Problem 81 - Chemistry: The Central Science

Question

A Cu electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and \(1.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+0.08 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\)

Step-by-Step Solution

Answer

The formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is 1.

Step 1: Write down the half-reactions

The redox half-reactions occurring in the cell are: 1. Standard hydrogen electrode: \[2\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \quad\text{(reduction)}\] 2. Cu electrode: \[\mathrm{Cu^{2+}} + 4\mathrm{NH}_{3} \leftrightarrow \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} + 2\mathrm{e}^{-} \quad \text{(reduction)}\]

Step 2: Determine the overall cell reaction

Now we combine both half-reactions (balancing the number of electrons) to obtain the overall cell reaction: \[2\mathrm{H}^{+} + \mathrm{Cu^{2+}} + 4\mathrm{NH}_{3} \leftrightarrow \mathrm{H}_{2} + \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\]

Step 3: Use the Nernst equation to relate cell potential to concentration

The Nernst equation relates the cell potential (E) to concentrations and the standard cell potential (E°): \[E = E^{\circ} - \frac{RT}{nF} \ln{Q}\] In this equation, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient. Since the standard hydrogen electrode has a potential of 0 V, the standard cell potential (E°) for this cell is equal to the reduction potential of the Cu electrode. We can rewrite the Nernst equation using the given information and simplifying it, knowing that the temperature T = 298K (room temperature), n = 2 (since there are two moles of transferred electrons), and E = +0.08 V: \[0.08 \mathrm{V} = E^{\circ}_{\mathrm{Cu}} - \frac{8.314\mathrm{J/mol\cdot K}(298\mathrm{K})}{2(96485\mathrm{C/mol})} \ln{Q}\] We can also find the reaction quotient, Q, as follows: \[Q = \frac{[\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}][\mathrm{H}_{2}]^{1/2}}{[\mathrm{H}^{+}]^{2}[\mathrm{Cu^{2+}}][\mathrm{NH}_{3}]^{4}}\] Since the concentrations are given at 1 M, the equation can be simplified to: \[Q = \frac{1}{1} = 1\]

Step 4: Determine the standard reduction potential of Cu electrode

Substitute Q = 1 into the Nernst equation: \[0.08 \mathrm{V} = E^{\circ}_{\mathrm{Cu}} - \frac{8.314\mathrm{J/mol\cdot K}(298\mathrm{K})}{2(96485\mathrm{C/mol})} \ln{1}\] Since the logarithm of 1 is 0, we have: \[E^{\circ}_{\mathrm{Cu}} = 0.08 \mathrm{V}\]

Step 5: Calculate the formation constant

Now, we use the equation for the formation constant (Kf) of the complex ion: \[K_{f} = \frac{[\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}]}{[\mathrm{Cu^{2+}}][\mathrm{NH}_{3}]^{4}}\] Since the concentration of the complex ion and the concentration of \(\mathrm{NH}_{3}\) are equal to 1 M, we can write: \[K_{f} = \frac{1}{1 \cdot 1^{4}} = 1\] Thus, the formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is 1.

Key Concepts

Nernst Equation

The Nernst Equation is an essential component of electrochemistry, allowing us to understand how the cell potential or voltage of an electrochemical cell changes with the concentration of reactants and products. It's represented mathematically as:\[ E = E^{\circ} - \frac{RT}{nF} \ln{Q} \]Where:

• \( E \) is the cell potential.
• \( E^{\circ} \) is the standard cell potential.
• \( R \) is the gas constant (8.314 J/mol·K).
• \( T \) is the absolute temperature in Kelvin.
• \( n \) is the number of moles of electrons exchanged in the reaction.
• \( F \) is the Faraday constant (96485 C/mol).
• \( Q \) is the reaction quotient, which compares the concentrations of products and reactants.

At room temperature (298K), the equation simplifies, allowing a direct way to predict or calculate cell potential based on concentrations. Understanding this equation is crucial because it bridges thermodynamics and electrochemistry, helping us gauge how concentration impacts the electrical potential of a chemical cell.

Formation Constant

The formation constant, often denoted as \( K_f \), is a measure of the stability of a complex ion in solution. It reflects the equilibrium between a metal ion and ligands forming a complex. For example, when a copper ion forms the complex \([\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}\), the equilibrium is expressed as:\[ \mathrm{Cu}^{2+} + 4\mathrm{NH}_3 \leftrightarrow [\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+} \]The formation constant is given by:\[ K_f = \frac{[\mathrm{Complex\amp;ion}]}{[\mathrm{Metal\,ion}][\mathrm{Ligand}]^4} \]A high \( K_f \) value indicates a stable complex, meaning it is less likely to dissociate back into metal ion and ligands. In the context of electrochemical cells, \( K_f \) can be used to understand how easy or difficult it is for such complexes to form, influencing the cell's potential and functionality. This is particularly important when evaluating reactions involving ligands like ammonia.

Half-Reaction

A half-reaction is one part of a redox (reduction-oxidation) reaction, which can be either the oxidation or reduction process. In a redox reaction, electrons are transferred between species, and a reaction can be split into two half-reactions:- **Oxidation Half-Reaction**: Involves the loss of electrons. For example, when copper oxidizes, it forms \( \mathrm{Cu}^{2+} \) and releases electrons.- **Reduction Half-Reaction**: Involves the gain of electrons. For instance, in a standard hydrogen electrode, protons \( \mathrm{H}^{+} \) gain electrons to become \( \mathrm{H}_2 \).These reactions can be represented as:

• Reduction: \( 2\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_2 \)
• Oxidation: \( \mathrm{Cu}^{2+} + 4\mathrm{NH}_3 \leftrightarrow \left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+} + 2\mathrm{e}^{-} \)

Half-reactions are crucial for balancing redox reactions and understanding how electrons transfer within a cell. This transfer of electrons generates the electromotive force which drives the cell's functionality.

Redox Reactions

Redox reactions are fundamental to analyzing many electrochemical processes. They involve the transfer of electrons between chemical species and can be split into their component half-reactions, involving an oxidation and a reduction process.In our context, the core of redox reactions:- **Oxidation**: The loss of electrons by a molecule, atom, or ion. For instance, the transformation of copper from \( \mathrm{Cu} \) to \( \mathrm{Cu}^{2+} \).
- **Reduction**: The gain of electrons, essential for electron flow within living organisms and technology applications such as batteries.The full redox reaction in an electrochemical cell is the sum of its half-reactions:\[ 2\mathrm{H}^{+} + \mathrm{Cu^{2+}} + 4\mathrm{NH}_3 \leftrightarrow \mathrm{H}_2 + [\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+} \]Redox principles are applied in many fields including biological energy cycles, industrial processes, and renewable energy technologies. Understanding this concept gives insight into how electrical energy is harnessed from chemical reactions, which is vital in developing efficient and effective electrochemical cells.