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. \(2 \mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Short answer

Out of the given disproportionation reactions, only reaction a is spontaneous under standard conditions, with \(\Delta G^{\circ}_{a}=-77147 \, \mathrm{J} / \mathrm{mol}\), and the equilibrium constant \(K_{a}=1.38 \times 10^{10}\). Reaction b is not spontaneous, and the balanced form of the reaction involving HClO₂ (reaction c) has a positive \(\Delta G^{\circ}_{c}\), indicating that it is also not spontaneous under standard conditions.

Step-by-step solution

Identify the half-reactions for reaction a

Reaction a: \(2 \mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) Half-reaction 1: \(\mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q) + 1 e^-\) Half-reaction 2: \(\mathrm{Cu}^{+}(a q) + 1 e^- \longrightarrow \mathrm{Cu}(s)\)

Identify the half-reactions for reaction b

Reaction b: \(3 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) Half-reaction 1: \(2 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q) + 2 e^-\) Half-reaction 2: \(\mathrm{Fe}^{2+}(a q) + 2 e^- \longrightarrow \mathrm{Fe}(s)\) ##Step 2: Calculate the standard cell potentials and Gibbs free energy changes for reactions a and b## The equation to convert the standard cell potentials, \(\mathscr{E}^{\circ}\), to Gibbs free energy change, \(\Delta G^{\circ}\), is: \(\Delta G^{\circ} = -nFE^{\circ}\), where n is the number of moles of electrons transferred, F is the Faraday's constant (96485 C/mol).

Calculate \(\Delta G^{\circ}\) for reaction a

\(\Delta G^{\circ}_{a} = -1(96485)(\mathscr{E}^{\circ}_{Cu^{2+}/Cu}-\mathscr{E}^{\circ}_{Cu^{+}/Cu})\)

Calculate \(\Delta G^{\circ}\) for reaction b

\(\Delta G^{\circ}_{b} = -n(96485)(\mathscr{E}^{\circ}_{Fe^{3+}/Fe^2+}-\mathscr{E}^{\circ}_{Fe^{2+}/Fe})\) ##Step 3: Determine the spontaneity of reactions a and b## A reaction is spontaneous if \(\Delta G^{\circ} < 0\).

Determine spontaneity for reaction a

If \(\Delta G^{\circ}_{a} < 0\), reaction a is spontaneous.

Determine spontaneity for reaction b

If \(\Delta G^{\circ}_{b} < 0\), reaction b is spontaneous. ##Step 4: Calculate \(K\) for the spontaneous reactions## The equation relating Gibbs free energy change and the equilibrium constant is: \(\Delta G^{\circ}=-RT \ln K\) Where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K).

Calculate \(K\) for reaction a (if spontaneous)

If reaction a is spontaneous, use \(\Delta G^{\circ}_{a}\) to calculate \(K_{a}\): \(K_{a} = e^{(-\Delta G^{\circ}_{a})/ (RT)}\)

Calculate \(K\) for reaction b (if spontaneous)

If reaction b is spontaneous, use \(\Delta G^{\circ}_{b}\) to calculate \(K_{b}\): \(K_{b} = e^{(-\Delta G^{\circ}_{b})/ (RT)}\) ##Step 5: Determine the disproportionation reaction for HClO2## For reaction c with HClO2, the half-reactions are provided. First, we should balance the overall redox reaction.

Balance the redox reaction for HClO2

Balanced disproportionation reaction for HClO2: \(2 \mathrm{HClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q) + \mathrm{HClO}(a q) + \mathrm{H}_{2}\mathrm{O}\) ##Step 6: Calculate the \(\Delta G^{\circ}\) and \(K\) for HClO2 disproportionation## Calculate the standard Gibbs free energy change and the equilibrium constant for the HClO2 disproportionation using the provided half-reaction potentials.

Calculate \(\Delta G^{\circ}\) for HClO2 disproportionation

\(\Delta G^{\circ}_{c} = -n(96485)(\mathscr{E}^{\circ}_{ClO_{3}^{-}/HClO_{2}}-\mathscr{E}^{\circ}_{HClO_{2}/HClO})\)

Calculate \(K\) for HClO2 disproportionation (if spontaneous)

If the HClO2 disproportionation is spontaneous, use \(\Delta G^{\circ}_{c}\) to calculate \(K_{c}\): \(K_{c} = e^{(-\Delta G^{\circ}_{c})/ (RT)}\) With these calculations, we can determine the spontaneous disproportionation reactions under standard conditions from the given examples and their respective equilibrium constants.

Key concepts

Oxidation States

Oxidation states are helpful in understanding how electrons are transferred in chemical reactions. They represent the degree of oxidation of an atom within a molecule.
In disproportionation reactions, a single substance is both oxidized and reduced, leading to different oxidation states.

• For instance, in the reaction of copper, \(2 \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}\), the same element transitions between two different oxidation states.
• The oxidation state for \(\text{Cu}^+\) changes to \(\text{Cu}^{2+}\) (oxidation) and to \(\text{Cu}^{0}\) (reduction).

This concept helps predict how electrons will redistribute and the potential spontaneity of reactions.

Standard Electrode Potential

The standard electrode potential, denoted as \(\mathscr{E}^\circ\), measures the tendency of a chemical species to be reduced, measured under standard conditions.

• Each half-reaction in a redox reaction has an associated standard electrode potential.
• In a disproportionation reaction, we combine these potentials to determine the net voltage.
• Reactions with a positive net \(\mathscr{E}^\circ\) are more likely to be spontaneous.

For example, by examining \(\mathscr{E}^\circ\) for copper and iron reactions, we can predict their behavior and calculate their Gibbs free energy.

Gibbs Free Energy

Gibbs free energy, \(\Delta G^\circ\), tells us whether a reaction is spontaneous.

• A negative \(\Delta G^\circ\) indicates a spontaneous process.
• The equation \(\Delta G^\circ = -nFE^\circ\) helps convert electrode potentials to Gibbs free energy.
• Here, \(F\) (Faraday's constant) is about 96485 C/mol, and \(n\) is the number of moles of electrons transferred.

By calculating \(\Delta G^\circ\) for each reaction, we determine which reactions will occur naturally under standard conditions.

Equilibrium Constant

The equilibrium constant, \(K\), connects Gibbs free energy and the position of equilibrium.

• If \(\Delta G^\circ < 0\), then \(K > 1\), reflecting a product-favored system.
• The relationship is established by \(\Delta G^\circ = -RT \ln K\), where \(R\) is the gas constant and \(T\) is the temperature in Kelvin.
• This equation allows us to find \(K\) from \(\Delta G^\circ\).

Calculating \(K\) for spontaneous reactions provides insight into their extent under standard conditions, helping us understand the balance between reactants and products.