Question

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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 state 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

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

• Oxidation state of \( O \) in \( O_2 \): \( 0 \)
• Oxidation state of \( O \) in \( CO \): \( -2 \)
• \( O \) is reduced (oxidation state decreases from \( 0 \) to \( -2 \)), so \( O_2 \) is an oxidizing agent.

Step 3: Analyze Reaction B

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

• Oxidation state of \( O \) in \( O_2 \): \( 0 \)
• Oxidation state of \( O \) in \( SO_2 \): \( -2 \)
• \( O \) is reduced (oxidation state decreases from \( 0 \) to \( -2 \)), so \( O_2 \) is an oxidizing agent.

Step 4: Analyze Reaction C

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

• Oxidation state of \( O \) in \( O_2 \): \( 0 \)
• Oxidation state of \( O \) in \( OF_2 \): \( +2 \) (since \( F \) is \( -1 \), and \( 2(-1) + x = 0 \Rightarrow x = +2 \))
• \( O \) is oxidized (oxidation state increases from \( 0 \) to \( +2 \)), so \( O_2 \) is a reducing agent (not an oxidizing agent here).

Step 5: Analyze Reaction D

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

• Oxidation state of \( O \) in \( O_2 \): \( 0 \)
• Oxidation state of \( O \) in \( Na_2O_2 \): \( -1 \) (since \( Na \) is \( +1 \), and \( 2(+1) + 2x = 0 \Rightarrow x = -1 \))
• \( O \) is reduced (oxidation state decreases from \( 0 \) to \( -1 \)), so \( O_2 \) is an oxidizing agent.

Step 6: Analyze Reaction E

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

• Oxidation state of \( O \) in \( O_2 \): \( 0 \)
• Oxidation state of \( O \) in \( MgO \): \( -2 \)
• \( O \) is reduced (oxidation state decreases from \( 0 \) to \( -2 \)), so \( O_2 \) is an oxidizing agent.

Answer:

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