Problem 49 - Chemistry: The Central Science

Question

Given the following reduction half-reactions: \(\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V}\) \(\mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V}\) \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V}\) (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}{\underline{\phantom{xx}}}^{2-}(a q)\), by \(\mathrm{N}_{2} \mathrm{O}(a q)\), and \(\mathrm{by} \mathrm{VO}_{2}{\underline{\phantom{xx}}}^{+}(a q)\). (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 equations for the oxidation of Fe²⁺(aq) by S₂O₆²⁻(aq), N₂O(aq), and VO₂⁺(aq) are: 1. \(2Fe^{2+}(aq) + S_2O_6^{2-}(aq) + 2H^+(aq) \rightarrow 2Fe^{3+}(aq) + 2H_2SO_3(aq)\) 2. \(2Fe^{2+}(aq) + N_2O(g) + 2H^+(aq) \rightarrow 2Fe^{3+}(aq) + N_2(g) + H_2O(l)\) 3. \(Fe^{2+}(aq) + VO_2^+(aq) + 2H^+(aq) \rightarrow Fe^{3+}(aq) + VO^{2+}(aq) + H_2O(l)\) The values of ΔG° for each reaction at 298 K are: 1. -32,752 J/mol 2. -490,171 J/mol 3. -22,297 J/mol The equilibrium constants K for each reaction at 298 K are: 1. \(1.94 \times 10^4\) 2. \(1.49 \times 10^{-26}\) 3. 177.6

Step 1: Balance the given half-reactions

To balance the given half-reactions, follow these steps: 1. Assign oxidation numbers to each element. 2. Identify the substance being oxidized and the substance being reduced. 3. Balance the atoms in the half-reactions, separately. 4. Balance the charges by adding electrons to the appropriate side. 5. Combine the half-reactions to obtain the balanced chemical equation.

Step 2: Balance the oxidation of Fe²⁺(aq) by S₂O₆²⁻(aq)

First reaction: Oxidation half-reaction: \(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-\) The balanced equation for the oxidation of Fe²⁺(aq) by S₂O₆²⁻(aq) is: \(2Fe^{2+}(aq) + S_2O_6^{2-}(aq) + 2H^+(aq) \rightarrow 2Fe^{3+}(aq) + 2H_2SO_3(aq)\)

Step 3: Balance the oxidation of Fe²⁺(aq) by N₂O(aq)

Second reaction: Oxidation half-reaction: \(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-\) The balanced equation for the oxidation of Fe²⁺(aq) by N₂O(aq) is: \(2Fe^{2+}(aq) + N_2O(g) + 2H^+(aq) \rightarrow 2Fe^{3+}(aq) + N_2(g) + H_2O(l)\)

Step 4: Balance the oxidation of Fe²⁺(aq) by VO₂⁺(aq)

Third reaction: Oxidation half-reaction: \(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-\) The balanced equation for the oxidation of Fe²⁺(aq) by VO₂⁺(aq) is: \(Fe^{2+}(aq) + VO_2^+(aq) + 2H^+(aq) \rightarrow Fe^{3+}(aq) + VO^{2+}(aq) + H_2O(l)\)

Step 5: Calculate the ΔG° for each reaction at 298 K

Using the equation \(\Delta G^{\circ} = -nFE^{\circ}\), where n is the number of electrons transferred and F is Faraday's constant (F = 96485 C/mol): 1. For the first reaction: \(ΔG^{\circ} = -(2)(96485)(0.77 - 0.60) = -32752 J/mol\) 2. For the second reaction: \(ΔG^{\circ} = -(2)(96485)(0.77 + 1.77) = -490171 J/mol\) 3. For the third reaction: \(ΔG^{\circ} = -(1)(96485)(1.00 - 0.77) = -22297 J/mol\)

Step 6: Calculate the equilibrium constant K for each reaction at 298 K

Using the equation \(\Delta G^{\circ} = -RT \ln{K}\), where R is the gas constant (R = 8.314 J/mol K): 1. For the first reaction: \(-32752 = -(\frac{8.314 * 298}{2})(\ln{K})\) \(K = e^{\frac{32752}{3999.5}} = 1.94 \times 10^4\) 2. For the second reaction: \(-490171 = -(\frac{8.314 * 298}{2})(\ln{K})\) \(K = e^{\frac{-490171}{3999.5}} = 1.49 \times 10^{-26}\) 3. For the third reaction: \(-22297 = -(8.314 * 298)(\ln{K})\) \(K = e^{\frac{22297}{2490}} = 177.6\)

Key Concepts

Reduction Half-Reactions

Reduction half-reactions are crucial in understanding electrochemical processes. In any redox reaction, which involves electron transfer, we can split the overall equation into two parts: a reduction half-reaction and an oxidation half-reaction. A reduction half-reaction shows the gain of electrons, while the oxidation half-reaction demonstrates the loss of electrons.

Thermoelectrochemical tables often list these reactions along with their standard reduction potentials, denoted as \( E_{\mathrm{red}}^{\circ} \). These values indicate the tendency of a species to gain electrons under standard conditions. The more positive the \( E_{\mathrm{red}}^{\circ} \), the higher the tendency for the species to be reduced, acting as a stronger oxidizing agent.

• For example, the reaction \( \mathrm{Fe}^{3+}(aq) + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(aq) \) has \( E_{\mathrm{red}}^{\circ} = +0.77 \mathrm{~V} \), suggesting it readily gains an electron.
• On the contrary, \( \mathrm{N}_{2}\mathrm{O}(g) + 2 \mathrm{H}^{+}(aq) + 2 \mathrm{e}^{-} \rightarrow \mathrm{N}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \) with \( E_{\mathrm{red}}^{\circ} = -1.77 \mathrm{~V} \) has a lower tendency to gain electrons.

Understanding and balancing these half-reactions are crucial for determining the cell potential and the feasibility of the chemical process.

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G^{\circ} \), is a thermodynamic function used to predict the spontaneity of a reaction. It's crucial in electrochemistry for determining whether a redox reaction will proceed.

The relationship between Gibbs Free Energy and the standard cell potential \( (E^{\circ}) \) is given by the equation:\[ \Delta G^{\circ} = -nFE^{\circ} \]where:

• \( n \) is the number of moles of electrons exchanged in the reaction.
• \( F \) is Faraday's constant, approximately \( 96485 \text{ C/mol} \).

A negative \( \Delta G^{\circ} \) means the reaction is spontaneous under standard conditions. For instance, in the oxidation of \( \mathrm{Fe}^{2+} \) by \( \mathrm{S}_{2}\mathrm{O}_{6}^{2-} \), the \( \Delta G^{\circ} \) calculated was \(-32,752 \text{ J/mol} \), indicating spontaneity.

Each redox couple has its own \( \Delta G^{\circ} \), providing insight into which reactions are more energetically favorable. This connection helps predict which direction a reaction will naturally proceed and is a powerful tool in electrochemical cell design.

Equilibrium Constant

The equilibrium constant, \( K \), is a vital concept in understanding how far a reaction proceeds before reaching equilibrium. In the context of electrochemical reactions, \( K \) can be calculated directly from Gibbs Free Energy via the equation:\[ \Delta G^{\circ} = -RT \ln K \]where:

• \( R \) is the gas constant, \( 8.314 \text{ J/mol K} \).
• \( T \) is the temperature in Kelvin.

This equation highlights that a large and positive \( \ln K \) corresponds to a large equilibrium constant, signifying a reaction that lies largely to the right with more products at equilibrium.

For example:

• The equilibrium constant calculated for the oxidation of \( \mathrm{Fe}^{2+} \) by \( \mathrm{S}_{2}\mathrm{O}_{6}^{2-} \) was \( K = 1.94 \times 10^4 \), suggesting a significant amount of products at equilibrium.
• Conversely, \( K \) values closer to zero indicate a reaction with lesser products.

Using these constants helps in forming practical applications in areas like battery design, where predicting how a reaction will behave at equilibrium is essential.