Question

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

Short answer

The standard reduction potential for the Na+/Na couple, \(E_{\mathrm{Na^+/Na}}^\circ = 2.29 \)V.

Step-by-step solution

Identify the Half-Reactions

For cell 1: \(\mathrm{Na_{(s)}} \rightarrow \mathrm{Na^{+}_(aq)} + e^-\) and For cell 2: \(2H_{(aq)}^{+}+2e^- \rightarrow H_{2(g)}\) and \(\mathrm{Na_{(s)}} \rightarrow \mathrm{Na^{+}_{(aq)} + e^-)\). The equations can be written as:Voltaic cell 1: \(\mathrm{Na_{(s)}} \rightarrow \mathrm{Na^{+}_(aq) + e^-}\)Voltaic cell 2: \( \mathrm{Na_{(s)}} \rightarrow \mathrm{Na^{+}_{(aq)} + e^-\) and \(2H_{(aq)}^{+} + 2e^- \rightarrow H_{2(g)} \)

Calculate \( \Delta G \) for the Half-Reactions

Applying the formula \( \Delta G = -nFE^{\circ} \), the Gibbs free energy change for the reactions in both cells can be estimated. Since both reactions involve the transfer of one electron, \( n = 1 \). The \( F \) (Faraday's constant) is approximately \( 96485 C/mol \). Using the equation: For voltaic cell 1, \( \Delta G_{1} = -(1)(96485 C/mol)(0.8453 V) = -81556.105 J/mol = -81.556 kJ/mol \)For voltaic cell 2, \( \Delta G_{2} = -(2)(96485 C/mol)(1.8673 V) = -360314.535 J/mol = -360.314 kJ/mol \)

Establish the Overall Reaction and its Gibbs Free Energy Change

The overall reaction is constructed by adding and cancelling the respective species on both sides of the equations obtained in step 1. For the overall reaction, \(\mathrm{Na_{(s)}} + 2H^{+}(aq) \rightarrow \mathrm{Na^{+}(aq)} +H_2(g)\), the overall change in Gibbs free energy is the sum of the values obtained in step 2, \( \Delta G_{\text{total}} = \Delta G_1 + \Delta G_2 = -81.556 kJ/mol - 360.314 kJ/mol = -441.87 kJ/mol \)

Provide the Overall Standard Cell Potential and Compare with the Tabulated Value

Using \( \Delta G_{\text{total}} = -nFE_{\text{cell}}^{\circ} \), the standard cell potential can be calculated. Since the overall reaction involves the transfer of two electrons, \( n = 2 \):\( E_{\text{cell}}^{\circ} = -\Delta G_{\text{total}}/(2F) = -(-441.87 kJ/mol)/(2*96485 J/C) = 2.29 V \). The standard reduction potential for the Na+/Na couple can be compared with the value provided in Appendix D.

Key concepts

Gibbs Free Energy Change

Understanding the concept of Gibbs Free Energy Change, denoted as \(\Delta G\), is quintessential when studying chemical reactions and processes. It represents the amount of energy available to do work during a reaction at constant temperature and pressure. If \(\Delta G\) is negative, the reaction occurs spontaneously, whereas a positive value indicates that it is non-spontaneous and requires input of energy.

In the context of the given exercise, the Gibbs Free Energy Change for the electrochemical reactions in the voltaic cells is determined using the formula \(\Delta G = -nFE^\circ\), where \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is Faraday's constant (approximately 96485 C/mol), and \(E^\circ\) is the standard cell potential. For example, for a reaction involving the transfer of one electron in cell 1, \(\Delta G\) was calculated as \(\Delta G_1 = -(1)(96485 C/mol)(0.8453 V)\), yielding \(\Delta G_1\) in units of energy (joules or kilojoules per mole).

By calculating the Gibbs Free Energy Change, students can predict whether reactions will occur spontaneously under standard conditions and understand the relationship between energy change and cell potential in electrochemical reactions.

Electrochemical Cells and Cell Potentials

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa. Two primary types of electrochemical cells are galvanic (or voltaic) cells, which spontaneously produce electricity, and electrolytic cells, which use electricity to drive non-spontaneous chemical reactions.

The cell potential, also known as electromotive force (EMF), is the measure of the voltage difference between two half-cells in an electrochemical cell. It's denoted as \(E_{cell}\) and is directly related to the ability of the cell to perform electrical work. The cell potential under standard conditions (\(E^\circ_{cell}\)) is a crucial factor in calculating the Gibbs free energy change and understanding the thermodynamics of the reaction.

In the exercise example, the cell potentials for two different electrochemical cells involving sodium have been given. These potentials play a key role in determining the overall Gibbs free energy change and, ultimately, the feasibility of the reactions within these cells.

Standard Electrode Potential

The Standard Electrode Potential, denoted as \(E^\circ\), is a measure of the intrinsic tendency of a reactant or ion to gain electrons (get reduced) or to lose electrons (get oxidized). It is measured under standard conditions, which implies a temperature of 298 K, a 1 M concentration for each ion participating in the reaction, and a pressure of 1 atm for any gases involved. These potentials are always measured relative to the standard hydrogen electrode (SHE),, which is assigned a potential of zero volts.

Understanding the concept of standard electrode potential is fundamental when studying electrochemistry because it allows us to predict the direction of electron flow in an electrochemical cell. The exercise involves calculating the standard electrode potential of sodium in a unique context by using experimental cell potentials and the Gibbs free energy formulation. The standard electrode potential is essential for comparison with tabulated values and understanding the behavior of specific electrodes in electrochemical cells.

Solving computational problems in electrochemistry often implies using the standard electrode potential values found in tables (like Appendix D in the exercise) to construct cell diagrams and calculate overall cell potentials.

Voltaic Cells and Reactions

Voltaic cells, also known as galvanic cells, are a class of electrochemical cells that spontaneously convert chemical energy into electrical energy through redox reactions. These cells consist of two different metals immersed in solutions of their respective ions, separated by a salt bridge or a porous disk to maintain electrical neutrality.

The reactions that occur in voltaic cells are split into half-reactions, each taking place at its respective electrode. The oxidation half-reaction occurs at the anode, and the reduction half-reaction occurs at the cathode. By understanding the half-reactions, as shown in the exercise steps for voltaic cells 1 and 2, students can write the balanced overall equation for the cell's reaction.

The exercise demonstrates how to analyze the reactions occurring within the voltaic cell's individual compartments and then how to combine those to reveal the overall reaction happening in the cell. This approach solidifies the concept of combining half-cell reactions to better understand the entire electrochemical process occurring within voltaic cells.