Question

U t3

Given the reaction: 4 Al(s) + 30₂(g) → 2Al₂O3(s)
31 Write the balanced oxidation half-reaction for this
oxidation-reduction reaction.
32 What is the oxidation number of oxygen in Al₂O3?
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Answer 1

31. To write the balance oxidation half-reaction, we need to determine which element is undergoing oxidation (losing electrons). In this reaction, aluminum (Al) is being oxidized because its oxidation state changes from 0 to +3. The oxidation half-reaction is:

4 Al(s) → 4 Al³⁺(aq) + 12 e⁻

Note that we balance the number of electrons transferred by multiplying the half-reaction by 3.

32. In Al₂O3, aluminum has an oxidation state of +3. The total charge of the compound is 0. Therefore, the oxidation state of oxygen can be calculated as:

2(oxidation state of Al) + 3(oxidation state of O) = 0

2(+3) + 3(x) = 0

6 + 3x = 0

3x = -6

x = -2

the oxidation number of oxygen in Al₂O3 is: -2.