Question

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

Short answer

The equilibrium constant (K) for the disproportionation reaction is 3.54 × 10^3, and the standard Gibbs free energy (ΔG°) for In+(aq) is 42.346 kJ/mol.

Step-by-step solution

Determine the overall reaction's standard cell potential (E°)

We will reverse and add the given half-reactions to get the disproportionation reaction: Reversed 1st half-reaction: \(In^+(aq) \longrightarrow In^{3+}(aq) + 2 e^-\) (E° = +0.444 V) 2 × 2nd half-reaction: \(2 In^+(aq) + 2 e^- \longrightarrow 2 In(s) \) (E° = 2 × -0.126 V) Now add the reversed 1st and 2 × 2nd half-reactions: \(3 In^+(aq) \longrightarrow 2 In(s) + In^{3+}(aq)\) (E° = 0.444 - 2 × 0.126 V) Calculating E° for the overall reaction: E° = 0.444 - 2 × 0.126 = 0.192 V

Determine the equilibrium constant (K) using Nernst equation

Using the Nernst equation for the equilibrium condition and under standard conditions (\(E = E°\), Q = 1), we get: \(E° = \frac{RT}{nF} \ln K\) Here, \(E° = 0.192\) V (calculated in step 1) R = 8.314 J/(mol·K) (gas constant) T = 298.15 K (standard temperature) n = 3 (moles of electrons involved in the overall reaction) F = 96,485 C/mol (Faraday's constant) Solving for K: \(K = e^{\frac{nFE°}{RT}}\) K = \(e^{\frac{3 \times 96,485 \times 0.192}{8.314 \times 298.15}}\) K = 3.54 × 10^3 So, the equilibrium constant (K) for the disproportionation reaction is 3.54 × 10^3. b. Finding the standard Gibbs free energy (ΔG°) for In+(aq):

Determine the standard Gibbs free energy (ΔG°) for the overall reaction

Using the relationship between ΔG° and E°: ΔG° = -nFE° Here, ΔG° = standard Gibbs free energy (J/mol) n = 3 (moles of electrons involved) F = 96,485 C/mol (Faraday's constant) E° = 0.192 V (calculated in part a's step 1) ΔG° = -3 × 96,485 × 0.192 ΔG° = -55,554 J/mol or -55.554 kJ/mol

Determine the standard Gibbs free energy (ΔG°) for In+(aq)

From the given information, the standard Gibbs free energy (ΔG°) for In3+(aq) is -97.9 kJ/mol. Let's consider ΔG°(In+(aq)) for In+(aq) and X for the overall reaction (ΔG° = -55.554 kJ/mol) Using Hess's law: ΔG°(In+(aq)) + ΔG°(In3+(aq)) = ΔG° Therefore, ΔG°(In+(aq)) = ΔG° - ΔG°(In3+(aq)) ΔG°(In+(aq)) = -55.554 - (-97.9) ΔG°(In+(aq)) = 42.346 kJ/mol So, the standard Gibbs free energy (ΔG°) for In+(aq) is 42.346 kJ/mol.

Key concepts

Standard Reduction Potential

Standard reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. It is denoted by \( \mathscr{E}^{\circ} \) and measured in volts. A higher \( \mathscr{E}^{\circ} \) value indicates a greater tendency for reduction to occur.
In the problem, we are provided with two standard reduction potentials for indium:

• \( \text{In}^{3+}(aq) + 2 e^- \rightarrow \text{In}^+(aq) \quad \mathscr{E}^{\circ} = -0.444 \text{ V} \)
• \( \text{In}^+(aq) + e^- \rightarrow \text{In}(s) \quad \mathscr{E}^{\circ} = -0.126 \text{ V} \)

We can use these half-reactions to construct redox reactions and find the overall reaction's \( \mathscr{E}^{\circ} \).
By calculating the overall cell potential for the reaction, we identify how effectively the redox process can take place. A positive \( \mathscr{E}^{\circ} \) suggests a spontaneous reaction under standard conditions.

Nernst Equation

The Nernst equation allows us to calculate the cell potential at any conditions other than standard state, leveraging the standard cell potential \( E^{\circ} \) and adjusting for the concentration of reactants and products. For a cell at equilibrium, the Nernst equation simplifies to involve the equilibrium constant \( K \), as no net electron flow is occurring.
The Nernst equation is expressed as:\[ E = E^{\circ} - \frac{RT}{nF} \ln \frac{[\text{products}]}{[\text{reactants}]} \]

• \( E \) = Cell potential at non-standard conditions
• \( R \) = Universal gas constant (8.314 J/mol·K)
• \( T \) = Temperature in Kelvin
• \( n \) = Number of electrons transferred
• \( F \) = Faraday's constant (96,485 C/mol)

At equilibrium \( E = 0 \), and the equation becomes:\[ E^{\circ} = \frac{RT}{nF} \ln K \]This form allows us to solve for the equilibrium constant \( K \), reflecting the system's stability under standard conditions. For instance, a higher \( K \) value indicates a strong tendency for the reaction to proceed in the forward direction, resulting in more products.

Equilibrium Constant

The equilibrium constant \( K \) quantifies the concentrations of reactants and products at equilibrium for a given chemical reaction. It provides insight into the position of equilibrium and dictates how far the reaction proceeds.
For the disproportionation reaction in question, where one species is both oxidized and reduced:

• \( 3 \text{In}^+(aq) \rightarrow 2 \text{In}(s) + \text{In}^{3+}(aq) \)

Given the calculated standard cell potential \( E^{\circ} = 0.192 \text{ V} \), using the simplified Nernst equation for equilibrium, we determined:\[ K = e^{\frac{nFE^{\circ}}{RT}} \]Where:

• \( n = 3 \) (as three electrons are transferred in the reaction)
• \( T = 298.15 \text{ K} \) (standard temperature)

The result, \( K = 3.54 \times 10^3 \), indicates a favorable condition for forming the products \( \text{In}(s) \) and \( \text{In}^{3+}(aq) \) under the given standard conditions.

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) is a thermodynamic potential that predicts the feasibility of a reaction under constant pressure and temperature. A negative \( \Delta G \) indicates a spontaneous reaction. The relationship between Gibbs free energy, cell potential, and the equilibrium state is shown in mathematical form:\[ \Delta G^{\circ} = -nFE^{\circ} \]

• \( \Delta G^{\circ} \) is the standard Gibbs free energy change
• \( n \) is the number of moles of electrons
• \( F \) is Faraday's constant
• \( E^{\circ} \) is the standard cell potential

In our example, using the \( E^{\circ} \) calculated for the overall reaction, we derived:\[ \Delta G^{\circ} = -3 \times 96,485 \times 0.192 \text{ J/mol} = -55,554 \text{ J/mol or } -55.554 \text{ kJ/mol} \]For \( \text{In}^{+}(aq) \), with \( \Delta G_{f}^{\circ} \) for \( \text{In}^{3+}(aq) = -97.9 \text{ kJ/mol} \), we further used Hess’s law to find:\[ \Delta G^{\circ}(\text{In}^{+}(aq)) = \Delta G^{\circ} - \Delta G_{f}^{\circ}(\text{In}^{3+}(aq)) \]Resulting in a \( \Delta G^{\circ} \) of \( 42.346 \text{ kJ/mol} \) for \( \text{In}^{+}(aq) \), signifying its stability compared to its oxides.