Question

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{~V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{cell}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{B}_{\text {coll }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=\) \(1.0 \times 10^{-5} M\)

Short answer

a) The overall cell reaction is \(\mathrm{Zn}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\), and the standard cell potential \(\mathscr{E}_{\mathrm{cell}}^{\circ}\) is 0.32 V. b) The standard Gibbs free energy change \(\Delta G^{\circ}\) is -61749.60 J/mol, and the equilibrium constant \(K\) is \(2.44 \times 10^{11}\). c) The cell potential \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when the concentrations of the Zn and Fe ions are given is 0.208 V.

Step-by-step solution

Determine the overall cell reaction

To determine the overall cell reaction, we must combine the given half-reactions. The half-cell with a more negative standard potential, Zn^(2+), will act as anode (oxidized). Thus: $$\begin{array}{ll}\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow\mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{~V}\\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{~V} \end{array}$$ We will reverse the anodic half-reaction and sum both half-reactions to obtain the overall cell reaction: $$\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$$ $$\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}$$ $$\mathrm{Zn}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$$

Calculate the standard cell potential

To calculate the standard cell potential \(\mathscr{E}_{\mathrm{cell}}^{\circ}\), we need to subtract the standard potential of the anode from the standard potential of the cathode: \[\mathscr{E}_{\mathrm{cell}}^{\circ} = \mathscr{E}_{\mathrm{cathode}}^{\circ} - \mathscr{E}_{\mathrm{anode}}^{\circ} = (-0.44) - (-0.76) = 0.32~\text{V}\]

Calculate standard Gibbs free energy change and equilibrium constant

The standard Gibbs free energy change \(\Delta G^{\circ}\) can be calculated using the formula: \[\Delta G^{\circ} = -nFE_{\mathrm{cell}}^{\circ}\] where \(n\) is the number of moles of electrons transferred (2 in this case), \(F\) is the Faraday constant (approximately \(96485~\text{C/mol}\)), and \(\mathscr{E}_{\mathrm{cell}}^{\circ}\) is the Standard cell potential. So the standard Gibbs free energy change is: \[\Delta G^{\circ} = -2 \times 96485 \times 0.32 = -61749.60 ~\text{J/mol}\] Next, we will calculate the equilibrium constant \(K\) using the relationship between \(\Delta G^{\circ}\) and the equilibrium constant: \[\Delta G^{\circ} = -RT \ln K\] Where \(R\) is the gas constant (\(8.314~\text{J/mol K}\)) and \(T\) is the temperature in Kelvin (298K). We can rearrange the equation and solve for \(K\): \[K = \text{e}^{\frac{-\Delta G^{\circ}}{RT}} = \text{e}^{\frac{61749.60}{(8.314)(298)}} = 2.44 \times 10^{11}\]

Calculate the cell potential at given concentrations

Finally, we will calculate the cell potential \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when the concentrations of the Zn and Fe ions are given. The Nernst equation helps us calculate the cell potential under non-standard conditions: \[\mathscr{E}_{\text {cell}} = \mathscr{E}_{\text {cell}}^{\circ}-\frac{RT}{nF} \ln Q\] Where \(Q\) is the reaction quotient: \[Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Fe}^{2+}]}\] Now, we can plug in the given concentration values to find \(Q\), and then use the Nernst equation to find the cell potential: \[Q = \frac{0.10}{1.0 \times 10^{-5}} = 10^4\] \[\mathscr{E}_{\text {cell}} = 0.32 - \frac{(8.314)(298)}{(2)(96485)} \ln 10^4 = 0.32-0.0296 \times 4 = 0.208~\text{V}\] So, the cell potential at given concentrations is 0.208 V.

Key concepts

Understanding Standard Electrode Potential

The standard electrode potential, denoted as \(\mathscr{E}^\circ\), is a measure of the intrinsic electric potential of an electrode compared to a standard hydrogen electrode at 25°C, 1 atmospheric pressure, and a solution concentration of 1M. In the context of galvanic cells, it predicts the tendency of a half-cell to be reduced. Higher (more positive) standard electrode potential means a greater tendency to gain electrons and be reduced, while a lower (more negative) standard potential indicates a greater tendency to lose electrons and be oxidized. Each half-cell in a galvanic cell has its standard electrode potential, and the overall cell potential, \(\mathscr{E}_{\mathrm{cell}}^\circ\), is calculated by subtracting the potential of the anode from that of the cathode. This concept is critical to understanding the driving force behind the flow of electrons from one electrode to another.

Gibbs Free Energy Change in Galvanic Cells

Gibbs free energy change, \(\Delta G^\circ\), represents the maximum amount of work that can be performed by a chemical reaction at constant temperature and pressure. For reactions in a galvanic cell, \(\Delta G^\circ\) is directly related to the cell potential and the amount of charge transferred during the redox reaction. \(\Delta G^\circ\) is calculated by the formula: \[\Delta G^\circ = -nFE_{\mathrm{cell}}^\circ\] where \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is the Faraday constant, and \(\mathscr{E}_{\mathrm{cell}}^\circ\) is the standard cell potential. A negative value of \(\Delta G^\circ\) implies that the reaction is spontaneous. This concept is valuable for determining not only the spontaneity but also the efficiency of the conversion of chemical energy into electrical energy in galvanic cells.

Equilibrium Constant and Its Significance

The equilibrium constant, denoted as \(K\), is a dimensionless value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. In galvanic cells, \(K\) provides insight into the position of equilibrium. It is related to the Gibbs free energy change by the equation \[\Delta G^\circ = -RT \ln K\] where \(R\) is the universal gas constant and \(T\) is the absolute temperature. A large value of \(K\) implies that at equilibrium, the reaction favors the formation of products, which corresponds to a strong driving force for the cell's redox reaction. Calculating the equilibrium constant gives us a quantitative understanding of the chemical reaction's completeness under standard conditions.

Nernst Equation and Cell Potential

The Nernst equation describes how the cell potential varies with temperature and concentration. It is an essential tool for calculating the actual cell potential \(\mathscr{E}_{\text {cell }}\) under non-standard conditions. The equation is given by: \[\mathscr{E}_{\text {cell}} = \mathscr{E}_{\text {cell}}^\circ - \frac{RT}{nF} \ln Q\] with \(Q\) being the reaction quotient which reflects the instantaneous ratio of product and reactant concentrations. The Nernst equation enables us to predict cell behavior in real-world applications where conditions deviate from standard ones, helping to simulate and understand the operation of batteries, sensors, and natural biological processes.