Question

Consider a cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ} &=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{E}^{\circ} &=0.77 \mathrm{~V} \end{aligned} $$ 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 \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). 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 \(25^{\circ} \mathrm{C}\).

Short answer

The equilibrium constant (K) for the reaction \(Au^{3+}(aq) + 4 Cl^{-}(aq) \rightleftharpoons AuCl_{4}^{-}(aq)\) at \(25^{\circ} \text{C}\) is approximately \(1.92 \times 10^{49}\).

Step-by-step solution

Identify the anode and the cathode reactions

In a galvanic cell, the negative electrode is the anode, and the positive electrode is the cathode. The half-reaction with the lower standard reduction potential will undergo oxidation (losing electrons) and become the anode. Given the half-reactions and their standard reduction potentials: 1. \(Au^{3+} + 3e^{-} \rightarrow Au \quad\quad \mathscr{E}^{\circ} = 1.50 V\) 2. \(Fe^{3+} + e^{-} \rightarrow Fe^{2+} \quad \mathscr{E}^{\circ} = 0.77 V\) Since has a lower standard reduction potential, it will undergo oxidation, and thus, the anode half-reaction will be: \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\) The cathode half-reaction remains the same: \(Au^{3+} + 3e^{-} \rightarrow Au\)

Draw the cell

Now we draw the cell with labeled components: - At the anode, write the oxidation half-reaction: \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\) - At the cathode, write the reduction half-reaction: \(Au^{3+} + 3e^{-} \rightarrow Au\) - Connect an external wire for electron flow from the anode to the cathode. - Place a salt bridge between the two half-cell compartments to balance the charge. - Indicate the direction of electron flow towards the cathode with an arrow. - Label the anode, cathode, and concentrations of \(Fe^{2+}\) and \(Au^{3+}\) as 1M under standard conditions. #b. Calculate the value of K for the reaction given the cell potential when [Cl^(-)]= 0.10 M#

Write the Nernst equation

The Nernst equation allows us to relate the cell potential at non-standard conditions (\(E_\text{cell}\)) to the standard cell potential (\(E_\text{cell}^{\circ}\)): \(E_\text{cell} = E_\text{cell}^{\circ} - \frac{0.0592}{n} \log Q\) Where \(n\) is the number of electrons transferred and \(Q\) is the reaction quotient.

Calculate the standard cell potential

The standard cell potential can be calculated using the difference in standard reduction potentials: \(E_\text{cell}^{\circ} = \mathscr{E}_\text{cathode}^{\circ} - \mathscr{E}_\text{anode}^{\circ} = 1.50 V - 0.77 V = 0.73 V\)

Use the Nernst equation to find K

First, determine the number of electrons transferred: \(n = 3\) Second, arrange the given reaction in the compartment containing gold: \(Au^{3+} + 4Cl^{-} \rightleftharpoons AuCl_{4}^{-}\) Next, find the corresponding reaction quotient: \(Q = \frac{[AuCl_{4}^{-}]}{[Au^{3+}][Cl^{-}]^{4}}\) Now substitute the given cell potential, standard cell potential, and reaction quotient into the Nernst equation: \(0.31 V = 0.73 V - \frac{0.0592}{3} \log Q\) Solve for the reaction quotient: \(\log Q = \frac{3(0.73 V - 0.31 V)}{0.0592} \Rightarrow Q = 10^{49.28}\) Finally, calculate the equilibrium constant, K: \(K = \frac{[AuCl_{4}^{-}]_{\text{eq}}}{[Au^{3+}]_{\text{eq}}[Cl^{-}]_{\text{eq}}^{4}} = Q\) Since \([Cl^{-}]_{\text{eq}} = 0.10 M\), we can substitute this value into the equation: \(K = \frac{[AuCl_{4}^{-}]_{\text{eq}}}{[Au^{3+}]_{\text{eq}}(0.10)^{4}}\) Now solve for K: \(K = 10^{49.28} \Rightarrow K \approx 1.92 \times 10^{49}\) So, the value of K for the reaction is approximately \(1.92 \times 10^{49}\) at \(25^{\circ} \text{C}\).

Key concepts

Galvanic Cell

A galvanic cell, also known as a voltaic cell, is a device that generates electrical energy from spontaneous chemical reactions. It is made up of two half-cells, each containing a different metal electrode immersed in a solution of its ions. These half-cells are connected by a salt bridge and an external circuit.

Key components include:

• Anode: The electrode where oxidation occurs. In this example, iron ( Fe^{2+} ightarrow Fe^{3+} + e^{-} ) serves as the anode. Electrons are produced here.
• Cathode: The electrode where reduction occurs. Gold ( Au^{3+} + 3e^{-} ightarrow Au ) is the cathode. Electrons flow to this electrode, causing the reduction reaction.
• Electron Flow: Electrons move from the anode to the cathode through an external wire.

The salt bridge helps maintain electrical neutrality by allowing ions to flow between the compartments, preventing the build-up of charge. These cells provide essential insights into electrochemical processes.

Understanding galvanic cells can help you comprehend how batteries work and the principles behind electrochemistry.

Nernst Equation

The Nernst equation describes how the potential of an electrochemical cell varies with the concentration of the ions involved in the reaction. It's crucial for understanding cells that are not at standard conditions. The equation is:
\[E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{n} \log Q\]

Where:

• \(E_{\text{cell}}\) is the cell potential under non-standard conditions.
• \(E_{\text{cell}}^{\circ}\) is the standard cell potential.
• \(n\) is the number of electrons transferred in the reaction.
• \(Q\) is the reaction quotient, reflecting the ratio of concentration of products to reactants.

The Nernst equation allows us to calculate how changes in concentration or pressure affect cell potential, making it a powerful tool for predicting cell behavior in real-world conditions.

In our example, when \([Cl^-] = 0.10 M\), the equation helps calculate the cell potential and equilibrium constant, reflecting the reaction's shift in equilibrium.

Standard Cell Potential

The standard cell potential (\(E_{\text{cell}}^{\circ}\)) is the difference in potential between two electrodes measured under standard conditions (1 M concentration, 1 atm pressure, 25°C). It indicates the driving force behind an electrochemical reaction.

It can be calculated using the formula:
\[E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\]

In the given exercise, the standard cell potential is:
\[E_{\text{cell}}^{\circ} = 1.50 V - 0.77 V = 0.73 V\]

This value provides insight into the cell's ability to perform work and indicates that the reaction is spontaneous under standard conditions. Positive values of \(E_{\text{cell}}^{\circ}\) suggest that the chemical reaction is likely to proceed without additional energy input.

Such calculations help in designing batteries and understanding different electrochemical systems, offering a foundational concept in electrochemistry.