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Mobile App Chapter 21: Problem 139
Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution: $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{Cr}^{3+}$$
Short Answer
Expert verified
3CH\textsubscript{3}CH\textsubscript{2}OH + 2Cr\textsubscript{2}O\textsubscript{7}\textsuperscript{2-} + 28H\textsuperscript{+} → 3CH\textsubscript{3}COOH + 4Cr\textsuperscript{3+} + 14H\textsubscript{2}O
Step by step solution
01
Write the Oxidation Half-Reaction
Identify the species being oxidized and write the oxidation half-reaction. Ethanol (CH\textsubscript{3}CH\textsubscript{2}OH) is being oxidized to acetic acid (CH\textsubscript{3}COOH).\[\text{CH}_{3}\text{CH}_{2}\text{OH} \rightarrow \text{CH}_{3}\text{COOH} + 4\text{H}^{+} + 4e^{-}\]
02
Write the Reduction Half-Reaction
Identify the species being reduced. Dichromate (\text{Cr}_2\text{O}_7^{2-}) is being reduced to \text{Cr}^{3+}.\[\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} + 6e^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\]
03
Balance Electrons in Both Half-Reactions
Ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.\[3(\text{CH}_{3}\text{CH}_{2}\text{OH} \rightarrow \text{CH}_{3}\text{COOH} + 4\text{H}^{+} + 4e^{-})\]\[2(\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} + 6e^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O})\]Which become:\[3\text{CH}_{3}\text{CH}_{2}\text{OH} \rightarrow 3\text{CH}_{3}\text{COOH} + 12\text{H}^{+} + 12e^{-}\]\[2\text{Cr}_{2}\text{O}_{7}^{2-} + 28\text{H}^{+} + 12e^{-} \rightarrow 4\text{Cr}^{3+} + 14\text{H}_2\text{O}\]
04
Add the Half-Reactions Together
Add the two balanced half-reactions together, canceling out the electrons and any other species that appear on both sides.\[3\text{CH}_{3}\text{CH}_{2}\text{OH} + 2\text{Cr}_{2}\text{O}_{7}^{2-} + 28\text{H}^{+} \rightarrow 3\text{CH}_{3}\text{COOH} + 4\text{Cr}^{3+} + 14\text{H}_2\text{O}\]
05
Verify the Balanced Equation
Ensure that both the atoms and the charges are balanced in the final equation. Both sides of the equation should contain the same number of each type of atom and the sum of the charges should be equal.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
oxidation half-reaction
In a redox reaction, the oxidation half-reaction is where a substance loses electrons. Oxidation involves an increase in oxidation state.
For the given exercise, ethanol (CH\textsubscript{3}CH\textsubscript{2}OH) undergoes oxidation to become acetic acid (CH\textsubscript{3}COOH).
When writing the oxidation half-reaction, we identify the species that loses electrons and show the process of electron loss.
The half-reaction includes charges to indicate the electrons involved: ewline \[ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \]
For the given exercise, ethanol (CH\textsubscript{3}CH\textsubscript{2}OH) undergoes oxidation to become acetic acid (CH\textsubscript{3}COOH).
When writing the oxidation half-reaction, we identify the species that loses electrons and show the process of electron loss.
The half-reaction includes charges to indicate the electrons involved: ewline \[ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \]
reduction half-reaction
Reduction is the gain of electrons by a molecule, atom, or ion. This process decreases the oxidation state of the species.
In this exercise, dichromate (\text{Cr}\textsubscript{2}\text{O}\textsubscript{7}\textsuperscript{2-}) is reduced to chromium ions (\text{Cr}\textsuperscript{3+}).
To write the reduction half-reaction, identify the species gaining electrons and show this electron gain. Here is the equation: ewline \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]
In this exercise, dichromate (\text{Cr}\textsubscript{2}\text{O}\textsubscript{7}\textsuperscript{2-}) is reduced to chromium ions (\text{Cr}\textsuperscript{3+}).
To write the reduction half-reaction, identify the species gaining electrons and show this electron gain. Here is the equation: ewline \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]
acidic solution
When balancing redox reactions in an acidic solution, we need to add H\textsuperscript{+} ions to help balance hydrogen atoms.
These hydrogen ions are used to balance the half-reactions.
For instance, in the reduction half-reaction: ewline \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \] the 14 H\textsuperscript{+} ions balance the oxygens with water molecules on the product side.
These hydrogen ions are used to balance the half-reactions.
For instance, in the reduction half-reaction: ewline \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \] the 14 H\textsuperscript{+} ions balance the oxygens with water molecules on the product side.
electron balance
Balancing electrons is crucial in redox reactions to ensure charge conservation between oxidation and reduction half-reactions.
Each electron lost by one reactant must be gained by another.
In our exercise, the oxidation half-reaction loses 4 electrons, and the reduction half-reaction gains 6 electrons: \[ \begin{align*} \text{Oxidation:} & \quad \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \ \text{Reduction:} & \quad \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \ \end{align*} \]
We multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons: \[ \begin{align*} 3\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} & \rightarrow 3\mathrm{CH}_{3}\mathrm{COOH} + 12\mathrm{H}^{+} + 12e^{-} \ 2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 28\mathrm{H}^{+} + 12e^{-} & \rightarrow 4\mathrm{Cr}^{3+} + 14\mathrm{H}_2\mathrm{O} \ \end{align*} \]
Each electron lost by one reactant must be gained by another.
In our exercise, the oxidation half-reaction loses 4 electrons, and the reduction half-reaction gains 6 electrons: \[ \begin{align*} \text{Oxidation:} & \quad \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \ \text{Reduction:} & \quad \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \ \end{align*} \]
We multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons: \[ \begin{align*} 3\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} & \rightarrow 3\mathrm{CH}_{3}\mathrm{COOH} + 12\mathrm{H}^{+} + 12e^{-} \ 2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 28\mathrm{H}^{+} + 12e^{-} & \rightarrow 4\mathrm{Cr}^{3+} + 14\mathrm{H}_2\mathrm{O} \ \end{align*} \]
dichromate reduction
Dichromate (\text{Cr}\textsubscript{2}\text{O}\textsubscript{7}\textsuperscript{2-}) reduction is an important process in balancing redox reactions.
Dichromate ions are reduced to Cr\textsuperscript{3+} ions.
In our reaction, dichromate accepts electrons during this reduction process: ewline \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]
I'm going to add water molecules (H\textsubscript{2}O) to balance the oxygens.
Dichromate ions are reduced to Cr\textsuperscript{3+} ions.
In our reaction, dichromate accepts electrons during this reduction process: ewline \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]
I'm going to add water molecules (H\textsubscript{2}O) to balance the oxygens.
ethanol oxidation
Ethanol oxidation is the process where ethanol (\text{CH}\textsubscript{3}\text{CH}\textsubscript{2}\text{OH}) is converted to acetic acid (\text{CH}\textsubscript{3}\text{COOH}).
This involves the loss of electrons and is shown in the half-reaction: ewline \[ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \]
Here, the ethanol molecule loses 4 electrons, and the oxidation state of carbon increases from -1 in ethanol to +3 in acetic acid.
This involves the loss of electrons and is shown in the half-reaction: ewline \[ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \]
Here, the ethanol molecule loses 4 electrons, and the oxidation state of carbon increases from -1 in ethanol to +3 in acetic acid.
