Question

question 3 (1 point)
identify the reducing agent in the reaction:
$\ce{al}(s) + \ce{o_2}(g) \
ightarrow \ce{al_2o_3}(s)$
$\ce{al}(s)$
$\ce{o_2}(g)$
$\ce{al_2o_3}(s)$
$\ce{al}(s)$ and $\ce{o_2}(g)$
there is no reducing agent

Explanation:

To identify the reducing agent, we use the concept of oxidation states. In the reaction \( \text{Al}(s) + \text{O}_2(g)
ightarrow \text{Al}_2\text{O}_3(s) \), the oxidation state of Al in \( \text{Al}(s) \) is 0, and in \( \text{Al}_2\text{O}_3 \), the oxidation state of Al is +3 (since O is -2, and \( 2x + 3(-2) = 0 \) gives \( x = +3 \)). So Al is oxidized (loses electrons). The reducing agent is the substance that is oxidized (donates electrons to reduce another substance). Since Al is oxidized, \( \text{Al}(s) \) is the reducing agent. \( \text{O}_2 \) is reduced (oxidation state of O goes from 0 to -2), so it is the oxidizing agent. \( \text{Al}_2\text{O}_3 \) is the product, not an agent.

Answer:

A. \( \text{Al}(s) \)