To determine which statement correctly describes a half-reaction in the given redox reaction, we need to examine the changes in oxidation states of the elements involved.
The overall reaction is:
[tex]\[ 2 \text{Al} (s) + 6 \text{HCl} (aq) \longrightarrow 2 \text{AlCl}_3 (aq) + 3 \text{H}_2 (g) \][/tex]
Let's break down the oxidation states of each element on both sides of the reaction.
### Reactants
- Aluminum (Al): In its elemental form, the oxidation state is 0.
- Hydrogen (H) in HCl: The oxidation state of hydrogen in HCl is +1.
- Chlorine (Cl) in HCl: The oxidation state of chlorine in HCl is -1.
### Products
- Aluminum (Al) in AlCl[tex]\(_3\)[/tex]: In [tex]\(\text{AlCl}_3\)[/tex], the oxidation state of aluminum is +3.
- Chlorine (Cl) in AlCl[tex]\(_3\)[/tex]: The oxidation state of chlorine in [tex]\(\text{AlCl}_3\)[/tex] is -1 (as conformable with the formula AlCl[tex]\(_3\)[/tex], where [tex]\( \text{Al} \)[/tex] is +3 and 3 chlorines sum up to -3).
- Hydrogen (H) in H[tex]\(_2\)[/tex]: In [tex]\(\text{H}_2\)[/tex], the oxidation state of hydrogen is 0.
### Analysis
1. Aluminum (Al):
- Changes from 0 in [tex]\( \text{Al} (s) \)[/tex] to +3 in [tex]\(\text{AlCl}_3 (aq)\)[/tex].
- This indicates that aluminum is oxidized (loss of electrons).
[tex]\[ \text{Half-reaction: } \text{Al} \rightarrow \text{Al}^{3+} + 3e^-. \][/tex]
3. Hydrogen (H):
- Changes from +1 in [tex]\(\text{HCl} (aq)\)[/tex] to 0 in [tex]\(\text{H}_2 (g)\)[/tex].
- This indicates that hydrogen is reduced (gain of electrons).
[tex]\[ \text{Half-reaction: } 2 \text{H}^+ + 2e^- \rightarrow \text{H}_2. \][/tex]
3. Chlorine (Cl):
- Remains at -1 in both [tex]\(\text{HCl} (aq)\)[/tex] and [tex]\(\text{AlCl}_3 (aq)\)[/tex].
- Chlorine does not change its oxidation state.
From the analysis, the correct statement describing a half-reaction correctly is:
[tex]\[ \text{Aluminum is oxidized from 0 to +3.} \][/tex]
Thus, the correct statement is:
[tex]\[ \text {Aluminum is oxidized from 0 to +3.}. \][/tex]
