Problem 85 - Chemistry and Chemical Reactivity

Question

A Balance the following equations involving organic compounds. $$\begin{aligned} &\text { (a) } \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{aq}) \rightarrow\\\ &&\mathrm{Ag}(\mathrm{s})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \quad \text { (acid solution) } \end{aligned}$$ $$\begin{aligned} &\text { (b) } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \rightarrow\\\ &&\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \quad+\mathrm{Cr}^{3+}(\mathrm{aq}) \quad \text { (acid solution) } \end{aligned}$$

Step-by-Step Solution

Answer

Reactions are balanced as follows: (a) \( 2\mathrm{Ag}^{+} + \mathrm{C}_6\mathrm{H}_5\mathrm{CHO} + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Ag} + \mathrm{C}_6\mathrm{H}_5\mathrm{CO}_2\mathrm{H} + 2\mathrm{H}^{+} \). (b) \(3\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{Cr}_2\mathrm{O}_7^{2-} + 8\mathrm{H}^{+} \rightarrow 3\mathrm{CH}_3\mathrm{CO}_2\mathrm{H} + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \).

Step 1: Identify Oxidation and Reduction Components (Reaction a)

In the reaction \( \mathrm{Ag}^{+} + \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \rightarrow \mathrm{Ag} + \mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}_2 \mathrm{H} \), \( \mathrm{Ag}^{+} \) is reduced to \( \mathrm{Ag} \), and \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \) is oxidized to \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}_2 \mathrm{H} \).

Step 2: Balance Atoms and Charges (Reaction a)

Each silver ion, \( \mathrm{Ag}^{+} \), gains one electron to form \( \mathrm{Ag} \) solid, so there is no need for further balancing. Then, add water to balance oxygen atoms and hydrogen ions to balance hydrogen atoms: \[2\mathrm{Ag}^{+} + \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Ag} + \mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}_2\mathrm{H} + 2\mathrm{H}^{+}\]

Step 3: Identify Oxidation and Reduction Components (Reaction b)

For \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} + \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H} + \mathrm{Cr}^{3+} \), ethanol (\( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \)) is oxidized to acetic acid (\( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H} \)), and \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \) is reduced to \( \mathrm{Cr}^{3+} \).

Step 4: Balance Atoms (Reaction b)

Balance the chromate ions by ensuring chromium atoms are the same on both sides.Start by making sure there are two chromium atoms on each side: \[\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+}\]

Step 5: Balance Charges and Add Electrons (Reaction b)

Add electrons to balance the equation. The dichromate ion gains 6 electrons:\[\mathrm{Cr}_2 \mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\]

Step 6: Balance Ethanol and Acetic Acid (Reaction b)

Oxidize \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \) to \( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H} \) by balancing the carbon and hydrogen atoms:\[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \rightarrow \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H} + 2\mathrm{H}^+ + 2e^-\]

Step 7: Combine and Balance Reaction b Overall

To balance the overall equation, multiply the oxidation equation by 3 to match with 6 electrons required by the reduction equation:\[3\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} + \mathrm{Cr}_2 \mathrm{O}_7^{2-} + 8\mathrm{H}^+\rightarrow 3\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H} + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2 \mathrm{O}\]

Step 8: Verify Mass and Charge Balance

Ensure that the number of atoms for each element and the total charge is balanced on both sides of each reaction.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between molecules. A defining feature of these reactions is the change in oxidation states of the reactants. Reduction refers to the gain of electrons, while oxidation means the loss of electrons. When we discuss redox reactions, we're exploring both these processes occurring simultaneously.

A unique aspect of redox reactions is their occurrence in pairs, meaning that for one species to be oxidized, another must be reduced. This transfer of electrons is what we use to determine the oxidation and reduction components of a chemical reaction. Understanding this principle is crucial when balancing redox equations, as it ensures both the conservation of mass and charge within the chemical equation.

Oxidation and Reduction

Oxidation and reduction are complementary processes that are central to redox reactions. In simple terms, oxidation involves the loss of electrons from a molecule. Conversely, reduction is the gain of electrons by a molecule. When we say a substance is oxidized, it means there's an increase in its oxidation state. Likewise, when a substance is reduced, its oxidation state decreases. These changes help balance chemical equations, especially when organic compounds are involved.

In our given exercises,

• reaction (a) involves the oxidation of benzaldehyde (\( \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \)) to form benzoic acid (\( \mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}_2 \mathrm{H} \)). Silver ions (\( \mathrm{Ag}^+ \)) are reduced to form silver metal (\( \mathrm{Ag} \)).
• In reaction (b), ethanol (\( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \)) is oxidized to acetic acid (\( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H} \)), and dichromate ions (\( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \)) are reduced to chromium (\( \mathrm{Cr}^{3+} \)).

By following these electron transfers, we can systematically balance the chemical equations.

Organic Chemistry

Organic chemistry focuses on the study of carbon-containing compounds, which are the building blocks of life. This branch of chemistry explains reactions like oxidation and reduction in the context of organic molecules. Understanding how different organic molecules react and transform into one another is essential for balancing equations in organic chemical reactions.

Ethanol and benzaldehyde, which appear in our exercises, are standard organic compounds that undergo redox reactions. The purpose of balancing such chemical equations is to ensure mass and charge are conserved. This often involves accounting for changes in both carbon-based molecules and auxiliary atoms like oxygen and hydrogen during the reaction. In organic chemistry, knowing how to manipulate these factors allows chemists to predict reaction products and design new reactions.

Chromate Ions

Chromate ions, represented as \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \), play a significant role in redox chemistry. They are strong oxidizing agents, meaning they have a high potential to accept electrons from other molecules. This property makes them useful in oxidizing organic compounds.

During redox reactions, chromate ions undergo reduction, transitioning from a higher oxidation state (as part of dichromate) to a lower one, typically forming chromium ions (\( \mathrm{Cr}^{3+} \)). The dichromate ion's ability to facilitate this electron transfer makes it notable in the oxidation of alcohols to carboxylic acids, such as the transformation of ethanol into acetic acid.

In the context of chemical equations, chromate ions not only help drive these reactions but also help balance the electron flow. Because they accept multiple electrons, they influence how we balance electron transfer equations to ensure stoichiometric accuracy.