Problem 137 - Chemistry An Atoms First Approach

Question

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{lr} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & 1.50 \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg} & -2.37 \end{array}$$ a. What is the standard potential for this cell? b. A nonstandard cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Mg}^{2+}\right]=\) \(1.00 \times 10^{-5} M .\) The cell potential is observed to be 4.01 V. Calculate \(\left[\mathrm{Au}^{3+}\right]\) in this cell. \mathscr{E}^{\circ}

Step-by-Step Solution

Answer

a. The standard potential for this cell is 3.87 V. b. The concentration of Au³⁺ in the nonstandard cell at 25°C is approximately 7.92 × 10⁻⁵ M.

Step 1: Write the cell reaction

To find the standard potential for this cell, first, write the overall cell reaction by combining the given half-reactions. The overall cell reaction will have both reduction and oxidation occurring simultaneously. $$\begin{array}{l} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg} \end{array}$$

Step 2: Determine the oxidation and reduction reactions

In this cell, Au³⁺ will be reduced since it has a higher standard reduction potential. Therefore, the Mg²⁺ half-reaction must be reversed as the oxidation half-reaction. The reverse half-reaction will be: $$\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}$$

Step 3: Calculate the standard cell potential

The standard cell potential is the sum of the standard reduction potential for the reduction half-reaction (Au³⁺) and the oxidation half-reaction (Mg). Since the oxidation half-reaction is the reverse of the given Mg²⁺ half-reaction, the sign of the standard reduction potential must be changed to positive for the oxidation half-reaction. Standard potential = \( \mathscr{E}^{\circ}_(\textrm{reduction}) - \mathscr{E}^{\circ}_(\textrm{oxidation}) \) Standard potential = \(1.50 - (-2.37)\) Standard potential = \(3.87 V\) So, the standard potential for this cell is 3.87 V. #b. Calculate [Au³⁺] in the nonstandard cell#

Step 1: Identify the Nernst equation

To find the concentration of Au³⁺ in the nonstandard cell, we can use the Nernst equation. The Nernst equation is given by: $$E_\mathrm{cell} = \mathscr{E}^\circ_\mathrm{cell} - \frac{RT}{nF} \ln(\mathrm{Q})$$ Where E_cell is the observed cell potential, \( \mathscr{E}^\circ_\mathrm{cell} \) is the standard cell potential, R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin, n is the number of electrons transferred in the balanced redox reaction (in this case, n = 6), and F is the Faraday constant (96,485 C/mol).

Step 2: Calculate the reaction quotient (Q)

The reaction quotient (Q) for this cell is given by: $$\mathrm{Q}=\frac{\left[\mathrm{Au}^{3+}\right]}{\left[\mathrm{Mg}^{2+}\right]^{2}}$$

Step 3: Solve the Nernst equation for [Au³⁺]

Rearrange the Nernst Equation to solve for the concentration of [Au³⁺]: $$\left[\mathrm{Au}^{3+}\right] = \left[\mathrm{Mg}^{2+}\right]^2 \mathrm{exp}\left( \frac{nF(E_{cell} - \mathscr{E}^\circ_{cell})}{RT} \right)$$ Now, plug in the values for temperature, cell potential, standard cell potential, and the concentration of Mg²⁺: $$\left[\mathrm{Au}^{3+}\right] = \left(1.00 \times 10^{-5}\right)^2 \mathrm{exp}\left( \frac{6 \times 96,485(4.01 - 3.87)}{8.314 \times 298} \right)$$

Step 4: Compute the concentration of [Au³⁺]

Now, compute the concentration of [Au³⁺]: $$\left[\mathrm{Au}^{3+}\right] \approx 7.92 \times 10^{-5} \,\mathrm{M}$$ The concentration of Au³⁺ in the nonstandard cell at 25°C is approximately 7.92 × 10⁻⁵ M.

Key Concepts

Standard Cell Potential

To understand the concept of standard cell potential, it's essential to know its role in electrochemical cells. The standard cell potential, denoted as \( \mathscr{E}^\circ \), is the potential difference between two electrodes of a cell under standard conditions, which are typically 1 M concentration for aqueous solutions, 1 atm pressure for gases, and a temperature of 25°C. This potential arises due to the tendency of a redox reaction to occur spontaneously. The higher the potential, the more the reaction will drive the electrons through the external circuit.

For a galvanic cell, which generates electrical energy from spontaneous chemical reactions, the standard cell potential can be calculated by adding the potentials of the cathodic (reduction) and anodic (oxidation) reactions. Usually, this is done by subtracting the standard reduction potential of the anode from that of the cathode. In our given exercise, the standard cell potential is calculated as follows:

- Reduction reaction: \( \mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \) with \( \mathscr{E}^\circ = 1.50 \, \text{V} \)
- Oxidation reaction: Reversing \( \mathrm{Mg}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mg} \) to \( \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2 \mathrm{e}^{-} \) which changes \( \mathscr{E}^\circ \) to \(+2.37 \, \text{V} \)
- \( \mathscr{E}^\circ_{\text{cell}} = 1.50 \, \text{V} - (-2.37 \, \text{V}) = 3.87 \, \text{V} \)

Nernst Equation

The Nernst equation plays a crucial role in connecting the electrical potential of a galvanic cell under nonstandard conditions to its concentration, pressure, and temperature. It helps us find the actual cell potential by considering deviations from standard conditions. The formula is:
\[E_\mathrm{cell} = \mathscr{E}^\circ_\mathrm{cell} - \frac{RT}{nF} \ln(Q)\]

Where:

• \( E_\mathrm{cell} \) is the cell potential at nonstandard conditions
• \( \mathscr{E}^\circ_\mathrm{cell} \) is the standard cell potential
• \( R \) is the universal gas constant \((8.314 \, \text{J/mol·K})\)
• \( T \) is the temperature in Kelvin
• \( n \) is the number of electrons transferred in the reaction
• \( F \) is the Faraday constant \((96,485 \, \text{C/mol})\)
• \( Q \) is the reaction quotient, a measure of the concentrations of chemical species

The Nernst equation can be rearranged to find unknown concentrations in nonstandard conditions, using known values for the cell potential and standard conditions.

Redox Reaction

A redox reaction is a fundamental concept in electrochemistry involving the transfer of electrons between two species. These reactions are composed of two half-reactions: reduction and oxidation. In each galvanic cell, one species donates electrons (oxidation), and another accepts electrons (reduction).

In the classroom example, the reaction involves:

• The reduction half-reaction: \( \mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \)
• The oxidation half-reaction: \( \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2 \mathrm{e}^{-} \) (reverse of the given reduction)

The term 'redox' stands for reduction and oxidation, highlighting the electron flow from one reactant to another. Balancing these reactions requires ensuring that the electrons lost in oxidation match the electrons gained in reduction. In this case, scaling the half-reactions to achieve electron balance is vital, often computed with the Least Common Multiple (LCM) of transferred electrons, resulting in 6 electrons cooperating in this reaction.

Understanding the exchange of electrons allows us to harness these reactions to power electronic devices, battery functioning, and more.

Concentration Calculation

Concentration calculation is essential for understanding how the Nernst equation is applied to real-world scenarios. By rearranging the Nernst equation, you can solve for the unknown concentration of reactants or products in a cell.

In the exercise, we calculate \([\mathrm{Au}^{3+}]\) using:
\[\left[\mathrm{Au}^{3+}\right] = \left[\mathrm{Mg}^{2+}\right]^2 \exp\left( \frac{nF(E_{\text{cell}} - \mathscr{E}^\circ_{\text{cell}})}{RT} \right)\]

This setup involves the concentration of \([\mathrm{Mg}^{2+}]\), the measured cell output \((E_{\text{cell}})\), and constants from the Nernst equation. With experimentally determined values, you can deduce the concentration of the gold ions, which is essential for converting theoretical concepts into practical applications.

Using this equation, students learn to predict and adjust concentrations needed to achieve desired cell potentials, crucial in constructing batteries and electronic devices where precise chemical balances matter.