91Ó°ÊÓ

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 \[ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{COOH} + 4\mathrm{H}^{+} + 4e^{-} \]

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 \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]

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.

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: \[ \begin{align*} 3\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} & \rightarrow 3\mathrm{CH}_{3}\mathrm{COOH} + 12\mathrm{H}^{+} + 12e^{-} \ 2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 28\mathrm{H}^{+} + 12e^{-} & \rightarrow 4\mathrm{Cr}^{3+} + 14\mathrm{H}_2\mathrm{O} \ \end{align*} \]

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.

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.