oxygen is acting as an oxidizing agent in all of the following reactions except a. $2 c(s) + o_2(g)…

oxygen is acting as an oxidizing agent in all of the following reactions except a. $2 c(s) + o_2(g) \\rightarrow 2 co(g)$ b. $s(s) + o_2(g) \\rightarrow so_2(g)$ c. $2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)$ d. $2 na(s) + o_2(g) \\rightarrow na_2o_2(s)$ e. $2 mg(s) + o_2(g) \\rightarrow 2 mgo(s)$

oxygen is acting as an oxidizing agent in all of the following reactions except a. $2 c(s) + o_2(g) \\rightarrow 2 co(g)$ b. $s(s) + o_2(g) \\rightarrow so_2(g)$ c. $2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)$ d. $2 na(s) + o_2(g) \\rightarrow na_2o_2(s)$ e. $2 mg(s) + o_2(g) \\rightarrow 2 mgo(s)$

Answer

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) \rightarrow 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) \rightarrow 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) \rightarrow 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) \rightarrow 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) \rightarrow 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) \rightarrow 2OF_2(g) )