Problem 100 - 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 approximately 1.997 x 10^{-3}.

Step 1: Write the balanced cell reaction and determine the states of species involved in the reaction.

First, we need to write the balanced cell reaction. The Cu electrode produces Cu2+ by losing 2 electrons, and the standard hydrogen electrode (SHE) produces H2 gas by gaining 2 electrons. The complex ion, \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\), is formed by the reaction of Cu2+ with 4 NH3 molecules. So, the balanced cell reaction is: \[ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} + \underbrace{2 \mathrm{H}^{+} + 2 \mathrm{e}^{-}}_{\text {Standard hydrogen electrode}} \rightarrow \mathrm{Cu}^{2+} + 4 \mathrm{NH}_{3}+ \underbrace{\mathrm{H}_{2}}_{\text {Standard hydrogen electrode}} \] Now, we can determine the states of species involved in the reaction: - Cu electrode: solid - \(\mathrm{NH}_{3}\): aqueous - \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\): aqueous - Cu2+: aqueous - H+: aqueous - H2: gas

Step 2: Applying the Nernst equation to find the reaction quotient, Q.

We know that the cell emf (E) is 0.08 V and the number of electrons transferred (n) is 2. The standard cell potential for the hydrogen electrode (E0) is 0 V. Now we can apply the Nernst equation to find the reaction quotient, Q: \(E = E^0 - \frac{0.05916}{n}\log Q\) \(0.08 = 0 - \frac{0.05916}{2}\log Q\) Solve for Q: \(Q = 10^{\frac{2(0.08)}{0.05916}}\)

Step 3: Calculate the reaction quotient, Q.

Now, we can calculate the value of Q: \(Q = 10^{\frac{2(0.08)}{0.05916}} = 10^{2.701} \approx 500.65\)

Step 4: Write the expression for K, the formation constant, and calculate its value.

Now that we have the reaction quotient, Q, we can write an expression for the formation constant, K, using the balanced cell reaction: \(K = \frac{[\mathrm{Cu}^{2+}][\mathrm{NH}_{3}]^4}{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}][\mathrm{H}^{+}]^2}\) Since the cell reaction occurs at 1 M concentration for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and NH3, and the cell emf = 0.08 V, we can assume that the concentrations of Cu2+ and H+ ions are also 1 M. Therefore, we can simplify the expression for K as: \(K = \frac{(1)(1)^4}{(500.65)(1)^2}\) Calculate the value of K: \(K = \frac{1}{500.65} \approx 1.997 \times 10^{-3}\) Thus, the formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is approximately 1.997 x 10^{-3}.

Key Concepts

Understanding Electrochemical Cells

Electrochemical cells are the foundation of batteries and various other electronics that utilize chemical reactions to produce electrical energy. An electrochemical cell comprises two electrodes, the anode and the cathode, immersed in an electrolyte solution. These components facilitate a redox reaction, wherein one electrode experiences oxidation (loss of electrons), and the other undergoes reduction (gain of electrons).

Through the cell's external circuit, the electrons flow from the anode to the cathode, generating an electric current. The ability of an electrochemical cell to do work, measured in volts (V), is its electromotive force (emf). The given exercise involves a copper electrode (Cu) and a standard hydrogen electrode (SHE), forming an electrochemical cell from which we can derive important chemical properties such as the formation constant.

The Nernst Equation Explained

The Nernst equation is a mathematical expression that relates the potential of an electrochemical cell to the concentration of the chemical species involved in the reaction. It is an important tool for understanding and predicting how changes in concentration impact a cell's emf.

The general form of the Nernst equation is:
\[ E = E^0 - \frac{0.05916}{n} \log Q \]
where \(E\) is the cell potential, \(E^0\) is the standard cell potential, \(n\) is the number of moles of electrons transferred, and \(Q\) is the reaction quotient, which indicates the ratio of the concentrations of products to reactants. In the context of the provided exercise, we use the Nernst equation to calculate \(Q\), giving us a path to find the formation constant for the complex ion.

Chemical Equilibrium and the Reaction Quotient

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical process are equal, resulting in no net change in the concentrations of reactants and products over time. At equilibrium, the concentrations of the chemical species are constant.

To quantify the position of an equilibrium, chemists use the reaction quotient, \(Q\), a ratio similar to the equilibrium constant but calculated using the concentrations of the reactants and products that are not at equilibrium. The reaction quotient is instrumental in determining the direction in which a reaction must proceed to achieve equilibrium.

In our exercise, the reaction quotient \(Q\) was calculated using the Nernst equation. This allowed us to then determine how far the system was from equilibrium and provided the necessary information to calculate the formation constant for the complex ion.

Complex Ion Formation and Its Constants

Complex ions are species formed from a central metal ion and one or more molecules or ions bonded to it, known as ligands. The formation of complex ions is an equilibrium process described by the formation constant \(K_f\), which gives a measure of the stability of the complex in solution.

The formation constant is determined by the ratio of the concentration of the complex ion to the product of the concentrations of the individual ions and ligands, taking their stoichiometric coefficients as exponents:
\[ K_f = \frac{[\text{complex ion}]}{[\text{metal ion}][\text{ligand}]^n} \]
In the provided exercise, we are asked to calculate the formation constant for the complex ion \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\), utilizing the equilibrium expression and the calculated reaction quotient. The process involves understanding that at equilibrium, the cell potential is directly related to the formation constant, reflecting the complex ion's stability in the solution.