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 \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each reaction.

Short answer

a. The balanced equations for the reactions are: 1. \(\mathrm{Mn}(s) + 2\mathrm{HNO}_3 (aq) \rightarrow \mathrm{Mn}^{2+} (aq) + 2\mathrm{NO}_2 (g) + 2\mathrm{H}_2 \mathrm{O} (l)\) 2. \(5\mathrm{Mn}^{2+} (aq) + 2\mathrm{IO}_{4}^{-} (aq) + 6\mathrm{H}^{+} (aq) \rightarrow 5\mathrm{MnO}_{4}^{-} (aq) + 2\mathrm{I}^{-} (aq) + 3\mathrm{H}_2 \mathrm{O} (l)\) b. The standard cell potentials and Gibbs free energy changes for the reactions are: 1. Dissolving steel in nitric acid: \(\mathscr{C}^{\circ} = 1.964V\), \(\Delta G^{\circ} = -378.9\ \mathrm{kJ/mol}\) 2. Oxidizing \(\mathrm{Mn}^{2+}\) ions to \(\mathrm{MnO}_{4}^{-}\) ions: \(\mathscr{C}^{\circ} = 0.09V\), \(\Delta G^{\circ} = -873.4\ \mathrm{kJ/mol}\)

Step-by-step solution

a. Balancing the equations

First, we need to balance the equations for the manganese and periodate ions. To make it easier, we will break down the process into two separate reactions: 1. Dissolving steel in nitric acid, producing \(\mathrm{Mn}^{2+}\) ions: Unbalanced equation: \(\mathrm{Mn}(s) + \mathrm{HNO}_3 (aq) \rightarrow \mathrm{Mn}^{2+} (aq) + \mathrm{NO}_2 (g) + \mathrm{H}_2 \mathrm{O} (l)\) Balanced equation: \(\mathrm{Mn}(s) + 2\mathrm{HNO}_3 (aq) \rightarrow \mathrm{Mn}^{2+} (aq) + 2\mathrm{NO}_2 (g) + 2\mathrm{H}_2 \mathrm{O} (l)\) 2. Oxidizing \(\mathrm{Mn}^{2+}\) ions to \(\mathrm{MnO}_{4}^{-}\) ions by periodate ion \(\left(\mathrm{IO}_{4}^{-}\right)\) in acid solution: Unbalanced equation: \(\mathrm{Mn}^{2+} (aq) + \mathrm{IO}_{4}^{-} (aq) + \mathrm{H}^{+} (aq) \rightarrow \mathrm{MnO}_{4}^{-} (aq) + \mathrm{I}^{-} (aq) + \mathrm{H}_2 \mathrm{O} (l)\) Balanced equation (by considering the change in oxidation number of the species involved): \(5\mathrm{Mn}^{2+} (aq) + 2\mathrm{IO}_{4}^{-} (aq) + 6\mathrm{H}^{+} (aq) \rightarrow 5\mathrm{MnO}_{4}^{-} (aq) + 2\mathrm{I}^{-} (aq) + 3\mathrm{H}_2 \mathrm{O} (l)\)

b. Calculate standard cell potential and Gibbs free energy change

For both reactions, we need to use the formula for cell potential and Gibbs free energy change to calculate the values. The formulas are: - Standard cell potential: \(\mathscr{C}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) - Gibbs free energy change: \(\Delta G^{\circ} = -nFE^{\circ}\) Let's calculate these values separately for both reactions. 1. Reaction involving the dissolution of steel in nitric acid: The anode half-reaction can be written as: Anode: \(\mathrm{Mn}(s) \rightarrow \mathrm{Mn}^{2+} (aq) + 2 e^{-}\); \(E_{anode}^{\circ} = -1.18V\) (oxidation) The cathode half-reaction can be rewritten as (since 2 electrons were produced in the anode half-reaction): Cathode: \(2\mathrm{NO}_2 (g) + 4\mathrm{H}^{+} (aq) + 2e^{-} \rightarrow 2\mathrm{HNO}_3 (aq)\); \(E_{cathode}^{\circ} = 0.784V\) (reduction) Standard cell potential for this reaction: \(\mathscr{C}^{\circ} = (0.784) - (-1.18) = 1.964V\) The number of electrons involved in this reaction, \(n = 2\) since 2 electrons were produced in the anode half reaction. Faraday’s constant, \(F = 96485\ \mathrm{C/mol}\). Gibbs free energy change: \(\Delta G^{\circ} = -(2)(96485\ \mathrm{C/mol})(1.964\ \mathrm{V}) = -378942\ \mathrm{J/mol} = -378.9\ \mathrm{kJ/mol}\) 2. Reaction involving the oxidation of \(\mathrm{Mn}^{2+}\) ions to \(\mathrm{MnO}_{4}^{-}\) ions: The anode half-reaction: Anode: \(5\mathrm{Mn}^{2+} (aq) \rightarrow 5\mathrm{MnO}_{4}^{-} (aq) + 10e^{-}\); \(E_{anode}^{\circ} = 1.51V\) (oxidation) The cathode half-reaction (after balancing with 10 electrons since 10 electrons were produced in the anode half-reaction): Cathode: \(2\mathrm{IO}_{4}^{-} (aq) + 12\mathrm{H}^{+} (aq) + 10e^{-} \rightarrow 2\mathrm{I}^{-} (aq) + 6\mathrm{H}_{2}\mathrm{O} (l)\); \(E_{cathode}^{\circ} = 1.60V\) (reduction) Standard cell potential for this reaction: \(\mathscr{C}^{\circ} = (1.60) - (1.51) = 0.09V\) The number of electrons involved in this reaction, \(n = 10\) since 10 electrons were produced in the anode half reaction. Gibbs free energy change: \(\Delta G^{\circ} = -(10)(96485\ \mathrm{C/mol})(0.09\ \mathrm{V}) = -873367.5\ \mathrm{J/mol} = -873.4\ \mathrm{kJ/mol}\) So, the standard cell potential and Gibbs free energy change for the two reactions are as follows: 1. Dissolving steel in nitric acid: \(\mathscr{C}^{\circ} = 1.964V\), \(\Delta G^{\circ} = -378.9\ \mathrm{kJ/mol}\) 2. Oxidizing \(\mathrm{Mn}^{2+}\) ions to \(\mathrm{MnO}_{4}^{-}\) ions: \(\mathscr{C}^{\circ} = 0.09V\), \(\Delta G^{\circ} = -873.4\ \mathrm{kJ/mol}\)

Key concepts

Oxidation and Reduction in Chemistry

Understanding oxidation and reduction is crucial in chemistry, especially in redox reactions which play a fundamental role in various chemical processes, including metabolism, corrosion, and industrial production. Oxidation involves the loss of electrons, or an increase in oxidation state by a molecule, an atom, or an ion, whereas reduction involves the gain of electrons or a decrease in oxidation state.

Oxidation and reduction always occur together; when a substance oxidizes, another reduces – which is why we call them redox reactions. This interaction is beautifully demonstrated in the determination of manganese content in steel, where manganese goes from being an ion in a lower oxidation state to a higher one as it becomes permanganate.

Standard Cell Potential

The standard cell potential, denoted as \( \mathscr{C}^{\circ} \), is an essential concept when discussing electrochemical cells. It is defined by the difference in electrode potentials of the cathode and anode under standard conditions, which implies a temperature of 25°C, a pressure of 1 atmos per gas involved, and 1 Molar concentrations for the solutions. In simple terms, it measures the ability of a chemical cell to produce an electric current with a potential difference between the electrodes.

The standard cell potential provides insight into the spontaneity of a reaction; a positive \( \mathscr{C}^{\circ} \) indicates a spontaneous reaction under standard conditions. This value is pivotal in the analytical determination of substances like manganese, as confirmed by the calculation in the exercise.

Gibbs Free Energy Change

Gibbs free energy change, or \( \Delta G^{\circ} \), is a thermodynamic quantity used to predict the direction of chemical reactions and to determine whether a process is spontaneous. It relates directly to the standard cell potential via the equation \( \Delta G^{\circ} = -nFE^{\circ} \), where \( n \) is the number of moles of electrons transferred in the reaction, \( F \) is Faraday's constant representing the charge per mole of electrons, and \( E^{\circ} \) is the standard cell potential. If \( \Delta G^{\circ} \) is negative, the reaction occurs spontaneously under standard conditions. This calculation is of particular importance when analyzing redox reactions like those in the manganese determination method.

Balancing Chemical Equations

Balancing chemical equations is a basic yet vital skill in chemistry; it ensures the law of conservation of mass is followed in a chemical equation. Each element must have the same number of atoms on both the reactant and product sides. This principle is applied to complex redox reactions by breaking them into half-reactions for oxidation and reduction, balancing each for mass and charge, then combining them back.

For the analytical determination of manganese, the key to balancing the equation is to ensure that the electron change (i.e., the number of electrons lost or gained) in the oxidation half-reaction matches that in the reduction half-reaction. This process is essential for all subsequent calculations, including cell potential and Gibbs free energy.

Analytical Determination of Manganese

The analytical determination of manganese in materials like steel involves several chemical processes. The first step is the dissolution of steel in nitric acid, which results in the formation of manganese ions, \( \mathrm{Mn}^{2+} \). The second step includes the oxidation of \( \mathrm{Mn}^{2+} \) to \( \mathrm{MnO}_{4}^{-} \) ions, which are visually identifiable due to their deep color—an illustration of a qualitative analytical method.

To obtain quantitative data, chemists calculate the standard cell potential and Gibbs free energy change, as outlined in the exercise. The accuracy in determining the manganese concentration in steel, or in any other sample, depends on the precise execution of these methodological steps, reflective of the chemistry's finesse and analytical prowess.