Consider the redox reaction below: 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) Which statement correctly describes a half-reaction that is taking place? A) Hydrogen is oxidized from +1 to 0. B) Chlorine is reduced from -1 to 0. C) Aluminum is oxidized from 0 to +3. D) Hydrogen is reduced from 0 to -1.
The Correct Answer and Explanation is:
Correct Answer: C) Aluminum is oxidized from 0 to +3.
Explanation:
In the given redox reaction:2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)2Al(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2(g)2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)
Redox reactions involve oxidation (loss of electrons) and reduction (gain of electrons). Let’s analyze the oxidation states of each element to determine what is being oxidized and what is being reduced.
- Aluminum (Al):
- On the reactant side, aluminum is in its elemental form: Al(s). The oxidation state of an element in its pure form is 0.
- On the product side, aluminum is part of the compound AlCl₃. Chlorine (Cl) typically has an oxidation number of -1, and since there are three Cl atoms, the total negative charge is -3. Therefore, Al must have an oxidation state of +3 to balance the -3 from Cl.
- Change in oxidation state for aluminum: 0 → +3
- This indicates oxidation, since aluminum loses electrons.
- Hydrogen (H):
- In HCl, hydrogen has an oxidation number of +1.
- In H₂ gas, hydrogen exists in its elemental form, with an oxidation number of 0.
- Change in oxidation state for hydrogen: +1 → 0
- This indicates reduction, since hydrogen gains electrons.
- Chlorine (Cl):
- Chlorine remains at -1 in both HCl and AlCl₃.
- No change in oxidation state for chlorine, so it is neither oxidized nor reduced.
Thus, the aluminum is oxidized from an oxidation state of 0 to +3, and the hydrogen is reduced from +1 to 0.
Therefore, the correct half-reaction involving oxidation is:Al0→Al3++3e−\text{Al}^0 \rightarrow \text{Al}^{3+} + 3e^-Al0→Al3++3e−
Correct option: C) Aluminum is oxidized from 0 to +3.
