Problem 15 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Sodium chromate \(\left(\mathrm{Na}_{2} \mathrm{CrO}_{4}\right)\) can be prepared by oxidizing a chromium(lil) salt with sodium peroxide (Na_O_) in alkaline solution. The \(\mathrm{Cr}^{3+}\) ions are oxidized to \(\mathrm{CrO}_{4}^{2-}\) ions. The \(\mathrm{O}_{2}^{2-}\) ions are reduced to OH' ions. Construct half equations and a balanced overall equation for the reaction. (Section \(1.4)\)

Step-by-Step Solution

Answer

2 \( \text{Cr}^{3+} \) + 3 \( \text{O}_2^{2-} \) \( \rightarrow \) 2 \( \text{CrO}_4^{2-} \) + 4 \( \text{OH}^- \)

Step 1: Identify Oxidation and Reduction Half-Reactions

Determine the species being oxidized and reduced. The chromium (Cr^{3+}) is oxidized to CrO_4^{2-} and the peroxide O_2^{2-} is reduced to hydroxide ions (OH^-). This establishes the overall redox change.

Step 2: Write the Oxidation Half-Reaction

For the oxidation process, chromium changes from Cr^{3+} to CrO_4^{2-}. The balanced half-reaction considering the change in oxidation state and addition of water molecules and hydrogen ions is:\[ ext{Cr}^{3+} + 4 ext{OH}^- ightarrow ext{CrO}_4^{2-} + 2 ext{H}_2 ext{O} + 3 ext{e}^-\]

Step 3: Write the Reduction Half-Reaction

For the reduction part, peroxide ions O_2^{2-} are reduced to hydroxide ions:\[ ext{O}_2^{2-} + 2 ext{H}_2 ext{O} + 2 ext{e}^- ightarrow 4 ext{OH}^-\]

Step 4: Balance Electrons in the Half-Reactions

Multiply the oxidation and reduction half-reactions by appropriate coefficients to equalize the number of electrons transferred. The oxidation reaction is multiplied by 2, and the reduction reaction is multiplied by 3 to get:\[2( ext{Cr}^{3+} + 4 ext{OH}^- ightarrow ext{CrO}_4^{2-} + 2 ext{H}_2 ext{O} + 3 ext{e}^-)\]\[3( ext{O}_2^{2-} + 2 ext{H}_2 ext{O} + 2 ext{e}^- ightarrow 4 ext{OH}^-)\]

Step 5: Combine and Simplify

Add the balanced half-reactions to form the overall balanced equation, ensuring all species not involved in the redox change (water and hydroxide ions) balance:\[2 ext{Cr}^{3+} + 8 ext{OH}^- + 3 ext{O}_2^{2-} + 6 ext{H}_2 ext{O} ightarrow 2 ext{CrO}_4^{2-} + 4 ext{H}_2 ext{O} + 12 ext{OH}^-\]

Step 6: Simplify and Finalize the Balanced Equation

Simplify by canceling out identical terms on both sides of the equation to reach the final balanced equation:\[2 ext{Cr}^{3+} + 3 ext{O}_2^{2-} ightarrow 2 ext{CrO}_4^{2-} + 4 ext{OH}^-\]

Key Concepts

Oxidation

In the world of chemistry, oxidation is a crucial process that involves the loss of electrons. This change often modifies the oxidation state of an element, typically increasing it. For example, in our exercise, the chromium ion (\(\mathrm{Cr^{3+}}\)) is oxidized to chromate (\(\mathrm{CrO_4^{2-}}\)).
During oxidation, the element undergoes a transformation in its structure and chemical properties due to electron loss. It's key to remember that oxidation accompanies an increase in the oxidation number, which indicates a gain in positive charge or a decrease in negative charge.
Understanding oxidation, especially in redox reactions, is imperative because:

• It helps us comprehend how different substances interact.
• It plays a vital role in balancing equations and predicting the outcome of a chemical reaction.
• Identifying oxidation in a reaction helps in determining reductions, since they always occur concurrently in redox reactions.

Reduction

Reduction happens alongside oxidation in any redox process, like a friendly dance of electron transfer. In reduction, a substance gains electrons, which means its oxidation state decreases — a key aspect to grasp in understanding chemical behavior.
For the reaction in our example, peroxide ions (\(\mathrm{O_2^{2-}}\)) are reduced to hydroxide ions (\(\mathrm{OH^-}\)).
Reduction is significant because:

• It provides a method to convert different ions, impacting how substances are structured and behave.
• Reduced forms of elements are typically more stable than their oxidized counterparts, affecting reaction dynamics.
• Recognizing reduction in reactions aids in predicting and balancing chemical equations, making it a cornerstone of chemistry education.

Balancing Equations

Balancing equations is like solving a puzzle in chemistry, ensuring both sides of a chemical equation reflect the same amount of substances. It's a fundamental skill to master for predicting and understanding chemical reactions.
In the problem provided, balancing involves first writing half-reactions for oxidation and reduction, then ensuring the same number of electrons are lost and gained.
Here's a concise strategy to tackle this:

• Write separate half-reactions for oxidation and reduction.
• Balance the atoms and charges in each half-reaction.
• Make sure the number of electrons lost equals those gained.
• Combine the half-reactions, canceling out identical species on both sides.

Balancing chemical equations is critical because it conserves mass and charge. This consistency is vital for precisely predicting how much of a substance will react with another.

Chemistry Education

In chemistry education, understanding redox reactions strengthens foundational knowledge crucial for many scientific disciplines. Redox reactions underlie many processes, like metabolic pathways in biology, industrial chemistry applications, and even environmental science.
Effective chemistry education involves not merely memorizing formulas but understanding connections between chemical processes and everyday applications.
Key aspects to emphasize in chemistry education for mastering these concepts include:

• Encouraging problem-solving skills by practicing writing and balancing redox equations.
• Incorporating real-world examples, such as rust formation or battery function, to illustrate redox concepts.
• Teaching the logic and reasoning behind steps taken to balance equations, not just the steps themselves.

This approach helps students grasp abstract concepts like electron transfer, relate them to familiar scenarios, and prepares them for advanced studies in chemistry and related fields.