Question

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 \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} .\)

Short answer

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 \([\mathrm{Fe}^{2+}]=1.0\times 10^{-5}\,\text{M}\).

Step-by-step solution

Identify the oxidation and reduction half-reactions

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: $$\begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \ (\text {Oxidation})\\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \ (\text {Reduction}) \end{array}$$

Balance the electron transfer

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

Add the half-reactions to find the overall cell reaction

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

Calculate E_cell

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 ΔG°

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: $$\Delta G^\circ = -2 \times 96485\,\mathrm{C\, mol}^{-1} \times 0.32\,\mathrm{V} = -61830\,\mathrm{J\, mol^{-1}}$$

Calculate K

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.#

Apply the Nernst Equation

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 for E_cell

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 \([\mathrm{Fe}^{2+}]=1.0\times 10^{-5}\,\text{M}\).

Key concepts

Galvanic Cell

Galvanic cells, also known as voltaic cells, are fascinating devices that transform chemical energy into electrical energy. They do this through redox reactions, where one species undergoes oxidation (loses electrons) while another undergoes reduction (gains electrons). This process occurs in two separate compartments called half-cells, each connected by a salt bridge allowing ions to migrate and complete the circuit.

• In a galvanic cell, the anode is where oxidation occurs, and electrons are released.
• The cathode is where reduction occurs, and electrons are gained.
• Electrons flow from the anode to the cathode through an external circuit, powering any connected devices.

This movement of electrons generates an electrical current, which is the primary aim of a galvanic cell. Understanding these processes helps illuminate the inner workings of many everyday batteries and power sources.

Nernst Equation

The Nernst Equation is essential in electrochemistry, describing how the voltage of a cell (electrical potential) changes with the concentration of the ions involved. It's like a tool that shows us how practical, real-world conditions affect the ideal voltage of a cell.Here's the basic Nernst Equation:\[\mathscr{E}_{\text{cell}} = \mathscr{E}_{\text{cell}}^{\circ} - \frac{RT}{nF} \ln\left(\frac{[\text{products}]}{[\text{reactants}]}\right)\]

• \(\mathscr{E}_{\text{cell}}\) is the cell potential under non-standard conditions.
• \(\mathscr{E}_{\text{cell}}^{\circ}\) is the standard cell potential.
• \(R\) is the universal gas constant, \(8.314 \,\text{J} \, \text{K}^{-1} \, \text{mol}^{-1}\).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of electrons exchanged in the reaction.
• \(F\) is the Faraday constant, \(96485\,\text{C} \, \text{mol}^{-1}\).
• \(\ln\left(\frac{[\text{products}]}{[\text{reactants}]}\right)\) is the natural log of the reaction quotient.

This equation explains why the cell potential isn't always the same and how concentration influences a cell's capability to do work.

Standard Electrode Potential

The standard electrode potential, symbolized as \(\mathscr{E}^{\circ}\), is a measure of the driving force behind a redox reaction. It indicates how strongly an element wants to gain (or lose) electrons when measured against a standard hydrogen electrode under standard conditions (1 M concentration, 1 atm pressure, and 25°C or 298 K).Some important insights about standard electrode potentials:

• A positive \(\mathscr{E}^{\circ}\) value suggests that the reaction is more favorable towards reduction, indicating it easily gains electrons.
• A negative \(\mathscr{E}^{\circ}\) value suggests that the reaction favors oxidation, meaning it tends to lose electrons.
• The comparison of different standard electrode potentials helps us predict which substances will oxidize or reduce when paired.

By knowing \(\mathscr{E}^{\circ}\) values, we can piece together possible cell reactions and feasibility, just as we saw in the case of zinc and iron.

Gibbs Free Energy

Gibbs Free Energy, noted as \(\Delta G\), is crucial in chemistry as it tells us about the spontaneity of a process. If \(\Delta G\) is negative, the process is spontaneous and can do work; if positive, the process is non-spontaneous.Here's how we relate Gibbs Free Energy to cell potential:\[\Delta G = -nF\mathscr{E}_{\text{cell}}\]

• \(\Delta G\) is the change in Gibbs free energy.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is the Faraday constant.
• \(\mathscr{E}_{\text{cell}}\) is the cell potential.

This equation shows that a positive cell potential (\(\mathscr{E}_{\text{cell}}\)) corresponds to negative \(\Delta G\), indicating a reaction that can proceed spontaneously. For our zinc-iron cell, a negative free energy highlights that energy is released, making the reaction favorable with little intervention.