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:

Step1: Recall the definition of an oxidizing agent

An oxidizing agent is a substance that gets reduced (its oxidation state decreases) during a reaction.

Step2: Analyze Reaction A

In \( 2C(s) + O_2(g)
ightarrow 2CO(g) \):

• The oxidation state of \( O \) in \( O_2 \) is \( 0 \).
• In \( CO \), the oxidation state of \( O \) is \( -2 \).
• Since the oxidation state of \( O \) decreases from \( 0 \) to \( -2 \), \( O_2 \) is reduced, so it acts as an oxidizing agent.

Step3: Analyze Reaction B

In \( S(s) + O_2(g)
ightarrow SO_2(g) \):

• The oxidation state of \( O \) in \( O_2 \) is \( 0 \).
• In \( SO_2 \), the oxidation state of \( O \) is \( -2 \).
• The oxidation state of \( O \) decreases from \( 0 \) to \( -2 \), so \( O_2 \) is reduced and acts as an oxidizing agent.

Step4: Analyze Reaction C

In \( 2F_2(g) + O_2(g)
ightarrow 2OF_2(g) \):

• The oxidation state of \( O \) in \( O_2 \) is \( 0 \).
• In \( OF_2 \), fluorine has an oxidation state of \( -1 \) (since fluorine is the most electronegative element). Let the oxidation state of \( O \) be \( x \). Then, \( x + 2(-1) = 0 \) (since the overall charge of \( OF_2 \) is \( 0 \)). Solving for \( x \), we get \( x = +2 \).
• The oxidation state of \( O \) increases from \( 0 \) to \( +2 \), so \( O_2 \) is oxidized (not reduced). Thus, \( O_2 \) is not acting as an oxidizing agent here.

Step5: Analyze Reaction D

In \( 2Na(s) + O_2(g)
ightarrow Na_2O_2(s) \):

• The oxidation state of \( O \) in \( O_2 \) is \( 0 \).
• In \( Na_2O_2 \), the oxidation state of \( O \) is \( -1 \) (since \( Na \) has an oxidation state of \( +1 \), and \( 2(+1) + 2x = 0 \) gives \( x = -1 \)).
• The oxidation state of \( O \) decreases from \( 0 \) to \( -1 \), so \( O_2 \) is reduced and acts as an oxidizing agent.

Step6: Analyze Reaction E

In \( 2Mg(s) + O_2(g)
ightarrow 2MgO(s) \):

• The oxidation state of \( O \) in \( O_2 \) is \( 0 \).
• In \( MgO \), the oxidation state of \( O \) is \( -2 \).
• The oxidation state of \( O \) decreases from \( 0 \) to \( -2 \), so \( O_2 \) is reduced and acts as an oxidizing agent.

Answer:

C. \( 2F_2(g) + O_2(g)
ightarrow 2OF_2(g) \)