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:

Step 1: Recall the definition of an oxidizing agent

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

Step 2: Analyze Reaction A

In \( \ce{2C(s) + O_{2}(g) -> 2CO(g)} \):

• The oxidation state of \( \ce{O} \) in \( \ce{O_{2}} \) is \( 0 \).
• In \( \ce{CO} \), the oxidation state of \( \ce{O} \) is \( -2 \).
• Since the oxidation state of \( \ce{O} \) decreases from \( 0 \) to \( -2 \), \( \ce{O_{2}} \) is reduced and acts as an oxidizing agent.

Step 3: Analyze Reaction B

In \( \ce{S(s) + O_{2}(g) -> SO_{2}(g)} \):

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

Step 4: Analyze Reaction C

In \( \ce{2F_{2}(g) + O_{2}(g) -> 2OF_{2}(g)} \):

• The oxidation state of \( \ce{O} \) in \( \ce{O_{2}} \) is \( 0 \).
• In \( \ce{OF_{2}} \), fluorine has an oxidation state of \( -1 \) (since fluorine is the most electronegative element). Let the oxidation state of \( \ce{O} \) be \( x \). Then, \( x + 2(-1) = 0 \) (since the compound is neutral), so \( x = +2 \).
• The oxidation state of \( \ce{O} \) increases from \( 0 \) to \( +2 \), so \( \ce{O_{2}} \) is oxidized and acts as a reducing agent, not an oxidizing agent.

Step 5: Analyze Reaction D

In \( \ce{2Na(s) + O_{2}(g) -> Na_{2}O_{2}(s)} \):

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

Step 6: Analyze Reaction E

In \( \ce{2Mg(s) + O_{2}(g) -> 2MgO(s)} \):

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

Answer:

C. \( \ce{2F_{2}(g) + O_{2}(g) -> 2OF_{2}(g)} \)