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:

Step1: Define oxidizing agent role

An oxidizing agent gains electrons, so its oxidation number decreases.

Step2: Find O oxidation number in reactants

In $\text{O}_2$, oxidation number of O is $0$.

Step3: Calculate O oxidation number in products for each reaction

Reaction A: $\text{CO}$

O oxidation number: $-2$ (decreases from $0$)

Reaction B: $\text{SO}_2$

O oxidation number: $-2$ (decreases from $0$)

Reaction C: $\text{OF}_2$

F is $-1$, so $x + 2(-1) = 0 \implies x = +2$ (increases from $0$)

Reaction D: $\text{Na}_2\text{O}_2$

Na is $+1$, so $2(+1) + 2x = 0 \implies x = -1$ (decreases from $0$)

Reaction E: $\text{MgO}$

O oxidation number: $-2$ (decreases from $0$)

Step4: Identify non-oxidizing agent case

Only in reaction C, O's oxidation number increases, so $\text{O}_2$ is a reducing agent here.

Answer:

C. $2 \text{F}_2(\text{g}) + \text{O}_2(\text{g})
ightarrow 2 \text{OF}_2(\text{g})$