Question

Given the following reduction half-reactions: $$ \begin{gathered} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) \quad E_{\text {red }}^{\infty}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \quad E_{\mathrm{ret}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \quad \mathrm{E}_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \quad E_{\mathrm{red}}^{0}=+1.00 \mathrm{~V} \end{gathered} $$ (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 \(\mathrm{VO}_{2}^{+}\)(aq). (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Short answer

To summarize: (a) Balanced chemical equations for the oxidation of \(Fe^{2+}(aq)\) are: 1. \(3 \ Fe^{2+}(aq)+ S_2O_6^{2−}(aq) + 4H^+(aq) \longrightarrow 3 \ Fe^{3+}(aq) + 2 \ H_2SO_3(aq)\) 2. \(2 \ Fe^{2+}(aq) + N_2O(g) + 2H^+(aq) \longrightarrow 2 \ Fe^{3+}(aq) + N_2(g) + H_2O(l)\) 3. \(Fe^{2+}(aq) + VO_2^+(aq) + 2H^+(aq) \longrightarrow Fe^{3+}(aq) + VO^{2+}(aq) + H_2O(l)\) (b) Standard Gibbs free energy change (\(\Delta G^{\circ}\)) for each reaction at 298K are: 1. \(32,847.9\,\text{J/mol}\) 2. \(490,456.2\,\text{J/mol}\) 3. \(-22,171.55\,\text{J/mol}\) (c) Equilibrium constant \(K\) for each reaction at 298K are: 1. \(K = 0.011\) 2. \(K \approx 0\) 3. \(K = 12,826\)

Step-by-step solution

Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\)

Since we are given reduction half-reactions, first, let's find the oxidation half-reactions for \(Fe^{2+}(aq)\), which has the given reduction half-reaction: $$ \mathrm{Fe}^{3+}(aq) + \mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(aq) \quad E_{\text {red }}^{\infty}=+0.77 \mathrm{V} $$ To find its corresponding oxidation half-reaction, simply reverse the above reaction and change the sign of its reduction potential: $$ \mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{e}^{-} \quad E_{\text {ox }}^{\infty}=-0.77 \mathrm{V} $$ Now we will combine each of the given half-reactions with the oxidation half-reaction for \(Fe^{2+}(aq)\). In each case, we need to ensure that the number of transferred electrons is equal, so we need to balance each reaction by multiplying if needed. (a) Balancing the oxidation of \(Fe^{2+}(aq)\) by \(S_2O_6^{2-}(aq)\): $$ S_2O_6^{2-}(aq)+4H^{+}(aq)+2e^− \longrightarrow 2H_2SO_3(aq) \quad E_\text{red}^{\circ}=+0.60V \\ Fe^{2+}(aq) \longrightarrow Fe^{3+}(aq) + e^− \quad E_\text{ox}^{\circ} = -0.77V $$ Multiply the second reaction by 2 and sum: $$ 3 \ Fe^{2+}(aq)+ S_2O_6^{2−}(aq) + 4H^+(aq) \longrightarrow 3 \ Fe^{3+}(aq) + 2 \ H_2SO_3(aq) $$ Similarly, balance the oxidation of \(Fe^{2+}(aq)\) by \(N_2O(aq)\) and \(VO_2^+(aq)\): (b) Balancing the oxidation of \(Fe^{2+}(aq)\) by \(N_2O(aq)\): $$ N_2O(g) + 2H^+(aq) + 2e^− \longrightarrow N_2(g) + H_2O(l) \quad E_\text{red}^{\circ}=-1.77V \\ Fe^{2+}(aq) \longrightarrow Fe^{3+}(aq) + e^− \quad E_\text{ox}^{\circ} = -0.77V $$ Multiply the second reaction by 2 and sum: $$ 2 \ Fe^{2+}(aq) + N_2O(g) + 2H^+(aq) \longrightarrow 2 \ Fe^{3+}(aq) + N_2(g) + H_2O(l) $$ (c) Balancing the oxidation of \(Fe^{2+}(aq)\) by \(VO_2^+(aq)\): $$ VO_2^+(aq) + 2H^+(aq) + e^− \longrightarrow VO^{2+} + H_2O(l) \quad E_\text{red}^{\circ}=+1.00V \\ Fe^{2+}(aq) \longrightarrow Fe^{3+}(aq) + e^− \quad E_\text{ox}^{\circ} = -0.77V $$ Just sum both reactions, as they are already balanced in terms of electron count: $$ Fe^{2+}(aq) + VO_2^+(aq) + 2H^+(aq) \longrightarrow Fe^{3+}(aq) + VO^{2+}(aq) + H_2O(l) $$

Calculate \(\Delta G^{\circ}\) for each reaction at 298K

We can calculate the standard Gibbs free energy change (\(\Delta G^{\circ}\)) for each reaction using the Nernst equation: $$ \Delta G^{\circ} = -nFE^{\circ} $$ Where \(n\) is the number of electrons transferred, \(F\) is the Faraday's constant (96,485 C/mol), and \(E^{\circ}\) is the standard cell potential. For each reaction, the standard cell potential, \(E^{\circ}\), can be calculated by summing the reduction potential of the reducing agent and the oxidation potential of the oxidizing agent: For the reaction of \(Fe^{2+}(aq)\) with \(S_2O_6^{2-}(aq)\), \(E^{\circ} = E_\text{red}^{\circ}(S_2O_6^{2-}) + E_\text{ox}^{\circ}(Fe^{2+}) = +0.60V - 0.77V = -0.17V\). \(n=2\) electrons are transferred in this reaction, so \(\Delta G^{\circ} = -2(96485C/mol)(-0.17V) = 32,847.9 J/mol\). Next, for the reaction of \(Fe^{2+}(aq)\) with \(N_2O(aq)\), \(E^{\circ} = E_\text{red}^{\circ}(N_2O) + E_\text{ox}^{\circ}(Fe^{2+}) = -1.77V - 0.77V = -2.54V\). \(n=2\) electrons are transferred, so \(\Delta G^{\circ} = -2(96485C/mol)(-2.54V) = 490,456.2 J/mol\). Finally, for the reaction of \(Fe^{2+}(aq)\) with \(VO_2^+(aq)\), \(E^{\circ} = E_\text{red}^{\circ}(VO_2^+) + E_\text{ox}^{\circ}(Fe^{2+}) = +1.00V - 0.77V = 0.23V\). \(n=1\) electron is transferred, so \(\Delta G^{\circ} = -(1)(96485C/mol)(0.23V) = -22,171.55 J/mol \).

Calculate the equilibrium constant \(K\) for each reaction at 298K

To calculate the equilibrium constant \(K\) for each reaction, we will use the following equation: $$ \Delta G^{\circ} = -RT \ln K $$ Where \(R\) is the gas constant (8.314 J/(mol·K)) and \(T\) is the temperature in Kelvin (298 K). We have determined the \(\Delta G^{\circ}\) values for each reaction in the previous step. Now let's calculate the equilibrium constants. For the reaction of \(Fe^{2+}(aq)\) with \(S_2O_6^{2-}(aq)\): $$ 32,847.9\,\text{J/mol} = - (8.314\,\text{J/(mol·K)}) (298\,\text{K}) \ln K \\ \ln K = -4.51 \\ K = \exp(-4.51) = 0.011 $$ For the reaction of \(Fe^{2+}(aq)\) with \(N_2O(aq)\): $$ 490,456.2\,\text{J/mol} = - (8.314\,\text{J/(mol·K)}) (298\,\text{K}) \ln K \\ \ln K = -86.83 \\ K = \exp(-86.83) \approx 0 $$ For the reaction of \(Fe^{2+}(aq)\) with \(VO_2^+(aq)\): $$ -22,171.55\,\text{J/mol} = - (8.314\,\text{J/(mol·K)}) (298\,\text{K}) \ln K \\ \ln K = 9.456 \\ K = \exp(9.456) = 12,826 $$ So, the equilibrium constants for the reactions involving \(Fe^{2+}(aq)\) with \(S_2O_6^{2-}(aq)\), \(N_2O(aq)\), and \(VO_2^+(aq)\) are 0.011, approximately 0, and 12,826, respectively.

Key concepts

Oxidation and Reduction

Understanding the concepts of oxidation and reduction is pivotal when studying chemistry, particularly, when it comes to balancing chemical reactions. Oxidation and reduction processes occur simultaneously in a type of reaction known as a redox reaction. In simple terms, oxidation refers to the loss of electrons, whereas reduction is the gain of electrons. An easy way to remember this is through the acronym 'OIL RIG': Oxidation Is Loss, Reduction Is Gain.

In a redox reaction, there is always a transfer of electrons from the species that gets oxidized to the one that gets reduced. As seen in the textbook exercise, \(\mathrm{Fe}^{2+}(aq)\) is oxidized by losing an electron, thus becoming \(\mathrm{Fe}^{3+}(aq)\), while the other species in each provided half-reaction is reduced by gaining electrons. Every redox process consists of two half-reactions: the oxidation half-reaction and the reduction half-reaction. These are often combined to reveal the overall reaction and to ensure that the number of electrons lost and gained is balanced. \

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• \For oxidation: \(\mathrm{Species} \longrightarrow \mathrm{Species}^+ + \mathrm{e}^-\)\<\/li>\
• \For reduction: \(\mathrm{Species}^+ + \mathrm{e}^- \longrightarrow \mathrm{Species}\)\<\/li>\<\/ul>\ Proper balancing of redox reactions is crucial, as it ensures the law of conservation of mass is upheld, reflecting the fact that matter cannot be created or destroyed in a chemical reaction.

Standard Gibbs Free Energy Change

The standard Gibbs free energy change (\(\Delta G^\circ\)) is a thermodynamic quantity used to predict the direction of chemical reactions and their spontaneity at standard conditions, which include a temperature of 298 K (approximately 25°C) and a pressure of 1 atmosphere.

The equation \(\Delta G^\circ = -nFE^\circ\) is instrumental in determining the free energy change, where \(n\) is the number of moles of electrons exchanged, \(F\) represents Faraday's constant (96,485 C/mol e-), and \(E^\circ\) is the standard cell potential, measured in volts. The sign of \(\Delta G^\circ\) provides vital information: a negative \(\Delta G^\circ\) indicates a spontaneous reaction, while a positive value suggests that the reaction is non-spontaneous under standard conditions.

In our textbook example, \(\Delta G^\circ\) is calculated for various reactions involving the oxidation of \(\mathrm{Fe}^{2+}(aq)\). Through the proper use of the Nernst equation, these calculations allow us to understand how likely a chemical reaction is to proceed without external energy input. It's essential to match the sign of the standard potential for oxidation and reduction correctly; otherwise, this would result in incorrect \(\Delta G^\circ\) values.

Equilibrium Constant Calculation

The equilibrium constant (\(K\)) is a fundamental concept in chemistry, signifying the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. It is a dimensionless quantity that offers valuable insight into the position of equilibrium—the extent to which reactants convert to products.

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation \(\Delta G^\circ = -RT \ln K\), where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin. A positive \(\Delta G^\circ\) value correlates to a small equilibrium constant, indicating that the reactants are favored at equilibrium. Conversely, a negative \(\Delta G^\circ\) signifies a large equilibrium constant, showing a preference for products at equilibrium.

In the exercise provided, the equilibrium constant is computed for each redox reaction involving the oxidation of \(\mathrm{Fe}^{2+}(aq)\), using the \(\Delta G^\circ\) values obtained. It's crucial to recognize that the natural logarithm (\(\ln\)) of the equilibrium constant is inversely proportional to \(\Delta G^\circ\), thus making it necessary to be precise while performing these calculations, as a mistake in determining the sign of \(\Delta G^\circ\) can lead to inaccuracies in the calculated values of \(K\).