Problem 86 - Chemistry An Atoms First Approach

Question

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Step-by-Step Solution

Answer

The spontaneity of the given reactions are as follows: a) Spontaneous, with \(\Delta G^{\circ} = - 35719.45\,\text{J/mol}\) and \(K \approx 1.93 \times 10^{6}\). b) Non-spontaneous. c) Non-spontaneous.

Step 1: Write the half-reactions and find the overall cell potential

The half-reactions are: - \(\mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{e}^{-};\quad \mathscr{E}_{1}^{\circ} = 0.15 \mathrm{V}\) - \(\mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Cu}(s);\quad \mathscr{E}_{2}^{\circ} = 0.52 \mathrm{V}\) The overall cell potential is \(\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{2}^{\circ} - \mathscr{E}_{1}^{\circ} = (0.52\,\text{V}) - (0.15\,\text{V}) = 0.37\,\text{V}\) Since the cell potential is positive, the reaction is spontaneous.

Step 2: Calculate Gibbs free energy change

Use the formula \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\), with \(n\) being the number of electrons transferred and \(F\) representing the Faraday's Constant (\(F = 96,485\,\text{C/mol}\), approximately). For this reaction, \(n = 1\). \(\Delta G^{\circ} = - (1)(96485\,\text{C/mol})(0.37\,\text{V}) = - 35719.45\,\text{J/mol}\). This is a negative value, confirming the reaction's spontaneity.

Step 3: Calculate the equilibrium constant

Use the formula \(\Delta G^{\circ} = -RT\ln{K}\), where \(R\) is the gas constant (\(R = 8.314\,\text{J/mol K}\)) and \(T\) is the temperature in Kelvins (in this case, \(T = 298\,\text{K}\)). Rearranging to find \(K\) gives: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) Substituting the known values: \(K = e^{-\frac{-35719.45\,\text{J/mol}}{(8.314\,\text{J/mol K})(298\,\text{K})}}\) \(K \approx 1.93 \times 10^{6}\) Now let's analyze part (b): b) For the reaction: \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\)

Step 1: Write the half-reactions and derive the overall cell potential

The half-reactions are: - \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 3 \mathrm{Fe}^{3+}(a q)+3 \mathrm{e}^{-};\quad \mathscr{E}_{1}^{\circ} = 0.77 \mathrm{V}\) - \(3 \mathrm{Fe}^{3+}(a q)+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(s);\quad \mathscr{E}_{2}^{\circ} = 0.036 \mathrm{V}\) The overall cell potential is \(\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{2}^{\circ} - \mathscr{E}_{1}^{\circ} = 0.036\,\text{V} - 0.77\,\text{V} = - 0.734\,\text{V}\) The cell potential is negative, so the reaction is non-spontaneous under standard conditions, and there is no need to calculate \(\Delta G^{\circ}\) and \(K\). c) For the unbalanced reaction: \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q)\) We will use the given half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Step 1: Balance the reaction and find the overall cell potential

By looking at the half-reactions, we can see that the balanced reaction is: \(2 \mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q)\) The overall cell potential is \(\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{1}^{\circ} - \mathscr{E}_{2}^{\circ} = (1.21\,\text{V}) - (1.65\,\text{V}) = - 0.44\,\text{V}\) Since the cell potential is negative, the reaction is non-spontaneous under standard conditions, and there is no need to calculate \(\Delta G^{\circ}\) and \(K\).

Key Concepts

Electrochemistry

Electrochemistry is a branch of chemistry that deals with the interrelation of electrical currents and chemical reactions, and the conversion of chemical energy into electrical energy and vice versa. In the context of disproportionation reactions, electrochemistry helps us understand how a single substance can undergo oxidation and reduction in a single, self-contained reaction. This process involves the transfer of electrons from the oxidized to the reduced form of the same substance.

For instance, when we consider the reaction where copper (I) ions in solution disproportionates into copper (II) ions and solid copper, the electrochemical principles allow us to write separate half-reactions for the oxidation and reduction processes. Then, we can calculate the overall cell potential, \(\mathscr{E}_{cell}^\circ\), by taking the difference between the standard electrode potentials of these half-reactions. A positive overall cell potential indicates a spontaneous reaction under standard conditions, which reflects the inherent tendency of the species to undergo the disproportionation.

Gibbs Free Energy

Gibbs Free Energy, symbolized as \(\Delta G\), is a thermodynamic quantity that represents the maximum amount of usable energy obtainable from a chemical reaction at constant pressure and temperature. It's a critical indicator of the spontaneity of reactions; a negative value of \(\Delta G\) signifies that a reaction is spontaneous, whereas a positive value implies that the reaction is non-spontaneous.

In our disproportionation example with copper ions, we applied the formula \(\Delta G^\circ = -nFE_{cell}^\circ\) where \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is Faraday's constant, and \(E_{cell}^\circ\) is the overall standard cell potential. This formula is derived from the direct relationship between electrochemical cell potential and the change in Gibbs Free Energy, reflecting how electrochemical concepts are grounded in fundamental thermodynamics.

Equilibrium Constant

The equilibrium constant, \(K\), is another important concept in both electrochemistry and thermodynamics, providing a quantitative measure of the extent of a reaction at equilibrium. Specifically, \(K\) relates to the concentrations of the reactants and products in a reversible reaction at equilibrium. A large value for \(K\) indicates that, at equilibrium, the products are favored; conversely, a small \(K\) suggests that the reactants are favored.

By using the relationship \(\Delta G^\circ = -RT\ln{K}\), where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin, we can calculate \(K\) from the Gibbs Free Energy. In the case of the copper disproportionation reaction, calculating \(K\) revealed a very large value, which indicates a strong product-favoring equilibrium under standard conditions. Such calculations are extremely useful for predicting the position of equilibrium in electrochemical reactions.

Standard Electrode Potentials

Standard electrode potentials (\(\mathscr{E}^\circ\)), measured in volts, are intrinsic properties of chemical species that provide a measure of the tendency of a substance to lose or gain electrons, thus serving as a reduction or oxidation agent. In an electrochemical cell, each half-reaction has an associated \(\mathscr{E}^\circ\), and the difference in these potentials between the anode and cathode gives the cell its driving force.

When assessing the spontaneity of disproportionation reactions, we utilize the standard electrode potentials to determine the overall potential for the cell. For a reaction to be spontaneous, \(\mathscr{E}_{cell}^\circ\) must be positive. In our example problems, we examined the standard electrode potentials for half-reactions involving copper and iron species, determining which reactions were spontaneous by comparing the potentials. This application showcases the essential role of standard electrode potentials in predicting reaction spontaneity in electrochemistry.