}\,\text{M}}\right)$$

Solve the equation to find the cell potential: $$\mathscr{E}_{\text{cell}}= 0.349\,\text{V}$$ In summary, the overall cell reaction is: $$\mathrm{Zn} + \mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+} + \mathrm{Fe}$$ with a standard cell potential of \(0.32\,\mathrm{V}\), a change in Gibbs free energy of \(-61830\,\mathrm{J \ mol^{-1}}\), an equilibrium constant of \(4.79 \times 10^{25}\), and a cell potential of \(0.349\,\mathrm{V}\) at \(25^\circ \text{C}\) when \([\mathrm{Zn}^{2+}]=0.10\,\text{M}\) and \([\mathrm{Fe}^{2+}]=1.0\times 10^{-5}\,\text{M}\).