Question

In which of the following reactions does the oxidation number of aluminum change?

1) 2Al(OH)3 → Al₂O₃ + 3H₂O
2) Al₂O₃ + 2NaOH +3H₂O → 2 NaAl(OH)4
3) NaAl(OH)4 → Al(OH)3 +NaOH
4) none

Answer 1

None of the given reactions involve a change in the oxidation number of aluminum; it remains +3 in all reactants and products.

Step-by-step explanation:

In the reactions presented, the oxidation number of aluminum changes in the reaction where aluminum is oxidized. This is seen in the formula Al becoming Al³⁺ + 3e⁻, indicating a loss of three electrons, hence an increase in its oxidation state from 0 to +3. To find out where the change occurs, we can analyze each reaction:

• 2Al(OH)3 → Al₂O₃ + 3H₂O: Aluminum's oxidation state does not change, it remains +3 in both the reactant and the product.
• Al₂O₃ + 2NaOH +3H₂O → 2 NaAl(OH)4: Aluminum goes from an oxidation state of +3 in Al₂O₃ to an oxidation state of +3 in NaAl(OH)4, so there is no change.
• NaAl(OH)4 → Al(OH)3 +NaOH: Again, aluminum maintains an oxidation state of +3, therefore there is no change.

So the answer to the student's question is none of these reactions involve a change in the oxidation number of aluminum.