Problem 51 - Chemistry The Central Science In Si Units

Question

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \end{aligned} $$ \(\mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \quad E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V}\) \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \quad E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V}\) (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q),\) by \(\mathrm{N}_{2} \mathrm{O}(a q),\) and by \(\mathrm{VO}_{2}^{+}(a q)\) (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Step-by-Step Solution

Answer

For each reaction, we obtain: \(2 \text{Fe}^{2+} + \text{S}_2\text{O}_6^{2-} + 4 \text{H}^+ \to 2 \text{Fe}^{3+} + 2 \text{H}_2\text{SO}_3 \), \(K \approx 3 \times 10^{46}\); \(2 \text{Fe}^{2+} + \text{N}_2\text{O} + 2 \text{H}^+ \to 2 \text{Fe}^{3+} + \text{N}_2 + \text{H}_2\text{O} \), \(K \approx 1 \times 10^{-34}\); \(\text{Fe}^{2+} + \text{VO}_2^+ + 2 \text{H}^+ \to \text{Fe}^{3+} + \text{VO}^{2+} + \text{H}_2\text{O} \), \(K \approx 9 \times 10^{29}\).

Step 1: Understanding the Concept of Redox Reactions

This exercise asks you to write balanced equations for the oxidation of \( \text{Fe}^{2+} \). This means \( \text{Fe}^{2+} \) will lose electrons (be oxidized) and the other compounds will gain electrons (be reduced). Each half-reaction given provides the reduction potential, which helps determine if the reaction will spontaneously proceed.

Step 2: Balancing the Reaction with \( \text{S}_2\text{O}_6^{2-} \)

1. Start by writing the two half-reactions: - \( \text{Fe}^{2+} \to \text{Fe}^{3+} + \text{e}^- \) with \( E^\circ = -0.77 \text{ V} \) - \( \text{S}_2\text{O}_6^{2-} + 4 \text{H}^+ + 2 \text{e}^- \to 2 \text{H}_2\text{SO}_3 \) with \( E^\circ = +0.60 \text{ V} \)2. Multiply the first half-reaction by 2 to balance the number of electrons: - \( 2\text{Fe}^{2+} \to 2\text{Fe}^{3+} + 2\text{e}^- \)3. The balanced equation: \( 2\text{Fe}^{2+} + \text{S}_2\text{O}_6^{2-} + 4 \text{H}^+ \to 2\text{Fe}^{3+} + 2\text{H}_2\text{SO}_3 \).

Step 3: Determining \( \Delta G^\circ \) for the Reaction with \( \text{S}_2\text{O}_6^{2-} \)

The formula for Gibbs free energy change is:\[ \Delta G^\circ = -nFE_{cell}^\circ \]Where \( n = 2 \) moles of electrons, \( F = 96,485 \text{ C/mol} \), and \( E_{cell}^\circ = 0.60 - (-0.77) = 1.37 \text{ V} \).Calculate:\[ \Delta G^\circ = -2 \times 96,485 \times 1.37 = -264,276 \text{ J/mol} \approx -264 \text{ kJ/mol} \]

Step 4: Calculating the Equilibrium Constant \( K \) for the Reaction with \( \text{S}_2\text{O}_6^{2-} \)

Use the relationship:\[ \Delta G^\circ = -RT \ln K \]Where \( R = 8.314 \text{ J/mol K} \) and \( T = 298 \text{ K} \).Rearrange to find:\[ K = e^{-\Delta G^\circ/RT} \]Substitute the values:\[ \ln K = \frac{264,276}{8.314 \times 298} \approx 106.5 \]\[ K = e^{106.5} \approx 3 \times 10^{46} \]

Step 5: Balancing the Reaction with \( \text{N}_2\text{O} \)

1. Write the oxidation half-reaction: - \( \text{Fe}^{2+} \to \text{Fe}^{3+} + \text{e}^- \)2. The reduction half-reaction is: - \( \text{N}_2\text{O} + 2\text{H}^+ + 2\text{e}^- \to \text{N}_2 + \text{H}_2\text{O} \) with \( E^\circ = -1.77 \text{ V} \)3. Multiply the oxidation half-reaction by 2, then add: - \( 2\text{Fe}^{2+} + \text{N}_2\text{O} + 2\text{H}^+ \to 2\text{Fe}^{3+} + \text{N}_2 + \text{H}_2\text{O} \)

Step 6: \( \Delta G^\circ \) and \( K \) for \( \text{N}_2\text{O} \) Reaction

1. \( E_{cell}^\circ = -1.77 - (-0.77) = -1.00 \text{ V} \)2. \( \Delta G^\circ = -2 \times 96,485 \times (-1.00) = 192,970 \text{ J/mol} = 192.97 \text{ kJ/mol} \)3. Calculate \( K \): - \( \ln K = \frac{-192,970}{8.314 \times 298} \approx -77.9 \) - \( K = e^{-77.9} \approx 1 \times 10^{-34} \)

Step 7: Balancing the Reaction with \( \text{VO}_2^+ \)

1. The oxidation half-reaction remains: - \( \text{Fe}^{2+} \to \text{Fe}^{3+} + \text{e}^- \)2. The reduction for \( \text{VO}_2^+ \): - \( \text{VO}_2^+ + 2\text{H}^+ + \text{e}^- \to \text{VO}^{2+} + \text{H}_2\text{O} \) with \( E^\circ = +1.00 \text{ V} \)3. Overall balanced equation: - \( \text{Fe}^{2+} + \text{VO}_2^+ + 2\text{H}^+ \to \text{Fe}^{3+} + \text{VO}^{2+} + \text{H}_2\text{O} \)

Step 8: \( \Delta G^\circ \) and \( K \) for \( \text{VO}_2^+ \) Reaction

1. \( E_{cell}^\circ = 1.00 - (-0.77) = 1.77 \text{ V} \)2. \( \Delta G^\circ = -1 \times 96,485 \times 1.77 = -170,786 \text{ J/mol} \approx -170.79 \text{ kJ/mol} \)3. Calculate \( K \): - \( \ln K = \frac{170,786}{8.314 \times 298} \approx 69.0 \) - \( K = e^{69.0} \approx 9 \times 10^{29} \)

Key Concepts

Reduction Half-Reactions

In chemistry, reduction half-reactions are crucial for understanding redox (reduction-oxidation) reactions. A reduction half-reaction involves the gain of electrons by a chemical species. It's represented separately to show the changes in oxidation states. This separation is essential for analyzing complex chemical processes.
When solving redox reactions, you will often see two half-reactions: one for oxidation and one for reduction. For example, with iron, the reduction half-reaction is:

• \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) with an electrode potential \( E_{\mathrm{red}}^{\circ} = +0.77 \text{ V} \)

The electrode potential indicates the ease or difficulty of the electron transfer process. Positive values mean a preference for reduction, while negative values might require extra energy.
Understanding these half-reactions allows you to balance overall redox reactions accurately, predict the direction of electron flow, and calculate energy changes in the system.

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G^{\circ} \), helps predict the spontaneity of reactions. It's the energy available to do useful work in chemical processes at constant temperature and pressure. In redox reactions, we find \( \Delta G^{\circ} \) with:

• \( \Delta G^{\circ} = -n F E_{\text{cell}}^{\circ} \)

Where \( n \) is the number of moles of electrons, \( F \) is Faraday's constant, and \( E_{\text{cell}}^{\circ} \) is the standard cell potential.
\( \Delta G^{\circ} \) values tell you if a reaction can spontaneously occur:

• Negative values mean the reaction is spontaneous and energy is released.
• Positive values mean the reaction is non-spontaneous and requires energy.

For example, in the reaction involving \( \mathrm{VO}_{2}^{+} \), \( \Delta G^{\circ} = -170.79 \text{ kJ/mol} \), indicating it happens spontaneously.

Equilibrium Constant

The equilibrium constant \( K \) of a reaction provides insights into the favorability of products over reactants once equilibrium is reached. We use it to understand the extent of reaction progress. For a redox process, you can relate it to Gibbs free energy using:

• \( \Delta G^{\circ} = -RT \ln K \)

Where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.
To solve for \( K \), rearrange to \( K = e^{-\Delta G^{\circ}/RT} \).
High \( K \) values indicate that products are heavily favored, while low values suggest reactants prevail at equilibrium.
For instance, when \( K \approx 9 \times 10^{29} \), as in the reaction with \( \mathrm{VO}_{2}^{+} \), it strongly favors product formation.

Oxidation-Reduction Equations

Oxidation-reduction (redox) equations represent reactions where electrons are transferred between species. These are important because they include electron charge balances and conservation of mass.
To craft a balanced redox equation, it's necessary to write the oxidation and reduction reactions separately, then combine and balance them by equalizing the lost and gained electrons.
For example, with \( \mathrm{Fe}^{2+} \) oxidizing:

• The oxidation reaction: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \)
• The reduction with \( \mathrm{VO}_{2}^{+} \): \( \text{VO}_2^+ + 2\text{H}^+ + \text{e}^- \rightarrow \text{VO}^{2+} + \text{H}_2O \)

Combine these:

• \( \text{Fe}^{2+} + \text{VO}_2^+ + 2\text{H}^+ \rightarrow \text{Fe}^{3+} + \text{VO}^{2+} + \text{H}_2O \)

This systematic approach ensures accuracy and understanding of electron flow and chemical reactions in complex systems.