Problem 146 - Chemistry An Atoms First Approach

Question

a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M},\) the cell potential is observed to be 0.31 V. Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

In summary, the equilibrium constant (K) for the reaction \(\mathrm{Au}^{3+} + 4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{AuCl}_{4}^{-}\) at 25°C is approximately \(K \approx 1.14 \times 10^{-21}\).

Step 1: Part a: Drawing the cell diagram

The electrochemical cell consists of two compartments (Au and NaCl). The Au compartment contains \(\mathrm{Au}^{3+}\) ions, and the compartment containing NaCl has \(\mathrm{Cl}^{-}\) ions. Here's a step-by-step break down on how to draw the cell diagram: 1. Draw two separate compartments (beakers), one for each participating substance. 2. Label the compartment on the left as the anode (oxidation occurs), which contains gold in this case. 3. Label the compartment on the right as the cathode (reduction occurs), which contains NaCl in this case. 4. Show the electron flow between the beakers with an arrow pointing from the anode to the cathode. 5. Place a salt bridge (an inverted U-tube containing an inert electrolyte) to connect the two compartments and allow the flow of ions between them. 6. For each compartment, label the concentrations of the species involved in the reaction. For example, for the Au compartment, you will label \(\mathrm{Au}^{3+}\) and for the NaCl compartment, you will label \(\mathrm{Cl^{-}}\).

Step 2: Part b: Calculating the equilibrium constant (K)

We are given the following information: - Reaction in equilibrium: \(\mathrm{Au}^{3+} + 4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{AuCl}_{4}^{-}\) - Cell potential (\(E_{cell}\)) at one point: 0.31 V - Concentration (M) of \(\mathrm{Cl}^{-}\) at one point: 0.10 M - Temperature: 25°C (298.15 K) Using this information and the Nernst equation, we can calculate the equilibrium constant (K) for the reaction. The Nernst equation relates the cell potential, the equilibrium constant, and the concentration of species at any given point. For the given reaction: n = 4 (number of electrons transferred) The Nernst equation is given as: \(E_{cell}=E^{\circ} - \frac{0.0592}{n} \log_{10} Q\) Where: - \(E_{cell}\) is the cell potential. - \(E^{\circ}\) is the standard cell potential at 25°C. - n is the number of electrons transferred during the reaction. - Q is the reaction quotient at a specific point. Under standard conditions, when the reaction is at equilibrium: \(E^{∘}_{cell} = E^{\circ}(\mathrm{cathode}) - E^{\circ}(\mathrm{anode})\) \(E^{\circ}_{cell} = 0\) \(E_{cell} = 0.31 V\) At equilibrium, Q = K, so the Nernst equation becomes: \(0.31 = 0 - \frac{0.0592}{4} \log_{10} K\) Now we can solve for K: 1. Move the constant term to the left side: \(0.31 = -0.0148 \log_{10} K\) 2. Divide both sides by -0.0148: \(-20.947 = \log_{10} K\) 3. Remove the logarithm using the base 10: \(K = 10^{-20.947}\) Solving for K, we get: \(K \approx 1.14 \times 10^{-21}\)

Key Concepts

Galvanic Cell

A galvanic cell, also known as a voltaic cell, is a device that converts chemical energy into electrical energy through a spontaneous redox reaction. It's crucial for students to visualize its components, as it aids in understanding the flow of electrons and the overall function of the cell.

An easy way to picture a galvanic cell is by imagining two separate beakers that represent the two half-cells, connected by a wire and a salt bridge. Within one beaker, there's the anode where oxidation occurs. It's considered the negative electrode as it donates electrons. In the other beaker, there's the cathode where reduction happens, acting as the positive electrode by accepting electrons.

The direction of electron flow is always from the anode to the cathode. The salt bridge, often a U-shaped tube filled with an inert electrolyte, maintains the electrical neutrality of the half-cells by allowing ions to flow between the compartments. Each half-cell contains an electrolyte with reactive ions, and the electrodes are typically made from metals or other conductive materials.

Cell Potential

Cell potential, symbolized as \(E_{cell}\), is the measure of the potential difference between two half-cells in an electrochemical cell. It's expressed in volts (V), reflecting how much voltage the cell can produce during its redox reaction. The difference in potential arises due to the reaction's drive to reach equilibrium by transferring electrons from the anode to the cathode.

The given exercise requires understanding that when a certain amount of \(\mathrm{NaCl}(s)\) is introduced to the cell, the cell potential changes. In our example, the addition of NaCl to the gold compartment results in a measured cell potential of 0.31 V. This observed potential is crucial to calculate other thermodynamic properties, such as the equilibrium constant, using the Nernst equation. It's important to realize that the cell potential gives insight into the driving force of the reaction; a higher potential means a greater tendency for the reaction to occur.

Equilibrium Constant

The equilibrium constant, represented as \(K\), is a number that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It provides a quantitative measure of the position of equilibrium. When the value of \(K\) is much greater than 1, the reaction favors the formation of products, and when it is much less than 1, the reaction prefers reactants.

In our exercise, the value of \(K\) is asked for a specific redox reaction involving \(\mathrm{Au}^{3+}\) and \(\mathrm{Cl}^{-}\). The very small equilibrium constant calculated, approximately \(1.14 \times 10^{-21}\), indicates that under standard conditions, the reaction largely favors the reactants over the products at 25°C. Understanding the magnitude of the equilibrium constant helps in predicting how a reaction will proceed and is also a key factor in the practical applications of electrochemical cells.

Nernst Equation

The Nernst equation is the cornerstone in electrochemistry for relating the cell potential to the reaction quotient and the concentrations of the reaction species. It allows us to determine the cell potential under non-standard conditions, factoring in temperature, concentrations, and the number of electrons transferred (n).

Here's the general form of the Nernst equation:
\(E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log_{10} Q\)
In this equation, \(E_{cell}\) is the measured cell potential, \(E^{\circ}_{cell}\) is the standard cell potential, \(n\) is the number of moles of electrons exchanged, and \(Q\) is the reaction quotient. During a reaction at equilibrium, \(Q\) equals the equilibrium constant \(K\).

In our case, the exercise uses the Nernst equation to find the \(K\) value, showing the direct connection between cell potential and the equilibrium state of a reaction. Simplified versions of the Nernst equation can also effectively teach students about the impact of temperature and reactant/product concentration on cell potential, further enhancing their understanding of electrochemical systems.