Question

oxygen is acting as an oxidizing agent in all of the following reactions except a. $2 c(s) + o_2(g) \
ightarrow 2 co(g)$ b. $s(s) + o_2(g) \
ightarrow so_2(g)$ c. $2 f_2(g) + o_2(g) \
ightarrow 2 of_2(g)$ d. $2 na(s) + o_2(g) \
ightarrow na_2o_2(s)$ e. $2 mg(s) + o_2(g) \
ightarrow 2 mgo(s)$

Explanation:

To determine in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation states of oxygen in each reaction. An oxidizing agent is reduced (its oxidation state decreases).

Step 1: Analyze Reaction A

In \( \ce{2C(s) + O2(g) -> 2CO(g)} \), the oxidation state of \( \ce{O} \) in \( \ce{O2} \) is \( 0 \), and in \( \ce{CO} \) it is \( -2 \). The oxidation state of \( \ce{O} \) decreases, so \( \ce{O2} \) is an oxidizing agent.

Step 2: Analyze Reaction B

In \( \ce{S(s) + O2(g) -> SO2(g)} \), the oxidation state of \( \ce{O} \) in \( \ce{O2} \) is \( 0 \), and in \( \ce{SO2} \) it is \( -2 \). The oxidation state of \( \ce{O} \) decreases, so \( \ce{O2} \) is an oxidizing agent.

Step 3: Analyze Reaction C

In \( \ce{2F2(g) + O2(g) -> 2OF2(g)} \), the oxidation state of \( \ce{O} \) in \( \ce{O2} \) is \( 0 \), and in \( \ce{OF2} \) it is \( +2 \). The oxidation state of \( \ce{O} \) increases, so \( \ce{O2} \) is oxidized (acting as a reducing agent), not an oxidizing agent.

Step 4: Analyze Reaction D

In \( \ce{2Na(s) + O2(g) -> Na2O2(s)} \), the oxidation state of \( \ce{O} \) in \( \ce{O2} \) is \( 0 \), and in \( \ce{Na2O2} \) (peroxide) it is \( -1 \). The oxidation state of \( \ce{O} \) decreases, so \( \ce{O2} \) is an oxidizing agent.

Step 5: Analyze Reaction E

In \( \ce{2Mg(s) + O2(g) -> 2MgO(s)} \), the oxidation state of \( \ce{O} \) in \( \ce{O2} \) is \( 0 \), and in \( \ce{MgO} \) it is \( -2 \). The oxidation state of \( \ce{O} \) decreases, so \( \ce{O2} \) is an oxidizing agent.

Answer:

C. \( \ce{2F2(g) + O2(g) -> 2OF2(g)} \)