Question

For the following reaction, identify the substances being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent.

CH4 + 2O2 yields CO2 + 2H2O

Answer 1

Not sure about this one but probably. I might be wrong because I often get confused in such questions. But i hope my answer is helpful.
CH4 will be the reducing agent because it is oxidizing it self by oxidizing the carbon and oxidizing the hydrogen while reducing the Oxygen.
in CH4 Carbon has the valency -4 and H has +1. in CO2 Carbon has the valency of +4 and Hydrogen in H2O has the valency of +2
in C: -4 to +4 is oxidation
in H: +1 to +2 is also oxidation
in O: -2 to to -4 is reduction ( in CO2)
concluding CH4 will be the reducing agent and O2 will be the oxidizing agent.
Always keep this key in mind: OIL RIG
OIL= Oxidation Is Loosing
RIG= Reduction is Gaining
Oxidizing agents always reduce them-self and oxidize other elements
and Reducing agents being the opposite.

Answer 2

CH4 gets oxidized to CO2, O2 gets reduced to H2O

CH4 is the reducing agent, O2 is the oxidizing agent

Step-by-step explanation:

Oxidation is a process which involves loss of electrons. Thus, the oxidation number of the substance getting oxidized will increase upon losing electrons. In contrast, reduction involves gain of electrons and substances that undergo reduction show an increase in the oxidation number.

Oxidizing agents are substances that get reduced whereas, reducing agents are ones that get oxidized.

In the given reaction:

CH4 + 2O2 → CO2 +2H2O

On the reactants side:

Oxidation number of C in CH4 = -4

H = +1 (there are 4 H , +4)

O2 = 0

On the products side:

C in CO2 = +4

H in H2O = +1

O = -2

Therefore, C gets oxidized from -4 to +4

O gets reduced from 0 to -2

CH4 is the reducing agent and O2 is the oxidizong agent