Problem 56 - Chemistry: The Central Science

Question

Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\). (e) Hydrazine reduces \(\mathrm{CrO}_{4}{\underline{\phantom{xx}}}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}{\underline{\phantom{xx}}}^{-}\)in base (hydrazine is oxidized to \(\mathrm{N}_{2}\) ).

Step-by-Step Solution

Answer

(a) Zn (s) + 2H⁺ (aq) → Zn²⁺ (aq) + H₂ (g) (b) S (s) + 6H⁺ (aq) + 6NO₃⁻ (aq) → SO₄²⁻ (aq) + 6NO₂ (g) + 4H₂O (l) (c) SO₂ (g) + 2H⁺ (aq) + NO₃⁻ (aq) → SO₄²⁻ (aq) + H₂O (l) + NO (g) (d) N₂H₄ (g) + 6F₂ (g) → 4NF₃ (g) (e) N₂H₄ (aq) + 2CrO₄²⁻ (aq) + 8OH⁻ (aq) → 2Cr(OH)₄⁻ (aq) + N₂ (g)

Step 1: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide.

1. Write down the reaction: Zn (s) + HNO₃ (aq) → Zn(NO₃)₂ (aq) + N₂O (g) + H₂O (l) 2. Balance the chemical equation: Zn (s) + 2HNO₃ (aq) → Zn(NO₃)₂ (aq) + N₂O (g) + H₂O (l) 3. Write down the net ionic equation: Zn (s) + 2H⁺ (aq) → Zn²⁺ (aq) + H₂ (g)

Step 2: (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide.

1. Write down the reaction: S (s) + HNO₃ (aq) → H₂SO₄ (aq) + NO₂ (g) 2. Balance the chemical equation: S (s) + 6HNO₃ (aq) → H₂SO₄ (aq) + 6NO₂ (g) + 2H₂O (l) 3. Write down the net ionic equation: S (s) + 6H⁺ (aq) + 6NO₃⁻ (aq) → SO₄²⁻ (aq) + 6NO₂ (g) + 4H₂O (l)

Step 3: (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide.

1. Write down the reaction: SO₂ (g) + HNO₃ (aq) → H₂SO₄ (aq) + NO (g) 2. Balance the chemical equation: 2SO₂ (g) + 2HNO₃ (aq) → 2H₂SO₄ (aq) + 2NO (g) 3. Write down the net ionic equation: SO₂ (g) + 2H⁺ (aq) + NO₃⁻ (aq) → SO₄²⁻ (aq) + H₂O (l) + NO (g)

Step 4: (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\).

1. Write down the reaction: N₂H₄ (g) + F₂ (g) → 2NF₃ (g) 2. Balance the chemical equation: N₂H₄ (g) + 6F₂ (g) → 4NF₃ (g) 3. Write down the net ionic equation (There are no spectator ions in this case so the full equation is used): N₂H₄ (g) + 6F₂ (g) → 4NF₃ (g)

Step 5: (e) Hydrazine reduces \(\mathrm{CrO}_{4}{ }^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}{ }^{-}\)in base (hydrazine is oxidized to \(\mathrm{N}_{2}\)).

1. Write down the reaction: N₂H₄ (aq) + CrO₄²⁻ (aq) + OH⁻ (aq) → Cr(OH)₄⁻ (aq) + N₂ (g) 2. Balance the chemical equation: N₂H₄ (aq) + 2CrO₄²⁻ (aq) + 8OH⁻ (aq) → 2Cr(OH)₄⁻ (aq) + N₂ (g) + 4H₂O (l) 3. Write down the net ionic equation: N₂H₄ (aq) + 2CrO₄²⁻ (aq) + 8OH⁻ (aq) → 2Cr(OH)₄⁻ (aq) + N₂ (g)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry. It involves making sure that the number of atoms for each element is the same on both sides of the equation. This principle is grounded in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. When you start balancing, begin by writing down the unbalanced equation.

Here are some tips to make the process easier:

• Identify each component of the reaction, i.e., reactants and products.
• Count the number of atoms of each element in the reactants and the products.
• Adjust the coefficients of the reactants and products to achieve the same number of atoms for each element on both sides.

Remember, you cannot change the subscripts in the chemical formula as it would change the substance itself.

For example, in the process of balancing the reaction between zinc metal and dilute nitric acid, we started with Zn (s) + HNO₃ (aq) → Zn(NO₃)₂ (aq) + N₂O (g) + H₂O (l). Upon balancing, the equation becomes Zn (s) + 2HNO₃ (aq) → Zn(NO₃)₂ (aq) + N₂O (g) + H₂O (l). By doing this, we've ensured that all atoms are accounted for on both sides.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often referred to as redox reactions, are chemical processes where the oxidation state of atoms changes through the transfer of electrons. These reactions are split into two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained.

Understanding redox reactions:

• Oxidizing Agent: The substance that gains electrons (is reduced).
• Reducing Agent: The substance that loses electrons (is oxidized).
• Track the change in oxidation states to identify which element undergoes oxidation and which undergoes reduction.

For instance, when concentrated nitric acid reacts with sulfur, sulfur is oxidized to sulfur dioxide, a process where sulfur changes from a neutral oxidation state to a positive one.

The balanced ionic equation for this example highlights the role each substance plays: S (s) + 6H⁺ (aq) + 6NO₃⁻ (aq) → SO₄²⁻ (aq) + 6NO₂ (g) + 4H₂O (l). Understanding how each component contributes to electron movement is key in mastering redox reactions.

Chemical Reactions in Aqueous Solutions

Chemical reactions in aqueous solutions are reactions that occur with reactants dissolved in water. These solutions are common in a variety of chemical processes because water is a versatile solvent. The interactions can lead to precipitation, displacement, and redox reactions.

Some important aspects to consider:

• Electrolytes: Substances that dissolve in water to form ions, contributing to electrical conductivity.
• Net Ionic Equations: Show only the species that actually participate in the reaction, omitting the spectator ions.
• Identifying strong acids, bases, and salts can often simplify predicting products and writing balanced equations.

For example, in reactions involving hydrazine reducing chromate, you might see a net ionic equation like N₂H₄ (aq) + 2CrO₄²⁻ (aq) + 8OH⁻ (aq) → 2Cr(OH)₄⁻ (aq) + N₂ (g). Here, the emphasis is on the ions actively participating, providing a clearer view of the chemistry taking place in solution.