91Ó°ÊÓ

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: $$ Fe^{2+}(aq) + VO_2^+(aq) + 2H^+(aq) \longrightarrow Fe^{3+}(aq) + VO^{2+}(aq) + H_2O(l) $$

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 \(\Delta G^{\circ} = -(1)(96485C/mol)(0.23V) = -22,171.55 J/mol \).

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.