Question

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

Short answer

The standard potential for this galvanic cell is 3.87 V. The concentration of Au³⁺ in the nonstandard cell at 25°C with a Mg²⁺ concentration of \(1.00 \times 10^{-5} M\) and a cell potential of 4.01 V is \(2.16 \times 10^{-4} M\).

Step-by-step solution

Calculate Standard Cell Potential

To calculate the standard cell potential (\(E^0_{cell}\)), we need to find the difference in potential (\(\Delta E^0\)) between the two half-reactions. First, identify which half-reaction serves as the anode (oxidation) and which one serves as the cathode (reduction). The reaction with the more positive potential will be the reduction (cathode) reaction, and the reaction with the less positive potential will be the oxidation (anode) reaction. Cathode: Au³⁺ + 3e⁻ → Au (\(E^0_{Au} = 1.50 V\)) Anode: Mg → Mg²⁺ + 2e⁻ (\(E^0_{Mg} = -2.37 V\)) Now calculate the standard cell potential (\(E^0_{cell}\)): \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\)

Determine the Standard Cell Potential

Substitute the given values of \(E^0_{Au}\) and \(E^0_{Mg}\) in the formula: \(E^0_{cell} = 1.50 - (-2.37) = 3.87 V\) The standard potential for this cell is 3.87 V.

Apply the Nernst Equation

To find the concentration of Au³⁺ in the nonstandard cell, we will use the Nernst equation: \(E_{cell} = E^0_{cell} - \frac{RT}{nF} \cdot ln(Q)\) Where: \(E_{cell}\) = cell potential \(E^0_{cell}\) = standard cell potential \(R\) = gas constant (8.314 J/mol K) \(T\) = temperature in Kelvin \(n\) = number of electrons transferred in the reaction (as calculated from the balanced chemical equation) \(F\) = Faraday’s constant (96485 C/mol) \(Q\) = reaction quotient For our problem, we have: \(E_{cell} = 4.01 V\) \(E^0_{cell} = 3.87 V\) \(T = 298 K\) ( Since \(25^\circ C = 298 K\)) \(n = 6\) (Considering the balanced reaction: 2Au³⁺ + 3Mg → 2Au + 3Mg²⁺)

Calculate the Reaction Quotient (Q)

Write the expression for the reaction quotient (Q) using the balanced chemical reaction: \(Q = \frac{([Mg^{2+}]^3)}{([Au^{3+}]^2)}\) We are given \([Mg^{2+}] = 1.00 × 10^{-5} M\), and we want to find \([Au^{3+}]\).

Solve for [Au³⁺]

Rearrange the Nernst equation to isolate ln(Q): \(ln(Q) = \frac{nFE_{cell}-nFE^0_{cell}}{RT} \) Now, substitute the values and solve for \(Q\): \(Q = e^{\frac{6 * 96485 (4.01 - 3.87)}{8.314 * 298}} = 103.95\) Next, solve for \([Au^{3+}]\) using the \(Q\) value and the given concentration of \([Mg^{2+}]\): \([Au^{3+}]^2 = \frac{([Mg^{2+}]^3)}{Q}\) \([Au^{3+}] = \sqrt{\frac{(1.00 × 10^{-5})^3}{103.95}} = 2.16 × 10^{-4} M\) The concentration of Au³⁺ in the nonstandard cell is \(2.16 × 10^{-4} M\).

Key concepts

Standard Cell Potential

Calculating the Standard Cell Potential is essential to understanding how a galvanic cell functions. A galvanic cell is comprised of two half-reactions, which occur in separate compartments known as electrodes. Each of these half-reactions has its own electrical potential. The standard cell potential, otherwise known as the electromotive force (EMF), is the difference between these two potentials.

To find this potential, first identify the cathode and anode in the reactions provided. The cathode is the electrode where reduction occurs, and it typically has a higher potential. In contrast, the anode is where oxidation occurs, featuring a lower potential. For the given exercise, the cathode reaction is the reduction of gold ions ( \(\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}\)), with a potential of \(1.50 \mathrm{V}\). The anode reaction, the oxidation of magnesium ( \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}\)), has a potential of \(-2.37 \mathrm{V}\).

The standard cell potential is calculated by subtracting the anode's potential from the cathode's potential, following this formula: \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\). For the given reactions, this calculation yields \(3.87 \mathrm{V}\). This represents the voltage the galvanic cell can ideally produce under standard conditions.

Nernst Equation

The Nernst Equation is a crucial tool for understanding how cell potential changes under non-standard conditions. While the standard cell potential gives us information at standard conditions, reactions in a real-world setting often occur under varying concentrations, pressures, and temperatures.

The Nernst Equation is stated as follows:\[E_{cell} = E^0_{cell} - \frac{RT}{nF} \cdot \ln(Q)\]- \(E_{cell}\) is the cell potential under non-standard conditions.- \(E^0_{cell}\) is the standard cell potential.- \(R\) is the gas constant, \(8.314 \mathrm{J}/\mathrm{mol} \cdot \mathrm{K}\).- \(T\) is the absolute temperature in Kelvin.- \(n\) is the number of moles of electrons exchanged.- \(F\) is Faraday's constant, \(96485 \mathrm{C}/\mathrm{mol}\).- \(Q\) is the reaction quotient, representing the concentration ratio of products to reactants.

In the previous example, the Nernst Equation helped calculate the cell potential when concentrations were not at standard state levels. Given the cell potential \(4.01 \mathrm{V}\), we used the standard potential \(3.87 \mathrm{V}\) and applied the equation to find other unknown variables pertinent to the system's chemistry.

Reaction Quotient

The Reaction Quotient, symbolized by \(Q\), provides a snapshot of a reaction's current state in terms of reactant and product concentrations. Calculating \(Q\) is essential for applying the Nernst equation to non-standard conditions.

This quotient is defined for any stage of the reaction and is expressed as the ratio of the concentrations of the products, each raised to the power of their stoichiometric coefficients, to the reactants, similarly raised to their respective powers. For the example given:\[Q = \frac{([Mg^{2+}]^3)}{([Au^{3+}]^2)}\]

In this balanced reaction between gold and magnesium, \(Q\) captures the dominance of products or reactants given their current concentrations. This ratio determines how the cell potential adjusts from the standard potential, as per the Nernst Equation. Changing \(Q\) shifts the reaction closer to or further from equilibrium, affecting both \(E_{cell}\) and \([Au^{3+}]\) in the system.

Half-Reactions

Understanding half-reactions is fundamental for analyzing galvanic cells. Each half-reaction occurs at a separate electrode and details either the oxidation or reduction process. In a galvanic cell, these half-reactions happen at the anode and cathode, respectively.

In the provided example, the half-reactions are:

• Reduction: \(\mathrm{Au}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Au}\)
• Oxidation: \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{e}^-\)

During these processes, electrons flow from the anode to the cathode. Each half-reaction has an associated standard potential, indicating the ease with which the process occurs spontaneously. Redox reactions are balanced by the movement of electrons, which hinges on these half-reactions.

Balancing these reactions also requires ensuring that net electron transfer is equal across the entire galvanic cell. This balancing act allows the galvanic cell to convert chemical energy into electrical energy efficiently. Mastering half-reactions helps us grasp the cell's electrochemical behavior, providing insights into the reactions' spontaneity and direction.