Problem 51 - Chemistry: The Central Science in SI Units, Global Edition

Question

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{aligned} $$ (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

The balanced chemical equations for the oxidation of Fe²⁺ by the given reactants are: (a) \(2\text{Fe}^{2+} + \text{S}_{2}\text{O}_{6}^{2-} + 4\text{H}^{+} \rightarrow 2\text{Fe}^{3+} + 2\text{H}_{2}\text{SO}_{3}\) (b) \(\text{Fe}^{2+} + \text{N}_{2}\text{O} + 2\text{H}^{+} \rightarrow \text{Fe}^{3+} + \text{N}_{2} + \text{H}_{2}\text{O}\) (c) \(\text{Fe}^{2+} + 2\text{VO}_{2}^{+} + 4\text{H}^{+} \rightarrow \text{Fe}^{3+} + 2\text{VO}^{2+} + 2\text{H}_{2}\text{O}\) The standard Gibbs free energy change (ΔG°) for each reaction at 298 K are: ΔG°₁ = -32,841 J/mol, ΔG°₂ = -245,099 J/mol, and ΔG°₃ = -22,193 J/mol. The equilibrium constants (K) for each reaction at 298 K are: K₁ = 5.37 × 10¹², K₂ = 8.68 × 10⁷³, and K₃ = 4.95 × 10⁷.

Step 1: Identify the redox couples

First, let's identify the redox couples involved in the given reactions. 1. Fe³⁺/Fe²⁺ 2. S₂O₆²⁻/H₂SO₃ 3. N₂O/N₂ 4. VO₂⁺/VO²⁺ We will write balanced chemical equations for the oxidation of Fe²⁺(aq) by S₂O₆²⁻(aq), N₂O(g), and VO₂⁺(aq).

Step 2: Balance each redox reaction

For each redox reaction, the half-reactions need to be combined, making sure that the electrons cancel out, and the substances correctly balanced. (a) Oxidation of Fe²⁺ by S₂O₆²⁻ Oxidation half-reaction: Fe²⁺ -> Fe³⁺ + e⁻ (E° = +0.77 V) Reduction half-reaction: S₂O₆²⁻ + 4H⁺ + 2e⁻ -> 2H₂SO₃ (E° = +0.60 V) Multiply the first half-reaction by 2 to cancel the electrons, and then combine both half-reactions: 2(Fe²⁺ -> Fe³⁺ + e⁻) + S₂O₆²⁻ + 4H⁺ + 2e⁻ -> 2H₂SO₃ Resulting reaction: 2Fe²⁺ + S₂O₆²⁻ + 4H⁺ -> 2Fe³⁺ + 2H₂SO₃ (b) Oxidation of Fe²⁺ by N₂O Oxidation half-reaction: Fe²⁺ -> Fe³⁺ + e⁻ (E° = +0.77 V) Reduction half-reaction: N₂O + 2H⁺ + 2e⁻ -> N₂ + H₂O (E° = -1.77 V) Combine both half-reactions (no need to multiply as the electrons are already equal): Fe²⁺ + N₂O + 2H⁺ -> Fe³⁺ + N₂ + H₂O (c) Oxidation of Fe²⁺ by VO₂⁺ Oxidation half-reaction: Fe²⁺ -> Fe³⁺ + e⁻ (E° = +0.77 V) Reduction half-reaction: VO₂⁺ + 2H⁺ + e⁻ -> VO²⁺ + H₂O (E° = +1.00 V) Multiply the second half-reaction by 2 to cancel the electrons, and then combine both half-reactions: Fe²⁺ -> Fe³⁺ + e⁻ + 2(VO₂⁺ + 2H⁺ + e⁻) -> 2VO²⁺ + 2H₂O Resulting reaction: Fe²⁺ + 2VO₂⁺ + 4H⁺ -> Fe³⁺ + 2VO²⁺ + 2H₂O

Step 3: Calculate ΔG° for each reaction

To calculate ΔG° for each reaction at 298 K, we need to utilize the following equation: ΔG° = -nFE° where n is the number of electrons transferred in the reaction, F is Faraday's constant (96485 C/mol), and E° is the standard cell potential for the reaction. ΔG° values for each reaction: (a) ΔG°₁ = -2 (96485 C/mol) (0.77 V - 0.60 V) = -32,841 J/mol (b) ΔG°₂ = -1 (96485 C/mol) (0.77 V - (-1.77 V)) = -245,099 J/mol (c) ΔG°₃ = -1 (96485 C/mol) (0.77 V - 1.00 V) = -22,193 J/mol

Step 4: Calculate the equilibrium constant K for each reaction

To determine the equilibrium constant K for each reaction at 298 K, we need to use the relationship between ΔG° and K: ΔG° = -RT ln K where R is the gas constant (8.314 J/(mol∙K)) and T is the temperature (298 K). We can solve for K by using the calculated ΔG° values. K values for each reaction: (a) K₁ = exp(-(-32,841 J/mol) / (8.314 J/(mol∙K) × 298 K)) = 5.37 × 10¹² (b) K₂ = exp(-(-245,099 J/mol) / (8.314 J/(mol∙K) × 298 K)) = 8.68 × 10⁷³ (c) K₃ = exp(-(-22,193 J/mol) / (8.314 J/(mol∙K) × 298 K)) = 4.95 × 10⁷ So, the equilibrium constants K at 298 K for each reaction are: K₁ = 5.37 × 10¹², K₂ = 8.68 × 10⁷³, and K₃ = 4.95 × 10⁷.

Key Concepts

Electrochemistry

Electrochemistry is a fascinating branch of chemistry that deals with the interaction between electrical energy and chemical changes. It plays a crucial role in many everyday processes, including battery operation and corrosion control.

Electrochemical reactions are either spontaneous or non-spontaneous. Spontaneous reactions can generate electrical energy, whereas non-spontaneous requires an input of electrical energy to occur. A classic example of electrochemical processes includes the redox reaction, where oxidation and reduction processes occur simultaneously.

• Oxidation: The loss of electrons from a chemical species.
• Reduction: The gain of electrons by a chemical species.

By balancing these reactions and combining them effectively, we get an overall cell reaction that produces electrical energy, which can be seen in the exercise provided. Understanding electrochemical principles is essential for calculating the energy changes associated with these reactions, establishing the foundation for more complex analyses, such as determining the Gibbs free energy and equilibrium constants.

Standard Reduction Potential

Standard reduction potential (\(E_{ ext{red}}^{ ext{°}}\)) is a measure of the tendency of a chemical species to gain electrons, i.e., to be reduced. It is expressed in volts and measured under standard conditions: 25°C, 1 atm pressure, and 1 M concentration.

Higher standard reduction potential values indicate a greater likelihood for reduction to occur. In the context of redox reactions, these values are used to predict the direction of electron flow between different species.

• A more positive potential implies a strong tendency to be reduced.
• A more negative potential indicates a tendency to lose electrons and therefore be oxidized.

In the given exercise, understanding and using these potentials is vital for calculating the cell potential and subsequently determining the Gibbs free energy (ΔG) of the reactions. When combining half-reactions, the standard reduction potentials are subtracted (\(E^{ ext{°}}_{ ext{cell}} = E^{ ext{°}}_{ ext{red, cathode}} - E^{ ext{°}}_{ ext{red, anode}}\)), offering insight into the reaction's feasibility.

Equilibrium Constant

The equilibrium constant (\(K\)) gives us insight into the extent of a chemical reaction under equilibrium conditions. It is a ratio of the concentrations of the products to the reactants, raised to their respective stoichiometric coefficients, at equilibrium.

In electrochemical cells, the equilibrium constant is linked to the Gibbs free energy change (\(ΔG^°\)) for the reaction. The relationship is expressed via the equation:\[ΔG^° = -RT \, \ln(K)\]where:

• \(R\)is the universal gas constant (8.314 J/(mol·K)).
• \(T\)is the temperature in Kelvin.

This means that knowing \(K\) allows us to deduce how far a reaction will proceed before reaching equilibrium. Large \(K\) values indicate the reaction lies far towards products, while small \(K\) values suggest minimal product formation at equilibrium. The calculation of \(K\)is crucial for understanding the feasibility and directionality of chemical processes, as seen in the reactions given in the exercise.

Gibbs Free Energy

Gibbs free energy (\(ΔG\)) is a thermodynamic quantity that reflects the amount of energy available to do work during a process at constant temperature and pressure. It serves as a powerful predictor of reaction spontaneity:

• \(ΔG < 0\): the process is spontaneous and can occur without energy input.
• \(ΔG = 0\): the process is at equilibrium.
• \(ΔG > 0\): the process is non-spontaneous and requires energy input.

In electrochemistry, \(ΔG^°\) relates to the cell potential (\(E^{ ext{°}}_{ ext{cell}}\)) as follows:\[ΔG^° = -nF \, E^{ ext{°}}_{ ext{cell}}\]where:

• \(n\)is the number of moles of electrons transferred.
• \(F\)is Faraday's constant (96485 C/mol).

Calculating \(ΔG^°\) gives a quantitative measure of the driving force behind a chemical reaction. In the exercise provided, understanding \(ΔG^°\)is essential to predict and evaluate the thermodynamic efficiency of the given redox reactions.