Problem 35 - General Chemistry: Principles and Modern Applications

Question

Complete and balance these half-equations. (a) \(\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (acidic solution) (b) \(\mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (acidic solution) (c) \(\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}\) (basic solution) Indicate whether oxidation or reduction is involved.

Step-by-Step Solution

Answer

a) \( \mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} + 2\mathrm{H}^{+} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + 2e^{-}\), oxidation \nb) \( \mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O} + \mathrm{H}_{2}\mathrm{O} + \mathrm{H}^{+} + 2e^{-}\), oxidation \nc) \( \mathrm{Al} + 4\mathrm{OH}^{-} + 3e^{-} \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}\), reduction

Step 1: Balancing the atoms other than H and O

(a) Both sides already have the same number of S atoms. \n(b) Both sides already have the same number of N atoms. \n(c) Both sides already have the same number of Al atoms.

Step 2: Balancing the oxygen atoms by adding waters

(a) Add 1 H2O on the left. Now we have \( \mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) \n(b) Add 1 H2O on the right. Now we have \( \mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O} + \mathrm{H}_{2}\mathrm{O} \) \n(c) There are already equal O atoms on both sides.

Step 3: Balancing the hydrogen atoms by adding protons or hydroxides

(a) As this is an acidic solution, add 2 H+ on the left. Now we have \( \mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} +2\mathrm{H}^{+} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) \n(b) As this is an acidic solution, add 1 H+ on the right. Now we have \( \mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O} + \mathrm{H}_{2}\mathrm{O} + \mathrm{H}^{+} \) \n(c) As this is a basic solution, add 4 OH- on the left. Now we have \( \mathrm{Al} + 4\mathrm{OH}^{-} \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}\)

Step 4: Balancing the charges by adding electrons

(a) Add 2 electrons on the right. The final balanced reaction is \( \mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} +2\mathrm{H}^{+} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + 2e^{-}\) \n(b) Add 2 electrons on the right. The final balanced reaction is \( \mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O} + \mathrm{H}_{2}\mathrm{O} + \mathrm{H}^{+} + 2e^{-} \) \n(c) Add 3 electrons on the left. The final balanced reaction is \( \mathrm{Al} + 4\mathrm{OH}^{-} + 3e^{-} \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}\)

Step 5: Determine whether oxidation or reduction

(a) As 2 electrons are produced, it's an oxidation. \n(b) As 2 electrons are produced, it's an oxidation. \n(c) As 3 electrons are consumed, it's a reduction.

Key Concepts

Oxidation-reduction

Oxidation-reduction reactions, or redox reactions, are chemical reactions that involve a transfer of electrons between two species. These reactions are fundamental in chemistry because they are part of many processes such as energy production, corrosion, and cellular respiration.

• Oxidation: This occurs when a substance loses electrons. In the exercise, oxidation is seen in the first two half-equations, where the reactants produce electrons.
• Reduction: This is the gain of electrons. In the final half-equation, where electrons are consumed, reduction takes place.

These processes always occur simultaneously — while one species undergoes oxidation, another undergoes reduction. This is important because it ensures the conservation of charge. The electrons lost by one species are gained by another, maintaining overall electron balance in the reaction.

Acidic and basic solutions

Understanding how to balance redox reactions in acidic and basic solutions is essential, as the medium in which a reaction occurs greatly influences the balancing method. Acidic solutions contain a higher concentration of hydrogen ions ( H^+ ) due to the presence of an acid, while basic solutions have more hydroxide ions ( OH^- ) typically due to a base.

• In acidic solutions, after balancing oxygen atoms with water, we balance hydrogen atoms by adding H^+ ions.
• Conversely, in basic solutions, hydrogen atoms are balanced by adding OH^- ions instead.

These ions help to balance the charges in half-equations. They can also influence the formation of products and are crucial in maintaining the reaction conditions appropriate for the substances involved. This step is critical in ensuring that all reactants and products react appropriately within their specific environment.

Redox reactions

Redox reactions are ever-present and vital processes both in the chemical industry and in everyday life. They include any chemical reaction in which the oxidation state of a substance changes. The balancing of redox equations, like the ones provided, allows us to predict how substances interact and transform.

• To balance a redox half-equation, follow these steps: balance atoms other than hydrogen and oxygen first, then balance oxygen with water, hydrogen with H^+ or OH^- , and finally balance charges with electrons.
• The electrons added in the final step (either on the reactants or products side) will help identify whether a half-equation involves oxidation or reduction, as seen in the original exercise.

By mastering these principles, you can solve complex chemical puzzles, making redox reactions an invaluable tool in scientific exploration and technological advancement. These steps encapsulate the systematic approach essential for successfully teaching and understanding chemistry.