The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing \(\mathrm{Mn}^{2+}\) ions. These ions are then oxidized to the deeply colored \(\mathrm{MnO}_{4}^{-}\) ions by periodate ion \(\left(\mathrm{IO}_{4}{ }^{-}\right)\) in acid solution. a. Complete and balance an equation describing each of the above reactions. b. Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each reaction.

Here, the steel is first dissolved in nitric acid, producing Mn²⁺ ions. The unbalanced equation can be written as: \( Mn + HNO_{3} \rightarrow Mn^{2+} + NO_{2} + H_{2}O \) To balance the reaction equation, follow the steps mentioned below: - Balance the manganese atoms: The manganese atoms are already balanced. - Balance the hydrogen atoms: There are 4 hydrogen atoms on the left and 2 on the right, so we need to have 2 molecules of water on the right side. - Balance the nitrogen atoms: There are 2 nitrogen atoms on the left, therefore, we need to have 2 molecules of nitric acid on the right side. - Balance the oxygen atoms: There are 6 oxygen atoms on the left and 6 on the right, so they are balanced. The balanced equation is: \( Mn + 2HNO_{3} \rightarrow Mn^{2+} + 2NO_{2} + 2H_{2}O \)

The unbalanced equation can be written as: \( Mn^{2+} + HIO_{4} \rightarrow MnO_{4}^{-} + H^{+} + IO_{3}^{-} \) To balance the reaction equation, the following steps need to be performed: - Balance the manganese atoms: They are already balanced. - Balance the iodine atoms: They are already balanced. - Balance the hydrogen atoms: Place 2 hydrogen atoms on the right side by adding 2 molecules of water and 4 hydrogen atoms on the left by adding 4 protons (H⁺). - Balance the oxygen atoms: Add 2 water molecules on the right side. - Balance the charge: Add 1 extra proton on the right side. The balanced equation is: \( Mn^{2+} + 2H_{2}IO_{4} + 6H^{+} \rightarrow MnO_{4}^{-} + 2IO_{3}^{-} + 2H_{2}O + 7H^{+} \)

To calculate the standard potential for each reaction, we can use the standard reduction potentials found in a table. For reaction 1, the half-reactions are: 1. Mn²⁺ +2e⁻ → Mn; ℰ⁰(Mn²⁺/Mn) = -1.18V 2. 2HNO₃ + 2e⁻ → 2NO₂ + 2H₂O; ℰ⁰(NO₂/HNO₃) = 0.8V To find the standard potential of the reaction, ℰ⁰_cell = ℰ⁰(Reduction) - ℰ⁰(Oxidation). \( ℰ_{1}^{\circ}=0.8V-(-1.18)V=1.98V \) For reaction 2, the half-reactions are: 1. Mn²⁺ → MnO₄⁻ + 5e⁻; ℰ⁰(MnO₄⁻/Mn²⁺) = 1.51V 2. HIO₄ + 6e⁻ → 2IO₃⁻ + 2H₂O; ℰ⁰(IO₃⁻/HIO₄) = 1.64V \( ℰ_{2}^{\circ}=1.51V-(1.64)V=-0.13V \)

The standard Gibbs free energy change can be calculated using the following formula: ΔG⁰ = -nFℰ⁰ where n is the number of moles of electrons involved in the reaction, F is the Faraday constant (96485 C/mol), and ℰ⁰ is the standard cell potential. For reaction 1 (oxidation), n = 2 mol of electrons. Therefore, \( ΔG_{1}^{\circ}=-2 \times 96485 C/mol \times 1.98V \) \( ΔG_{1}^{\circ}=-383923J/mol \) For reaction 2 (reduction), n = 5 mol of electrons. Therefore, \( ΔG_{2}^{\circ}=-5 \times 96485 C/mol \times (-0.13V) \) \( ΔG_{2}^{\circ}=-62879J/mol \) Final Answers: a. Balanced equations: \( Mn + 2HNO_{3} \rightarrow Mn^{2+} + 2NO_{2} + 2H_{2}O \) \( Mn^{2+} + 2H_{2}IO_{4} + 6H^{+} \rightarrow MnO_{4}^{-} + 2IO_{3}^{-} + 2H_{2}O + 7H^{+} \) b. ℰ⁰ and ΔG⁰: Reaction 1: ℰ⁰: 1.98V ΔG⁰: -383923 J/mol Reaction 2: ℰ⁰: -0.13V ΔG⁰: -62879 J/mol