Question

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.

Short answer

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

Step-by-step solution

Balance the dissolution of manganese in steel equation

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: \( Mn + 2HNO_{3} \rightarrow Mn^{2+} + 2NO_{2} + 2H_{2}O \)

Balance the oxidation of Mn²⁺ to MnO₄⁻ by the periodate ion equation

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: \( Mn^{2+} + 2H_{2}IO_{4} + 6H^{+} \rightarrow MnO_{4}^{-} + 2IO_{3}^{-} + 2H_{2}O + 7H^{+} \)

Calculate ℰ⁰ for each reaction

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 \( ℰ_{2}^{\circ}=1.51V-(1.64)V=-0.13V \)

Calculate ΔG⁰ for each reaction

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

Key concepts

Redox Reactions

In chemistry, redox reactions are essential because they describe how electrons are transferred between substances. "Redox" stands for reduction and oxidation, two key processes in these reactions. Reduction refers to the gain of electrons by a molecule, atom, or ion, while oxidation involves the loss of electrons. This transfer of electrons changes the oxidation state of the participating substances, leading to a chemical reaction.

Redox reactions are crucial in various applications, such as energy production, corrosion, and even biological processes. In the problem we are exploring, the conversion of manganese ions to permanganate ions, facilitated by the periodate ion, is a classic redox process. Each element involves distinct half-reactions where electrons are either gained or lost. This movement of electrons enables the progression of the chemical reaction, converting the manganese to a more oxidized state.

Oxidation State

The oxidation state, or oxidation number, is a concept that helps us keep track of electrons during redox reactions. It is an imaginary charge that an atom would have if all bonds to atoms of different elements were fully ionic. In other words, it tells us about the gain or loss of electrons.

Determining the oxidation state can seem tricky, but it follows a set of rules:

• The oxidation state of a free element is always 0.
• For a simple ion, the oxidation state is equal to the net charge.
• Hydrogen generally has an oxidation state of +1, and oxygen usually has -2.
• In compounds, the sum of the oxidation states must equal the net charge of the compound.

In our exercise, manganese starts as Mn²⁺, indicating an oxidation state of +2. Through the reaction, it is transformed into MnO₄⁻ with an oxidation state of +7 for manganese. Understanding these states clarifies how electron transfer takes place during these chemical transformations.

Balancing Equations

Balancing chemical equations is essential for accurately describing chemical reactions. It ensures the conservation of mass, indicating that matter cannot be created or destroyed in a reaction. The numbers of atoms for each element need to be equal on both sides of the equation.

The process involves several steps:

• First, write down the unbalanced equation.
• Balance elements that appear only once on each side first.
• Use coefficients to balance the overall number of atoms for each element on both sides of the equation.
• Finally, ensure the charges are also balanced, if dealing with ionic compounds.

In our specific task, we needed to balance reactions involving Mn and its conversion through oxidation. This is done by adjusting coefficients to balance the number of atoms and electrons exchanged during the reactions. Doing so accurately reflects the real-life stoichiometry of the chemical processes.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G^{\circ} \)) is a thermodynamic property that helps predict the feasibility of a reaction. It tells us whether a process can occur spontaneously without energy input. If \( \Delta G^{\circ} \) is negative, the reaction proceeds spontaneously, while a positive \( \Delta G^{\circ} \) indicates that the reaction is non-spontaneous and requires energy.

Calculating \( \Delta G^{\circ} \) involves the equation:\( \Delta G^{\circ} = -nF\mathscr{E}^{\circ} \), where \( n \) is the number of electrons transferred, \( F \) is the Faraday constant (96,485 C/mol), and \( \mathscr{E}^{\circ} \) is the standard cell potential. Through this, we understand how energy changes during redox reactions, which reflects in the electrons transfer process.

For instance, the exercise problem shows that the oxidation of manganese to MnO₄⁻ by the periodate ion has a slightly negative \( \Delta G^{\circ} \), indicating a non-spontaneous reaction under standard conditions. Conversely, manganese dissolution in nitric acid reflects a negative \( \Delta G^{\circ} \) value, depicting it as spontaneous.