Problem 60 - Chemistry Principles and Practice

Question

Disproportionation is a type of redox reaction in which the same species is simultaneously oxidized and reduced. One species that undergoes disproportionation is \(\mathrm{Cu}^{+}(\) aq \()\). $$ 2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) $$ If the half-reactions are $$ \begin{array}{ll} \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+}(\text {aq }) & E^{\circ}=0.153 \mathrm{~V} \\ \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) & E^{\circ}=0.521 \mathrm{~V} \end{array} $$ what are \(E^{\circ}, \Delta G^{\circ},\) and \(K_{\mathrm{eq}}\) for the overall reaction?

Step-by-Step Solution

Answer

\(E^{\circ}_{\text{cell}} = 0.674 \mathrm{~V}\), \(\Delta G^\circ = -65.02 \text{ kJ/mol}\), \(K_{eq} \approx 3.96 \times 10^{11}\).

Step 1: Identify the Half-Reactions

The overall reaction given is \(2 \mathrm{Cu}^{+} (\mathrm{aq}) \rightarrow \mathrm{Cu} (\mathrm{s}) + \mathrm{Cu}^{2+} (\mathrm{aq})\). Identify the two individual half-reactions: 1. \( \mathrm{Cu}^{2+} (\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+} (\mathrm{aq})\) with \(E^{\circ}_1 = 0.153 \mathrm{~V} \) 2. \( \mathrm{Cu}^{+} (\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Cu} (\mathrm{s})\) with \(E^{\circ}_2 = 0.521 \mathrm{~V} \).

Step 2: Determine the Overall Standard Cell Potential

For the overall reaction, use the standard reduction potentials. The reaction can be broken into: - The oxidation half-reaction: \( \mathrm{Cu}^{+} (\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+} (\mathrm{aq}) + \mathrm{e}^{-} \), corresponding to \( -E^{\circ}_1 = -0.153 \mathrm{~V} \).- The reduction half-reaction: \( \mathrm{Cu}^{+} (\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Cu} (\mathrm{s}) \) with \(E^{\circ}_2 = 0.521 \mathrm{~V}\).The standard cell potential \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduction}} - E^{\circ}_{\text{oxidation}} = 0.521 \mathrm{~V} - (-0.153 \mathrm{~V}) = 0.674 \mathrm{~V}\).

Step 3: Calculate the Standard Gibbs Free Energy Change, \(\Delta G^\circ\)

The relationship between Gibbs free energy change \(\Delta G^\circ\) and the standard cell potential \(E^{\circ}_{\text{cell}}\) can be described as: \[ \Delta G^\circ = -nFE^{\circ}_{\text{cell}} \] where \(n\) is the number of moles of electrons exchanged. Here, \(n = 1\). The Faraday constant \(F\) is \(96485 \text{ C/mol}\).So, \( \Delta G^\circ = -(1)(96485)(0.674) = -65016 \text{ J/mol} \) or \(-65.02 \text{ kJ/mol}\).

Step 4: Calculate the Equilibrium Constant, \(K_{eq}\)

The relationship between the equilibrium constant \(K_{eq}\) at standard conditions and \(\Delta G^\circ\) is given by: \[ \Delta G^\circ = -RT \ln K_{eq} \] where \(R\) is the universal gas constant (\(8.314 \text{ J/mol·K}\)) and \(T\) is the temperature in Kelvin, which is typically 298 K at standard conditions.Rearranging for \(K_{eq}\), we have:\[ K_{eq} = e^{-\Delta G^\circ / RT} = e^{65016 / (8.314 \times 298)} \]Calculating this gives: \( K_{eq} \approx 3.96 \times 10^{11} \).

Key Concepts

Disproportionation

In redox chemistry, disproportionation is an intriguing concept, where a single element undergoes simultaneous oxidation and reduction. In a disproportionation reaction, a species that is in an intermediate oxidation state is transformed into both a higher and a lower oxidation state.
For example, when analyzing the reaction \( 2 \mathrm{Cu}^{+} (\mathrm{aq}) \rightarrow \mathrm{Cu} (\mathrm{s}) + \mathrm{Cu}^{2+} (\mathrm{aq}) \), we notice that copper is oxidized and reduced at the same time. The

• Reduction half-reaction: \( \mathrm{Cu}^{+} (\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Cu} (\mathrm{s}) \)
• Oxidation half-reaction: \( \mathrm{Cu}^{+} (\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+} (\mathrm{aq}) + \mathrm{e}^{-} \)

show how copper transforms differently in each half-reaction. Such reactions are common in transition metal chemistry, where elements often have accessible multiple oxidation states, and understanding these reactions can be crucial for grasping more complex redox processes.

Gibbs Free Energy

Gibbs Free Energy, often denoted as \(\Delta G\), tells us whether a chemical reaction will be spontaneous under standard conditions. If \(\Delta G\) is less than zero, the reaction will proceed spontaneously.
The relation involves calculating the Gibbs Free Energy change using the formula:\[ \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \]Here:

• \(n\) represents the number of moles of electrons exchanged in the reaction. For the copper reaction, \(n = 1\).
• \(F\) is the Faraday constant, approximately \(96485 \text{ C/mol}\).
• \(E^{\circ}_{\text{cell}}\), the standard cell potential, provides the voltage difference in the reaction, given as \(0.674 \mathrm{~V}\) in this scenario.

Given these values in the copper reaction, we find \(\Delta G^{\circ}\) to be \(-65.02 \text{ kJ/mol}\). This negative sign emphasizes that the process is energetically favorable under standard conditions.

Equilibrium Constant

The equilibrium constant \(K_{eq}\) helps us understand the ratio of concentrations of products to reactants at equilibrium in a reaction. For redox reactions, \(K_{eq}\) relates to \(\Delta G^{\circ}\) through the expression:\[ \Delta G^{\circ} = -RT \ln K_{eq} \]where:

• \(R\) is the universal gas constant, \(8.314 \text{ J/mol·K}\).
• \(T\) is the temperature in Kelvin, typically \(298 \text{ K}\) at standard conditions.
• This equation often re-arranges to find \(K_{eq}\) using the exponential function as: \[ K_{eq} = e^{-\Delta G^{\circ} / RT} \]

By substituting our calculated \(\Delta G^{\circ} = -65.02 \text{ kJ/mol}\) into this equation, we derive that \(K_{eq} \approx 3.96 \times 10^{11}\). This large \(K_{eq}\) indicates that the reaction strongly favors the formation of products at equilibrium.