Question

Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ 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 cell under standard conditions can be represented as: Anode: Fe, where the oxidation reaction is \(Fe^{3+} + e^- \to Fe^{2+}\), and the concentration of [Fe³⁺] = 1 M and [Fe²⁺] = 1 M. Cathode: Au, where the reduction reaction is \(Au^{3+} + 3e^- \to Au\), and the concentration of [Au³⁺] = 1 M. Using the Nernst equation with the given cell potential of 0.31 V and the reaction \(Au^{3+}(aq) + 4\mathrm{Cl}^-(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)\), the equilibrium constant K is calculated to be approximately \(6.23 \times 10^{-11}\).

Step-by-step solution

Draw the cell under standard conditions

To draw the cell under standard conditions, we first need to identify the half-reactions that occur at the anode and the cathode based on their reduction potentials. $$\begin{aligned}\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}\ & \ \mathscr{E}^{\circ} = 1.50 V \\\ \mathrm{Fe}^{3+} + \mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \ \mathscr{E}^{\circ} = 0.77 V \end{aligned}$$ Since the reduction potential of Au³⁺ is greater than that of Fe³⁺, Au³⁺ will be reduced at the cathode, and Fe³⁺ will be oxidized at the anode. The cell representation is as follows: Anode | Cathode -|- \(Fe^{3+} + e^- \to Fe^{2+}\) | \(Au^{3+} + 3e^- \to Au\) Fe | Au Now, label the direction of electron flow, and the concentrations: - Anode: Fe, oxidation occurs, electrons flow from the anode to the cathode - Cathode: Au, reduction occurs, electrons flow to the cathode from the anode Concentrations: - Anode compartment: [Fe³⁺] = 1 M, [Fe²⁺] = 1 M (standard conditions) - Cathode compartment: [Au³⁺] = 1 M (standard conditions)

Calculate the value of K for the given reaction

Use the Nernst equation and the given cell potential (0.31 V) to find the equilibrium constant K for the reaction: $$\mathrm{Au}^{3+}(aq) + 4\mathrm{Cl}^-(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)$$ The Nernst equation is: $$\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF} \ln{Q}$$ Where: - \(\mathscr{E}\) is the cell potential under non-standard conditions - \(\mathscr{E}^{\circ}\) is the cell potential under standard conditions - R is the gas constant (8.314 J/mol·K) - T is the temperature in Kelvin (298 K for 25°C) - n is the number of electrons transferred (3 in this case) - F is the Faraday constant (96485 C/mol) - Q is the reaction quotient Since \(\mathrm{Au}^{3+}\) is being reduced in the cell, the overall reaction in the cell can be expressed as: $$\mathrm{Au}^{3+} + 3\mathrm{e}^{-} + 4\mathrm{Cl}^{-} \rightleftharpoons \mathrm{Au} + \mathrm{AuCl}_{4}^{-}$$ Let the cell potential under non-standard conditions be 0.31 V. Notice that only 0.10 M of Cl⁻ ions are added to the Au³⁺ compartment for the non-standard cell potential. Now, we can use the Nernst equation to find the reaction quotient, Q: $$0.31 = 1.50 - \frac{8.314 \times 298}{3 \times 96485} \ln{Q}$$ Solve for Q: $$Q = \exp{\left(\frac{3(0.31-1.50) \times 96485}{8.314 \times 298}\right)} \approx 6.23 \times 10^{-11}$$ For the given reaction, at equilibrium, \(Q = K\), and so the equilibrium constant K is: $$K \approx 6.23 \times 10^{-11}$$

Key concepts

Half-Reactions

In electrochemical cells, the reactions occurring at the electrodes are often split into two separate processes called half-reactions. Each half-reaction represents either an oxidation or a reduction process.
For the provided cell, we have:

• The cathode reaction: \(\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}\), with a standard reduction potential of 1.50 V.
• The anode reaction: \(\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\), with a standard reduction potential of 0.77 V.

Understanding half-reactions helps us determine which elements are oxidized or reduced.
In this exercise, gold (Au) is reduced, while iron (Fe) is oxidized. The electrons flow from the anode, where the oxidation occurs, to the cathode, where the reduction takes place. This flow is crucial for generating electric current in the cell.

Nernst Equation

The Nernst equation allows us to calculate the cell potential under non-standard conditions.
The general form is:\[\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF}\ln{Q}\]Here,

• \(\mathscr{E}\): cell potential at non-standard conditions.
• \(\mathscr{E}^\circ\): standard cell potential.
• \(R\): universal gas constant (8.314 J/mol·K).
• \(T\): temperature in Kelvin (298 K in this case).
• \(n\): number of moles of electrons exchanged (3 for this reaction).
• \(F\): Faraday's constant (96485 C/mol).
• \(Q\): reaction quotient.

In this problem, the Nernst equation helps find \(Q\) when the non-standard cell potential \(\mathscr{E}\) is given as 0.31 V. This equation captures how changes in concentration and reaction conditions affect the cell potential.

Cell Potential

Cell potential (or electromotive force, EMF) is the measure of the energy available from a redox reaction occurring within an electrochemical cell.
It's determined by the difference in reduction potentials between the cathode and the anode.
Under standard conditions, it can be calculated as:\[\mathscr{E}^\circ_{\text{cell}} = \mathscr{E}^\circ_{\text{cathode}} - \mathscr{E}^\circ_{\text{anode}}\]For the cell in the exercise:

• Cathode (Au³⁺ to Au): \(1.50\, \text{V}\).
• Anode (Fe³⁺ to Fe²⁺): \(0.77\, \text{V}\).

Thus, \(\mathscr{E}^\circ_{\text{cell}} = 1.50\, \text{V} - 0.77\, \text{V} = 0.73\, \text{V}\).
These calculations allow us to predict how much voltage a cell can produce, aiding in designing batteries or understanding their efficiency under varying conditions.

Equilibrium Constant

The equilibrium constant \(K\) gives us the ratio of concentrations of products to reactants at equilibrium for a reversible reaction.
It's a crucial parameter in determining reaction extents in chemical systems.
In our case, we found the \(K\) value using the reaction where \(\mathrm{Au}^{3+}(aq) + 4\mathrm{Cl}^{-}(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)\)
By applying the Nernst equation and substituting in the given cell potential (0.31 V), we calculated \(Q\), and subsequently \(K\), reflecting the system's equilibrium status.
This value \(K \approx 6.23 \times 10^{-11}\) indicates how the formation of \(\mathrm{AuCl}_{4}^{-}\) is influenced by the added \(\mathrm{Cl}^-\) ions. A small \(K\) value means that at equilibrium, very few products form, emphasizing the dominance of reactants, which can profoundly impact reaction predictability in practical applications.