Problem 113 - Chemistry The Molecular Science

Question

In the past, devices for testing a driver's breath for alcohol depended on this reaction: $$\begin{array}{r} 3 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq})+2 \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+8 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \\ 3 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+2 \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})+ \\ 2 \mathrm{~K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+11 \mathrm{H}_{2} \mathrm{O}(\ell) \end{array}$$ (a) Write the net ionic equation for this reaction. (b) What oxidation numbers are changing in the course of this reaction? (c) Which substances are being oxidized and reduced? (d) Which substance is the oxidizing agent and which is the reducing agent?

Step-by-Step Solution

Answer

(a) Net ionic: \(3 \text{C}_2\text{H}_5\text{OH} + 2 \text{Cr}_2\text{O}_7^{2-} + 16 \text{H}^+ \rightarrow 3 \text{CH}_3\text{COOH} + 4 \text{Cr}^{3+} + 11 \text{H}_2\text{O}\). (b) C: -1 to +3; Cr: +6 to +3. (c) Ethanol oxidized, \(\text{Cr}_2\text{O}_7^{2-}\) reduced. (d) \(\text{Cr}_2\text{O}_7^{2-}\) oxidizing agent, ethanol reducing agent.

Step 1: Write the Complete Ionic Equation

To determine the net ionic equation, first write the complete ionic equation by identifying which compounds dissociate into ions in the solution. In this reaction, \(\text{C}_2\text{H}_5\text{OH}\) and \(\text{CH}_3\text{COOH}\) do not dissociate, while \(\text{K}_2\text{Cr}_2\text{O}_7\), \(\text{H}_2\text{SO}_4\), \(\text{Cr}_2(\text{SO}_4)_3\), and \(\text{K}_2\text{SO}_4\) dissociate into ions.

Step 2: Cancel Spectator Ions

Identify and cancel out the spectator ions from both sides of the complete ionic equation. The spectator ions, which are not involved in the reaction, are \(\text{K}^+\) and \(\text{SO}_4^{2-}\). Remove them from both sides to form the net ionic equation.

Step 3: Write the Net Ionic Equation

After cancelling out the spectator ions, write the net ionic equation. By simplifying, the net ionic equation is: \[3 \text{C}_2\text{H}_5\text{OH} + 2 \text{Cr}_2\text{O}_7^{2-} + 16 \text{H}^+ \rightarrow 3 \text{CH}_3\text{COOH} + 4 \text{Cr}^{3+} + 11 \text{H}_2\text{O}\]

Step 4: Determine Oxidation Number Changes

Determine the oxidation numbers of the atoms in the reactants and products to see which elements change oxidation states. In \(\text{C}_2\text{H}_5\text{OH}\), carbon changes from -1 to +3, and in \(\text{Cr}_2\text{O}_7^{2-}\), chromium changes from +6 to +3.

Step 5: Identify Oxidized and Reduced Substances

The substance being oxidized is the one whose oxidation number increases. Here, ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) is oxidized. The substance being reduced is the one whose oxidation number decreases. Here, \(\text{Cr}_2\text{O}_7^{2-}\) is reduced.

Step 6: Determine Oxidizing and Reducing Agents

The oxidizing agent is the substance that gets reduced, and the reducing agent is the substance that gets oxidized. In this reaction, \(\text{Cr}_2\text{O}_7^{2-}\) is the oxidizing agent because it gains electrons, and \(\text{C}_2\text{H}_5\text{OH}\) is the reducing agent because it loses electrons.

Key Concepts

Oxidation Numbers

Oxidation numbers are a helpful way to keep track of electrons in chemical reactions. They represent the charge an atom would have if the compound was ionic. Think of them like a bookkeeping tool.
In the given reaction, ethanol \((\text{C}_2\text{H}_5\text{OH})\) and dichromate \((\text{Cr}_2\text{O}_7^{2-})\) are central to understanding the changes. For ethanol, carbon has an oxidation number of \-1\, and it changes to \(+3\) in acetic acid \((\text{CH}_3\text{COOH})\).
Meanwhile, chromium in dichromate goes from +6 to +3. These changes tell us which atoms are involved in electron transfer.

• Carbon in ethanol: \-1\ to \(+3\)
• Chromium in dichromate: \(+6\) to \(+3\)

By identifying these changes, we can understand which atoms are oxidized and reduced.

Redox Reactions

Redox reactions are processes where reduction and oxidation occur simultaneously. They're essential in many chemical and biological systems.
In a redox reaction, one substance loses electrons (oxidation), and another gains them (reduction). The oxidation number changes are a clear indication of these processes. In our reaction, the increase in the oxidation number of carbon from \-1\ to \(+3\) in ethanol shows oxidation. Conversely, the decrease from +6 to +3 in chromium reflects reduction.

• Oxidation: Ethanol \((\text{C}_2\text{H}_5\text{OH})\) loses electrons, carbon's oxidation number increases.
• Reduction: \((\text{Cr}_2\text{O}_7^{2-})\) gains electrons, chromium's oxidation number decreases.

Understanding these changes helps identify the reaction components.

Oxidizing and Reducing Agents

Oxidizing and reducing agents are crucial in redox reactions.
The oxidizing agent is the substance that gets reduced. It's the one that accepts electrons. In this case, dichromate \((\text{Cr}_2\text{O}_7^{2-})\) acts as the oxidizing agent. It undergoes reduction as its chromium atoms gain electrons.
The reducing agent, on the other hand, is the substance that gets oxidized. It's the one that donates electrons. Ethanol \((\text{C}_2\text{H}_5\text{OH})\) is the reducing agent. It loses electrons as its carbon gets oxidized.

• Oxidizing Agent: Dichromate \((\text{Cr}_2\text{O}_7^{2-})\)
• Reducing Agent: Ethanol \((\text{C}_2\text{H}_5\text{OH})\)

Recognizing these agents is key to predicting how redox reactions will proceed.