Question

The following standard reduction potentials have been det mined for the aqueous chemistry of indium: $$ \begin{array}{cl} \mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V} \end{array} $$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q) ?\)

Short answer

The equilibrium constant (K) for the disproportionation reaction is \(9.7 \times 10^7\). The standard Gibbs free energy of formation, ΔG°\(_\text{f}\), for In+(aq) is 102.6 kJ/mol.

Step-by-step solution

Write the disproportionation reaction

We are given the disproportionation reaction: $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ In this reaction, In+ ions are both oxidized and reduced.

Determine the overall redox reaction

To determine the overall redox reaction, we need to find an oxidation and reduction reaction that will give the disproportionation reaction when both are combined. Based on the given reduction reactions: 1. Reduction reaction: $\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q)$ 2. Oxidation reaction (reverse of the second given reaction): \(\mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s) + \mathrm{e}^{-}\) Now, we need to balance these two half-reactions to combine them properly. To do this, we multiply the reduction reaction by 2 and the oxidation reaction by 3: 1. Balanced reduction reaction: $2[\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q)]$ 2. Balanced oxidation reaction: \(3[\mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s) + \mathrm{e}^{-}]\) When we combine the two half-reactions, we get the disproportionation reaction: $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$

Calculate the standard cell potential, E°, using standard reduction potentials

To calculate the standard potential E°, subtract the oxidation reaction's standard reduction potential from the reduction reaction's standard reduction potential: $$ \mathscr{E}^{\circ}_{cell}=\mathscr{E}^{\circ}_{red} - \mathscr{E}^{\circ}_{ox} $$ Insert the values of the standard reduction potential: $$ \mathscr{E}^{\circ}_{cell}=(-0.444 \mathrm{~V}) - (-0.126 \mathrm{~V})=-0.318 \mathrm{~V} $$

Calculate the equilibrium constant using the Nernst equation

Now, we will use the Nernst equation to find the equilibrium constant, K: $$ \mathscr{E}^{\circ}= \frac{-RT}{nF} \ln K $$ Where: - \(\mathscr{E}^{\circ}\) is the standard cell potential - R is the gas constant, 8.314 J/(mol·K) - T is the temperature in Kelvin, 298K - n is the total number of moles of electrons transferred. In this case, n = 6. - F is the Faraday constant, 96485 C/mol Plugging the values into the equation: $$ -0.318V = \frac{-8.314 J/(mol \cdot K) \cdot 298K}{6 \cdot 96485 C/mol} \ln K $$ Now, we can solve for K: $$ K = e^{\frac{-0.318V \cdot 6 \cdot 96485 C/mol}{8.314 J/(mol \cdot K) \cdot 298K}}=9.7 \times 10^7 $$ So, the equilibrium constant, K, for the disproportionation reaction is \(9.7 \times 10^7\).

Calculate ΔG°\(_\text{f}\) for In+(aq) using ΔG°\(_\text{f}\) for In³+(aq)

We are given the standard Gibbs free energy of formation for In³+(aq), ΔG°\(_\text{f}\) = -97.9 kJ/mol. We can use the following relationship between ΔG°, standard cell potential, and Faraday constant: $$ \Delta G^{\circ}=-n F \mathscr{E}^{\circ} $$ First, we calculate the overall ΔG° of the disproportionation reaction: $$ \Delta G^{\circ}_{disproportionation} = -6 \cdot 96485 C/mol \cdot (-0.318V) = 184.2 \mathrm{~kJ/mol} $$ Now, let's write the ΔG°\(_\text{f}\) for the overall reaction: $$ \Delta G^{\circ}_{disproportionation} = 2 \Delta G^{\circ}_{\text{f}}(\mathrm{In}^{3+}(aq)) - 3 \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) $$ Plug in the value of ΔG°\(_\text{f}\)(In³+(aq)) and ΔG°\(_\text{f}\): $$ 184.2 \mathrm{~kJ/mol} = 2(-97.9 \mathrm{~kJ/mol}) - 3 \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) $$ Now, solve for ΔG°\(_\text{f}\)(In+(aq)): $$ \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) = \frac{2(-97.9 \mathrm{~kJ/mol}) - 184.2 \mathrm{~kJ/mol}}{-3} = 102.6 \mathrm{~kJ/mol} $$ So, the standard Gibbs free energy of formation, ΔG°\(_\text{f}\), for In+(aq) is 102.6 kJ/mol.

Key concepts

Disproportionation Reaction

A disproportionation reaction is a special type of redox reaction where the same chemical species undergoes both oxidation and reduction simultaneously. This means one part of the compound is oxidized, losing electrons, while another part is reduced, gaining electrons.
In the exercise provided, we see this phenomenon with the reaction:

• 3 \( ext{In}^+\)(aq) \(\longrightarrow\) 2 \( ext{In}(s)\) + \( ext{In}^{3+}(aq)\)

Here, \( ext{In}^+\) serves as both the reducing agent (since it gets oxidized to \( ext{In}^{3+}\)) and the oxidizing agent (since it gets reduced to \( ext{In}\) solid form). This unique dual role in the reaction makes it a classic example of disproportionation.

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G \), is a thermodynamic function used to predict whether a chemical reaction will occur spontaneously at constant temperature and pressure.
In simple terms, a negative \( \Delta G \) indicates that the reaction will happen naturally, while a positive \( \Delta G \) suggests it requires energy input. For the formation of substances, the standard Gibbs free energy of formation \( \Delta G_f^{\circ} \) tells us the energy change when 1 mole of a compound forms from its elements.
The task outlines calculating \( \Delta G \) for In\(^+\) (aq) using the known \( \Delta G \) values for In\(^{3+}\) (aq) and applying the basic formula:
\[ \Delta G^{\circ}_{disproportionation} = 2 \Delta G^{\circ}_{\text{f}}(\mathrm{In}^{3+}(aq)) - 3 \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) \]
Once values are substituted, we can uncover that \( \Delta G^{\circ}_{\text{f}}(\mathrm{In}^{+}(aq)) = 102.6 \mathrm{~kJ/mol} \).

Nernst Equation

The Nernst equation allows us to find the equilibrium constants of redox reactions by relating them to standard reduction potentials. It becomes handy when calculating cell potential at any condition, not just standard ones.\[ \mathscr{E}^{\circ}= \frac{-RT}{nF} \ln K \]
Here, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred, and \( F \) is the Faraday constant. Knowing the standard cell potential \( \mathscr{E}^{\circ}_{cell} \), we can rearrange the Nernst equation to solve for the equilibrium constant \( K \).
In this scenario, by substituting into the formula, we calculate \( K \) to determine the equilibrium status of the disproportionation reaction as \( 9.7 \times 10^7 \). This high value of \( K \) implies a strong tendency for the reaction to move spontaneously towards forming the products.

Redox Reaction

Redox reactions are chemical processes involving the transfer of electrons between two substances. These changes in oxidation states occur with one species undergoing oxidation (electron loss) and another experiencing reduction (electron gain).
In the case provided:

• The reduction process occurs when \( \text{In}^{3+} \)(aq) gains electrons to form \( \text{In}^+ \)(aq).
• Conversely, the oxidation occurs when \( \text{In}^+ \)(aq) gives electrons, resulting in \( \text{In} \)(s).

Recognizing these half-reactions helps us understand the balance required in redox reactions, where the electron exchange must be equal on both sides. By comprehending individual components like oxidation state changes and electron transfers, students get a clearer view of how these reactions drive and signify energy changes within a system.