To convert the given redox reaction to its oxidation half-reaction, we first need to identify the species that is being oxidized.
The balanced reaction is:
\[ 2 \text{Al} + 3 \text{Fe}^{3+} \rightarrow 2 \text{Al}^{3+} + 3 \text{Fe} \]
In this reaction, aluminum (Al) is being oxidized to aluminum ions (\(\text{Al}^{3+}\)), and iron ions (\(\text{Fe}^{3+}\)) are being reduced to solid iron (\(\text{Fe}\)).
Step 1: Identify Oxidation
Oxidation involves the loss of electrons. The oxidation half-reaction for aluminum can be written as follows:
Oxidation Half-Reaction:
\[ \text{Al} \rightarrow \text{Al}^{3+} + 3 \text{e}^- \]
Step 2: Balance the half-reaction
For clarity and context, the oxidation half-reaction is balanced in terms of mass and charge:
- Mass Balance: 1 Al on both sides.
- Charge Balance: 0 on the left (neutral Al) and +3 on the right (\(\text{Al}^{3+}\)) and 3 electrons (\(3 \times -1 = -3\)) complete the equation on the right side.
Final Oxidation Half-Reaction:
Thus, the oxidation half-reaction for the given overall reaction is:
\[ 2 \text{Al} \rightarrow 2 \text{Al}^{3+} + 6 \text{e}^- \]
If you multiply the reaction by 2 to match the original balanced equation. This indicates that two moles of aluminum are oxidized, resulting in two moles of aluminum ions and six electrons released in total.