How can you identify the substance being oxidized in the following reaction: #CH_4 + O_2 -> CO_2 + H_2O#?

1 Answer
Jul 11, 2016

Answer:

Here's what I got.

Explanation:

The first thing to notice here is that the chemical equation given to you is not balanced, so let's try to balance it using oxidation numbers once we identify the substance that is being oxidized.

So, assign oxidation numbers to the atoms that take part in the reaction

#stackrel(color(blue)(-4))("C") stackrel(color(blue)(+1))("H") _ (4(g)) + stackrel(color(blue)(0))("O")_ (2(g)) -> stackrel(color(blue)(+4))("C") stackrel(color(blue)(-2))("O")_ (2(g)) + stackrel(color(blue)(+1))("H")_ 2stackrel(color(blue)(-2))("O")_ ((l))#

Now, you're looking for elements that have different oxidation states on the reactants' side and on the products' side.

Notice that carbon's oxidation number went from #color(blue)(-4)# on the reactants' side to #color(blue)(+4)# on the products' side. An increase in the oxidation number tells you that the element in question is being oxidized.

Similarly, the oxidation number of oxygen goes from #color(blue)(0# on the reactants' side to #color(blue)(-2)# on the products' side. A decrease in the oxidation number tells you that the element is question is being reduced.

Therefore, you can say that carbon is being oxidized and oxygen is being reduced.

The oxidation half-reaction looks like this

#stackrel(color(blue)(-4))("C") -> stackrel(color(blue)(+4))("C") + 8"e"^(-)#

The reduction half-reaction looks like this

#stackrel(color(blue)(0))("O") _ 2 + 4"e"^(-) -> 2stackrel(color(blue)(-2))("O")""^(2-)#

Here each oxygen atom is gaining #2# electrons, so #2# oxygen atoms will gain a total of #4# electrons.

In order for a redox reaction to take place, you need equal numbers of electrons lost in the oxidation half-reaction and gained in the reduction half-reaction.

To get that to happen, multiply the reduction half-reaction by #2# and add the two half-reactions

#{(color(white)(aaaaaa)stackrel(color(blue)(-4))("C") -> stackrel(color(blue)(+4))("C") + 8"e"^(-)), (stackrel(color(blue)(0))("O") _ 2 + 4"e"^(-) -> 2stackrel(color(blue)(-2))("O")""^(2-)" "| xx 2) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#"CH"_ 4 + 2"O"_ 2 + color(red)(cancel(color(black)(8"e"^(-)))) -> "CO"_ 2 + color(red)(cancel(color(black)(8"e"^(-)))) + 2"H"_ 2 "O"#

You will thus have

#"CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((l))#

So, carbon is being oxidized and oxygen is being reduced. You can thus say that methane is acting as a reducing agent because it reduces oxygen gas.

Similarly, oxygen gas acts an oxidizing agent because it oxidizes methane to carbon dioxide.

Keep in mind that the substance that is being oxidized acts as a reducing agent and the substance that is being reduced acts as an oxidizing agent.