Question

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ 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 \(\mathrm{In}^{+}(a q)\) 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 approximately \(8.25 \times 10^{11}\), and the standard Gibbs free energy of formation, \(\Delta G_{f}^{\circ}\), for In+(aq) is approximately -183.7 kJ/mol.

Step-by-step solution

Write the disproportionation half-reactions

Since indium is both oxidized and reduced in the disproportionation reaction, we need to divide the reaction into two half-reactions. For the disproportionation reaction: \(3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)\) We can split it into half-reactions as: Reduction half-reaction: \(2\operatorname{In}^{+}(a q) + e^{-} \longrightarrow 2\operatorname{In}(s)\) Oxidation half-reaction: \(\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}^{3+}(a q) + e^{-}\)

Calculate the standard potential for a disproportionation reaction

First, multiply the oxidation half-reaction by 2 to double the number of electrons. This gives: \(\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}^{3+}(a q) + e^{-}\) Then multiply by 2: \(2\operatorname{In}^{+}(a q) \longrightarrow 2\operatorname{In}^{3+}(a q) + 2e^{-}\) Next, subtract the reduction half-reaction from the oxidation half-reaction: \((2\operatorname{In}^{+}(a q) \longrightarrow 2\operatorname{In}^{3+}(a q) + 2e^{-}) - (2\operatorname{In}^{+}(a q) + e^{-} \longrightarrow 2\operatorname{In}(s))\) Which results in the whole reaction: \(3 \operatorname{In}^{+}(a q) \longrightarrow 2\operatorname{In}(s)+\operatorname{In}^{3+}(a q)\) Now, the standard potential for the disproportionation reaction, \(\mathscr{E}_{dis}^{\circ}\), can be calculated by subtracting the reduction half-reaction potential, \(\mathscr{E}_1^{\circ}\), from the oxidation half-reaction potential, \(\mathscr{E}_2^{\circ}\), obtained from the given reduction potentials. Therefore: \(\mathscr{E}_{dis}^{\circ} = \mathscr{E}_2^{\circ} - \mathscr{E}_1^{\circ} = (-0.444 V) - (-0.126 V) = -0.318 V\)

Calculate the equilibrium constant

Now, we will use the Nernst Equation to calculate the equilibrium constant, K, for the disproportionation reaction: \(\mathscr{E}_{dis}^{\circ} = - \frac{RT}{2F} \ln K\) In this equation, R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (assume 298 K as the standard temperature), and F is the Faraday constant (96485 C/mol). Then we can solve for K: \(-0.318 V = - \frac{8.314 J \cdot 298 K}{2 \cdot 96485 C/mol} \ln K\) Solve for K: \(K \approx 8.247 \times 10^{11}\) So, the equilibrium constant for the disproportionation reaction is approximately \(8.25 \times 10^{11}\). b. Calculate the standard Gibbs free energy of formation for In+(aq):

Calculate the Gibbs free energy for the reduction process of In3+(aq) to In+(aq)

We will use the relationship between Gibbs free energy and the standard reduction potential: \(\Delta G_{red}^{\circ} = -nF\mathscr{E}_{red}^{\circ}\) Where \(\Delta G_{red}^{\circ}\) is the Gibbs free energy change for the reduction of In3+(aq) to In+(aq), n is the number of electrons involved in the reaction (in this case, n=2), F is the Faraday constant (96485 C/mol), and \(\mathscr{E}_{red}^{\circ} = -0.444V\) (the potential given for the reduction of In3+(aq) to In+(aq)): \(\Delta G_{red}^{\circ} = -(2)(96485 C/mol)(-0.444 V) = 85764 J/mol = 85.8 kJ/mol\)

Calculate the standard Gibbs free energy of formation of In+(aq)

Since we are given \(\Delta G_{f}^{\circ}\) for In3+(aq) to be -97.9 kJ/mol, and we have just calculated the Gibbs free energy change for the reduction process as 85.8 kJ/mol, we can find the standard Gibbs free energy of formation for In+(aq), \(\Delta G_{f, \rm{In}^+}^{\circ}\), using the following equation: \(\Delta G_{f, \rm{In}^+}^{\circ} = \Delta G_{f, \rm{In}^{3+}}^{\circ} - \Delta G_{red}^{\circ} = -97.9\, \mathrm{kJ/mol} - 85.8\, \mathrm{kJ/mol} = -183.7\, \mathrm{kJ/mol}\) So, the standard Gibbs free energy of formation, \(\Delta G_{f}^{\circ}\), for In+(aq) is approximately -183.7 kJ/mol.

Key concepts

Equilibrium Constant

An equilibrium constant, denoted as \( K \), provides insight into the balance between reactants and products at equilibrium in a chemical reaction. For redox reactions, the standard reduction potential can be used to determine the equilibrium constant. In our example, we calculated the equilibrium constant for the disproportionation of \( \mathrm{In}^{+} \). This was achieved using the Nernst equation, which relates standard potential and equilibrium constant through the formula: \(E_{dis}^{\circ} = - \frac{RT}{nF} \ln K\). Here, \( R \) is the gas constant (8.314 J/(mol·K)), \( T \) is temperature (in Kelvin), \( n \) is the number of electrons, and \( F \) is Faraday's constant (96485 C/mol). After solving the equation, the equilibrium constant for the disproportionation reaction was found to be \(8.25 \times 10^{11}\). This large \( K \) indicates the reaction heavily favors the formation of products.

Gibbs Free Energy

Gibbs free energy, \( \Delta G \), is a thermodynamic quantity that measures the maximum reversible work that a system can perform at constant temperature and pressure. It's a crucial concept in determining spontaneity. If \( \Delta G \) is negative, a reaction will proceed spontaneously. The Gibbs free energy and standard reduction potentials are related by the equation \( \Delta G^{\circ} = -nF\mathscr{E}_{red}^{\circ}\). In this instance, we used this relation to find \( \Delta G \) for the reduction of \( \mathrm{In}^{3+} \) to \( \mathrm{In}^{+} \). With the provided \( \mathscr{E}_{red}^{\circ} = -0.444\, \text{V} \), and knowing \( n = 2 \), the calculated Gibbs free energy change was \( 85.8 \text{ kJ/mol} \). Consequently, using the Gibbs free energy of formation for \( \mathrm{In}^{3+} \) and the reduction process, the standard Gibbs free energy of formation for \( \mathrm{In}^{+} \) was found to be \(-183.7 \text{ kJ/mol} \).

Standard Reduction Potential

The standard reduction potential, \( \mathscr{E}^{\circ} \), is a measure of a substance's tendency to gain electrons and be reduced. It's expressed in volts and is measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C). Each half-reaction in a redox process has an associated \( \mathscr{E}^{\circ} \). Substances with higher \( \mathscr{E}^{\circ} \) are better oxidizing agents. In our exercise, the standard reduction potentials for converting \( \mathrm{In}^{3+} \) to \( \mathrm{In}^{+} \) and \( \mathrm{In}^{+} \) to \( \mathrm{In} \) were \(-0.444 \mathrm{~V} \) and \(-0.126 \mathrm{~V} \) respectively. By determining the potential difference, the direction and magnitude of the reaction can be assessed, which is crucial for calculating other properties like \( \Delta G \) and \( K \).

Nernst Equation

The Nernst equation is fundamental for understanding how cell potentials change with varying conditions. The equation is expressed as: \( \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q \), where \( \mathscr{E} \) is the cell potential at non-standard conditions, \( Q \) is the reaction quotient, and other symbols carry their usual meanings. The Nernst equation allows us to calculate equilibrium constants or potentials in real-time, showing how systems deviate from standard conditions. In our exercise, it was instrumental for converting between \( \mathscr{E}_{dis}^{\circ} \) and \( K \). Using \( T = 298 \text{ K} \), we were able to determine the equilibrium constant for the reaction using the equation reformulating it as: \( \mathscr{E}^{\circ} = - \frac{RT}{nF} \ln K \), reinforcing how potential relates to equilibrium properties.