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) in a redox reaction.

Step 2: Analyze Reaction A: \( 2C(s) + O_2(g)

ightarrow 2CO(g) \)

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

Step 3: Analyze Reaction B: \( S(s) + O_2(g)

ightarrow SO_2(g) \)

• In \( O_2 \), \( O \) has an oxidation state of \( 0 \).
• In \( SO_2 \), \( O \) has an oxidation state of \( -2 \).
• Oxidation state of \( O \) decreases, so \( O_2 \) is reduced (oxidizing agent).

Step 4: Analyze Reaction C: \( 2F_2(g) + O_2(g)

ightarrow 2OF_2(g) \)

• In \( O_2 \), \( O \) has an oxidation state of \( 0 \).
• In \( OF_2 \), \( F \) is more electronegative than \( O \), so \( F \) has an oxidation state of \( -1 \). Let the oxidation state of \( O \) be \( x \). For \( OF_2 \): \( x + 2(-1) = 0 \Rightarrow x = +2 \).
• The oxidation state of \( O \) increases from \( 0 \) to \( +2 \), so \( O_2 \) is oxidized (not an oxidizing agent here).

Step 5: Analyze Reaction D: \( 2Na(s) + O_2(g)

ightarrow Na_2O_2(s) \)

• In \( O_2 \), \( O \) has an oxidation state of \( 0 \).
• In \( Na_2O_2 \) (peroxide), \( O \) has an oxidation state of \( -1 \).
• Oxidation state of \( O \) decreases from \( 0 \) to \( -1 \), so \( O_2 \) is reduced (oxidizing agent).

Step 6: Analyze Reaction E: \( 2Mg(s) + O_2(g)

ightarrow 2MgO(s) \)

• In \( O_2 \), \( O \) has an oxidation state of \( 0 \).
• In \( MgO \), \( O \) has an oxidation state of \( -2 \).
• Oxidation state of \( O \) decreases, so \( O_2 \) is reduced (oxidizing agent).

Answer:

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