Problem 81 - Chemistry and Chemical Reactivity

Question

The following reaction can be used to prepare iodine in the laboratory. $$\begin{aligned}2 \mathrm{NaI}(\mathrm{s}) &+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{s}) \rightarrow \\\& \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnSO}_{4}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\end{aligned}$$ (a) Determine the oxidation number of each atom in the equation. (b) What is the oxidizing agent, and what has been oxidized? What is the reducing agent, and what has been reduced? (c) Is the reaction observed product-favored or reactant-favored? (d) Name the reactants and products.

Step-by-Step Solution

Answer

(a) Assign oxidation numbers; (b) Iodine oxidized, Mn reduced; (c) Product-favored; (d) NaI, H₂SO₄, MnO₂, Na₂SO₄, MnSO₄, I₂, H₂O.

Step 1: Assign Oxidation Numbers

First, identify the oxidation numbers for each atom in the reaction based on common oxidation states. Sodium (Na) has an oxidation number of +1, iodine (I) has -1 in NaI, H has +1 and S has +6 in H₂SO₄, Mn in MnO₂ has +4, and O has -2. In the products, sodium (Na) has +1, sulfur (S) maintains +6 in Na₂SO₄ and MnSO₄, iodine (I) is zero in I₂, and H in H₂O is +1, while O is -2.

Step 2: Identify Oxidation and Reduction

Iodine's oxidation changes from -1 in NaI to 0 in I₂, meaning iodine is oxidized. Manganese changes from +4 in MnO₂ to +2 in MnSO₄, meaning manganese is reduced. Therefore, NaI is the reducing agent, and MnO₂ is the oxidizing agent.

Step 3: Assess Reaction Favorability

Determine if the reaction is product-favored or reactant-favored. Since iodine is produced, a less stable and more reactive form of the element, the reaction tends to be product-favored as it results in the formation of elemental iodine, a desired product.

Step 4: Name Reactants and Products

Reactants: Sodium iodide (NaI), sulfuric acid (H₂SO₄), and manganese dioxide (MnO₂). Products: Sodium sulfate (Na₂SO₄), manganese(II) sulfate (MnSO₄), iodine (I₂), and water (H₂O).

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is a fundamental step in mastering redox reactions. Oxidation numbers, also known as oxidation states, are assigned to atoms to indicate the degree of oxidation or reduction they experience during a chemical reaction.
Here are some guidelines to follow when assigning oxidation numbers:

• For elements in their pure form (like O₂, Na, or I₂), the oxidation number is always 0.
• The oxidation number of a monoatomic ion is equal to the charge of the ion. For example, the Na⁺ ion has an oxidation number of +1.
• Hydrogen generally has an oxidation number of +1, and oxygen usually has -2, with some exceptions (such as in peroxides where oxygen is -1).
• In compounds, the sum of the oxidation numbers of all atoms must be equal to the overall charge of the compound or ion.

In the example of the reaction provided, each element has been assigned its oxidation number based on these rules. By comparing oxidation numbers in the reactants and products, one can determine which species are oxidized and which are reduced.

Oxidizing and Reducing Agents

The concepts of oxidizing and reducing agents are central to understanding redox reactions. An oxidizing agent, or oxidant, is a substance that gains electrons in the chemical reaction, causing another substance to lose electrons or become oxidized. Conversely, a reducing agent, or reductant, is a substance that loses electrons, causing another substance to gain them and become reduced.
Here’s how you can identify them:

• The species that is reduced (gains electrons) is the oxidizing agent because it "oxidizes" another species.
• The species that is oxidized (loses electrons) is the reducing agent as it "reduces" the other element.

In the provided example, sodium iodide (NaI) acts as the reducing agent because iodine is oxidized from a -1 state in NaI to a 0 oxidation state in I₂. Manganese dioxide (MnO₂) is the oxidizing agent as manganese is reduced from a +4 state in MnO₂ to a +2 state in MnSO₄. Understanding this role of electron transfer can help you predict and balance chemical reactions.

Chemical Equation Balancing

Balancing chemical equations is crucial to ensure that the conservation of mass is respected. Each element should have the same number of atoms on both the reactant and product sides of the equation. For redox reactions, balancing requires extra care to ensure both mass and charge balance.
For balancing simple chemical equations:

• Begin with the most complex molecule.
• Balance elements one at a time, adjusting coefficients as needed.
• Make sure to balance coefficients, meaning whole numbers that serve as multipliers before molecule formulas.

For redox reactions specifically:

• Divide the equation into two half-reactions, one for oxidation and one for reduction. Balance atoms and charges in each half.
• Add electrons to balance charges if necessary. For example, if a particular element in the reactants loses electrons, the balanced equation may include electrons on the product side.
• Then, merge the half-reactions, ensuring that the electrons cancel out, leading to an overall balanced equation for both atoms and charges.

In the example given, each side of the equation maintains the same number of each type of atom, confirming a balanced chemical reaction. This balance is vital for accurately representing what occurs during a reaction.