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 \(298 \mathrm{~K}\).

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).

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

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

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.