Problem 81 - Chemistry 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 (Kf) for \(\mathrm{[Cu(NH_3)_4]^{2+}}\) is \( K_f = 1.15 \times 10^{-4} \).

Step 1: Determine the balanced equation for the cell reaction.

The overall balanced equation for the cell reaction, considering the Cu electrode and the standard hydrogen electrode can be written as follows: \[ \mathrm{Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}} \]

Step 2: Use the Nernst equation.

The Nernst equation is given by \[ E = E^\circ - \frac{RT}{nF} \ln Q \] where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred, F is the Faraday's constant, and Q is the reaction quotient. In this case, we have E = +0.08 V, E° (standard hydrogen electrode) = 0 V, R = 8.314 J/mol·K, T = 298 K, n = 2 (as Cu²⁺ is reduced to Cu), and F = 96,485 C/mol e⁻.

Step 3: Write the expression for the reaction quotient, Q.

For the given reaction, the expression for the reaction quotient, Q, is \[ Q = \frac{[\mathrm{[Cu(NH_3)_4]^{2+}}]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4} \]

Step 4: Plug in the known values and solve for Q.

We are given that the concentrations of \(\mathrm{[Cu(NH_3)_4]^{2+}}\) and \(\mathrm{NH_3}\) are both 1 M. As we are looking for the formation constant when the complex \(\mathrm{[Cu(NH_3)_4]^{2+}}\) is formed, we can assume that the concentration of free \(\mathrm{Cu^{2+}}\) ions is negligible compared to the concentration of \(\mathrm{NH_3}\). Thus, we can simplify the expression for Q as: \[ Q = \frac{[\mathrm{[Cu(NH_3)_4]^{2+}}]}{[\mathrm{NH_3}]^4} \] Now, we can plug in the known values into the Nernst equation and solve for Q: \[ 0.08 \mathrm{V} = 0 \mathrm{V} - \frac{(8.314\,\mathrm{J/mol\cdot K})(298\,\mathrm{K})}{(2)(96,485\,\mathrm{C/mol})} \ln Q \] Solving for Q, we get: \[ Q = 1.15 \times 10^{-4} \]

Step 5: Find the formation constant.

Since we have the value of Q, we can now plug it into the expression for the formation constant Kf, which is the same as Q for our simplified reaction: \[ K_f = \frac{[\mathrm{[Cu(NH_3)_4]^{2+}}]}{[\mathrm{NH_3}]^4} \] \[ K_f = \frac{1}{(1)^4} \times 1.15 \times 10^{-4} \] Hence, the formation constant (Kf) for \(\mathrm{[Cu(NH_3)_4]^{2+}}\) is: \[ K_f = 1.15 \times 10^{-4} \]

Key Concepts

Electrochemistry

Electrochemistry deals with the intersection between electricity and chemical reactions. It focuses on processes where electrons are transferred between chemical species. One of the most interesting aspects of electrochemistry is the study of electrochemical cells, which generate electrical energy from chemical reactions.
These cells are composed of electrodes submerged in electrolyte solutions, facilitating electron flow between them.

• Galvanic cells – These are electrochemical cells that produce electrical energy through spontaneous redox reactions.
• Electrolytic cells – These cells use electricity to induce non-spontaneous chemical reactions.

In this problem, we are working with a galvanic cell where a copper electrode is immersed in a solution containing a complex ion. The cell's potential is determined by how the reaction progresses.

Formation Constant

The formation constant, often written as \(K_f\), is crucial in understanding the stability of complex ions in solution. It quantifies the equilibrium state for the formation of a complex from its components, indicating the tendency for a metal ion to bind with ligands.
A higher \(K_f\) value indicates a more stable complex, as it signifies a strong bond between the metal and the attached molecules.

• Complex formation involves a central metal ion and surrounding ligands.
• Equilibrium concentration tells us how much of the initial reactants remain unreacted when equilibrium is reached.

In the exercise, we're asked to find the \(K_f\) for the complex \([\mathrm{Cu(NH_3)_4}]^{2+}\), based on the given electrode reactions and measured cell potential.

Complex ion equilibrium

Complex ion equilibrium refers to the delicate balance between complex ions and their constituent ions within a solution. These systems often involve intricate interactions dictated by both the chemical nature of the central metal ion and that of the ligands.
As the concentrations of components shift, the position of equilibrium adjusts accordingly.

• In this exercise, the equilibrium involves \([\mathrm{Cu^{2+}}]\), \(\mathrm{NH_3}\), and the complex \([\mathrm{Cu(NH_3)_4}]^{2+}\).
• The equilibrium equation helps in deriving expressions for both the formation constant and the reaction quotient \(Q\).

Understanding this equilibrium is vital for predicting how various changes in concentration can impact the overall stability and formation of the complexes.

Cell Potential Calculations

Calculating a cell's potential involves using the Nernst Equation, a fundamental formula in electrochemistry. This equation relates the cell potential to its standard electrode potential, temperature, and reaction quotient \(Q\):\[ E = E^\circ - \frac{RT}{nF} \ln Q \]Here:

• \(E\) is the measured cell potential, in this problem given as \(+0.08 \,\text{V}\).
• \(E^\circ\) is the standard cell potential. For the hydrogen electrode, it's \(0 \,\text{V}\).
• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged—here it is 2.
• \(F\) is the faraday constant.

By solving the Nernst Equation with these values, we can calculate the reaction quotient \(Q\), and thus deduce the formation constant \(K_f\), which helps explain the equilibrium chemistry of the cell.