A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

Step by step solution

01

Determine overall cell potential for Reaction a.

For reaction a, we have the following disproportionation: \(2\,Cu^+(aq) \rightarrow Cu^{2+}(aq) + Cu(s)\) First, we identify the two half-reactions: 1. \(Cu^+(aq) \rightarrow Cu^{2+}(aq) + e^-\): Oxidation 2. \(Cu^+(aq) + e^- \rightarrow Cu(s)\): Reduction Using the standard reduction potential table, we find that: 1. \(\mathscr{E}^\circ_{1} = 0.153\,\text{V}\) 2. \(\mathscr{E}^\circ_{2} = -0.153\,\text{V}\) Now, sum the two half-reactions to find the overall cell potential: \(\mathscr{E}^\circ_{rxn} = \mathscr{E}^\circ_{1} + \mathscr{E}^\circ_{2} = 0.153\,\text{V} + (-0.153\,\text{V}) = 0\,\text{V}\)

02

Calculate \(\Delta G^\circ\) and \(K\) for Reaction a.

Since the cell potential is 0 V, this reaction is not spontaneous. Therefore, there's no need to calculate \(\Delta G^\circ\) or \(K\) in this case.

03

Determine overall cell potential for Reaction b.

For reaction b, we have the following disproportionation: \(3\,Fe^{2+}(aq) \rightarrow 2\,Fe^{3+}(aq) + Fe(s)\) We identify the two half-reactions: 1. \(2\,Fe^{2+}(aq) \rightarrow 2\,Fe^{3+}(aq) + 2e^-\): Oxidation 2. \(Fe^{2+}(aq) + 2e^- \rightarrow Fe(s)\): Reduction From the standard reduction potential table, we know that: 1. \(-2\mathscr{E}^\circ_{1} = 2\times(0.771\,\text{V}) = 1.542\,\text{V}\) 2. \(\mathscr{E}^\circ_{2} = -0.771\,\text{V}\) Now, we sum the two half-reactions to get the cell potential: \(\mathscr{E}^\circ_{rxn} = -2\mathscr{E}^\circ_{1} + \mathscr{E}^\circ_{2} = 1.542\,\text{V} - 0.771\,\text{V} = 0.771\,\text{V}\)

04

Calculate \(\Delta G^\circ\) and \(K\) for Reaction b.

Now that we have the overall cell potential for reaction b, we can calculate the Gibbs free energy change: \(\Delta G^\circ = -nF\mathscr{E}^\circ = -2(96,485\,\text{C/mol})(0.771\,\text{V}) = -148,515\,\text{J/mol}\) Since \(\Delta G^\circ\) is negative, the reaction is spontaneous. Now, let's calculate the equilibrium constant \(K\): \(\Delta G^\circ = -RT\ln K \Rightarrow K = e^{(-\Delta G^\circ / RT)} = e^{(148,515\,\text{J/mol}/(8.314\,\text{J/(mol·K)}\times298.15\,\text{K}))} = 2.02\times10^{31}\)

05

Determine overall cell potential for reaction c.

Reaction c has an unbalanced chemical equation. The balanced disproportionation for reaction c is: \(2\,HClO_2(aq) \rightarrow ClO_3^-(aq) + HClO(aq) + H_2O\) Now, we're given the two relevant half-reactions, so we can write the overall redox reaction as: \(HClO_2 + 2H^+ + 2e^- \rightarrow HClO + H_2O\qquad \mathscr{E}^\circ_1 = 1.65\,\text{V}\) \(2HClO_2 \rightarrow ClO_3^-+ 3H^+ + 2e^-\qquad\, \mathscr{E}^\circ_2 = -1.21\,\text{V}\) Now, we sum the two half-reactions to get the cell potential: \(\mathscr{E}^\circ_{rxn} = \mathscr{E}^\circ_1 + \mathscr{E}^\circ_2 = 1.65\,\text{V} - 1.21\,\text{V} = 0.44\,\text{V}\)

06

Calculate \(\Delta G^\circ\) and \(K\) for Reaction c.

Now that we have the overall cell potential for reaction c, we can calculate the Gibbs free energy change: \(\Delta G^\circ = -nF\mathscr{E}^\circ = -2(96,485\,\text{C/mol})(0.44\,\text{V}) = -85,069\,\text{J/mol}\) Since \(\Delta G^\circ\) is negative, the reaction is spontaneous. Now, let's calculate the equilibrium constant \(K\): \(\Delta G^\circ = -RT\ln K \Rightarrow K = e^{(-\Delta G^\circ / RT)} = e^{(85,069\,\text{J/mol}/(8.314\,\text{J/(mol·K)}\times298.15\,\text{K}))} = 2.45\times10^{12}\) In summary, reactions b and c are spontaneous under standard conditions, with the following properties: - Reaction b: \(\Delta G^\circ = -148,515\,\text{J/mol}\) and \(K = 2.02\times10^{31}\) - Reaction c: \(\Delta G^\circ = -85,069\,\text{J/mol}\) and \(K = 2.45\times10^{12}\)

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