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QUESTION IMAGE

oxygen is acting as an oxidizing agent in all of the following reaction…

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 \(2C(s)+O_2(g)
ightarrow2CO(g)\):

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

Step 3: Analyze Reaction B

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

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

Step 4: Analyze Reaction C

In \(2F_2(g)+O_2(g)
ightarrow2OF_2(g)\):

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

Step 5: Analyze Reaction D

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

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

Step 6: Analyze Reaction E

In \(2Mg(s)+O_2(g)
ightarrow2MgO(s)\):

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

Answer:

C. \(2F_2(g)+O_2(g)
ightarrow2OF_2(g)\)