Problem 30 - Chemistry: Principles and Reactions

Question

Using Table \(20.4\), calculate, for the disproportionation of \(\mathrm{Au}^{+}\), (a) \(K\). (b) the concentration of \(\mathrm{Au}^{+}\) in equilibrium with \(0.10 \mathrm{M} \mathrm{Au}^{3+}\).

Step-by-Step Solution

Answer

The equilibrium constant (K) for the disproportionation reaction is approximately 162. The concentration of Au+ ions in equilibrium with 0.10 M of Au3+ ions is approximately 0.017 M.

Step 1: Write the balanced chemical equation for the disproportionation of \(\mathrm{Au}^{+}\) ions

The disproportionation of \(\mathrm{Au}^{+}\) ions can be described by the following balanced chemical equation: \begin{equation} 3\mathrm{Au}^{+} \rightarrow \mathrm{Au} + 2\mathrm{Au}^{3+} \end{equation} This equation tells us that three \(\mathrm{Au}^{+}\) ions will react to form one \(\mathrm{Au}\) atom and two \(\mathrm{Au}^{3+}\) ions.

Step 2: Determine the relationship between the standard cell potential and the equilibrium constant (K)

According to the Nernst Equation, we can relate the standard cell potential (\(E^\circ\)) and the equilibrium constant (K) as follows: \begin{equation} E^\circ = \frac{0.0592 \,V}{n} \log K \end{equation} Here, \(n\) is the number of electron transfer involved in the reaction, and in our case, n = 2, since two electrons are transferred when one \(\mathrm{Au}^{3+}\) ion is produced from one \(\mathrm{Au}^{+}\) ion. The standard potential (\(E^\circ\)) can be found using the Table \(20.4\) provided. We will find the difference in the standard potentials for Au(III) and Au(I) to get the overall standard potential for this reaction.

Step 3: Calculate the standard potential for the disproportionation reaction

From Table \(20.4\), we know the standard reduction potential for the following half-cell reactions: \begin{equation} \mathrm{Au}^{3+} + 3\mathrm e^- \rightarrow \mathrm{Au} \quad E^\circ_{Au(III) \rightarrow Au} = 1.498 \, V \end{equation} \begin{equation} \mathrm{Au}^{+} + \mathrm e^- \rightarrow \mathrm{Au} \quad E^\circ_{Au(I) \rightarrow Au} = 1.692 \, V \end{equation} To calculate the overall standard potential for the disproportionation reaction, take the difference between these two standard potentials: \begin{equation} E^\circ = E^\circ_{Au(III) \rightarrow Au} - E^\circ_{Au(I) \rightarrow Au} = 1.498 - 1.692 = -0.194 \, V \end{equation}

Step 4: Calculate the equilibrium constant (K) using the standard potential

Now, use the value of \(E^\circ\) and the Nernst equation to calculate the equilibrium constant (K): \begin{equation} K = 10^{\frac{-n E^\circ}{0.0592 \,V}} = 10^{\frac{-2 (-0.194 \,V)}{0.0592 \,V}} \approx 162 \end{equation} So, the equilibrium constant (K) for the disproportionation reaction is approximately 162.

Step 5: Determine the concentration of \(\mathrm{Au}^{+}\) in equilibrium with \(0.10 \mathrm{M}\) of \(\mathrm{Au}^{3+}\) ions

We can now apply the equilibrium expression for the reaction: \begin{equation} K = \frac{[\mathrm{Au}^{3+}]^2}{[\mathrm{Au}^{+}]^3} \end{equation} Substituting the provided concentration of \(\mathrm{Au}^{3+}\) and the value of K, we have: \begin{equation} 162 = \frac{(0.10)^2}{[\mathrm{Au}^{+}]^3} \end{equation} To solve for the concentration of \(\mathrm{Au}^{+}\), we will rearrange the equation and take the cube root: \begin{equation} [\mathrm{Au}^{+}] = \sqrt[3]{\frac{(0.10)^2}{162}} \approx 0.017 \, M \end{equation} So, the concentration of \(\mathrm{Au}^{+}\) ions in equilibrium with \(0.10 \mathrm{M}\) of \(\mathrm{Au}^{3+}\) ions is approximately \(0.017 \mathrm{M}\). To summarize: (a) The equilibrium constant (K) is approximately 162. (b) The concentration of \(\mathrm{Au}^{+}\) at equilibrium with \(0.10 \mathrm{M}\) of \(\mathrm{Au}^{3+}\) ions is approximately \(0.017 \mathrm{M}\).

Key Concepts

Disproportionation

In electrochemistry, a unique type of redox reaction is known as disproportionation. This reaction involves a single element undergoing both oxidation and reduction simultaneously. In simpler terms, one species transforms into two distinct oxidation states.

For example, let's look at the disproportionation of \(\mathrm{Au}^{+}\). In this reaction, the gold ion \(\mathrm{Au}^{+}\) serves as both a reducing agent and an oxidizing agent.

• Three \(\mathrm{Au}^{+}\) ions participate in this reaction.
• The outcome is one atom of gold (\(\mathrm{Au}\)) and two \(\mathrm{Au}^{3+}\) ions.

The chemical equation for this reaction is formalized as:

\[3\mathrm{Au}^{+} \rightarrow \mathrm{Au} + 2\mathrm{Au}^{3+}\]
This showcases that \(\mathrm{Au}^{+}\) is disproportionate by undergoing oxidation to form \(\mathrm{Au}^{3+}\) while being reduced to form metallic gold \(\mathrm{Au}\). Such fascinating reactions emphasize the dual nature of certain elements.

Chemical Equilibrium

Chemical equilibrium refers to a state where the rate of forward and reverse reactions is balanced. Consequently, the concentrations of reactants and products remain constant over time. This concept is vital in understanding how reactions behave in closed systems.

For the disproportionation reaction involving \(\mathrm{Au}^{+}\), equilibrium is achieved when the concentrations of all species become steady. The equilibrium constant, \(K\), quantifies this balance:

• It is defined using the concentrations of products and reactants at equilibrium.
• The formula to express this for the gold reaction is \[K = \frac{[\mathrm{Au}^{3+}]^2}{[\mathrm{Au}^{+}]^3}\]

This equilibrium constant helps assess the extent of the reaction. A larger \(K\) value, like 162, indicates the reaction favors the product side (i.e., more \(\mathrm{Au}\) and \(\mathrm{Au}^{3+}\) ions).

This concept is integral for predicting chemical behavior and the ability to control reaction conditions in electrochemical processes.

Nernst Equation

The Nernst Equation allows chemists to link the electrochemical cell potential under non-standard conditions with the reaction quotient and temperature. This connection is fundamental for practical electrochemistry when conditions deviate from the standard.

For reactions like the disproportionation of \(\mathrm{Au}^{+}\), it enables us to calculate equilibrium properties from the standard cell potential \(E^\circ\) and number of electrons transferred \(n\). Here's how it looks:

\[E = E^\circ - \frac{0.0592}{n} \log Q\]

• Here, \(E^\circ\) is the standard cell potential.
• \(n\) refers to the electron exchange number. In this case, it's 2.
• \(Q\) denotes the reaction quotient, calculated similarly to \(K\).

This equation provides insights into how different concentrations impact the cell voltage and overall dynamics of the cell. Through comprehensive understanding, the equation aids in calculating non-standard potentials and aids in analyzing equilibrium conditions.

Standard Potential

Standard potential, often symbolized as \(E^\circ\), is a measure of the intrinsic ability of a species to gain or lose electrons under standard conditions (25°C, 1M, 1 atm). It's an essential concept that indicates how readily a half-cell can either gain or lose electrons.

In Table 20.4 of Au(III) to Au(I) reactions, standard potentials are provided:

• \(E^\circ\) for \(\mathrm{Au}^{3+} + 3\mathrm{e^-} \rightarrow \mathrm{Au}\) is 1.498 V.
• \(E^\circ\) for \(\mathrm{Au}^+ + \mathrm{e^-} \rightarrow \mathrm{Au}\) is 1.692 V.

Calculating the difference between these provided potentials delivers the overall standard potential for a reaction.

For the disproportionation reaction, the calculation facilitates understanding of the direction and feasibility of the reaction. Observantly, a negative \(E^\circ\) value, as calculated (-0.194 V), suggests the reaction, at standard conditions, does not spontaneously favor product formation, but can proceed when equilibrium tweaks like concentration changes occur. This calculation underscores the role of standard potentials in determining the likelihood and extent of redox processes.