In the reaction #CH_4 + 2O_2 -> 2H_2O# #+# energy #+ CO_2#, which is reduced and which is oxidized?

1 Answer
Jun 21, 2018

Answer:

Just to retire this question...

Explanation:

CLEARLY the carbon of methane is OXIDIZED, and the dioxygen is reduced....

We could assign oxidation states to the BALANCED combustion equation...

#stackrel(""^(-IV))Cstackrel(""^(+I))H_4+2stackrel(""^0)O_2(g) rarr stackrel(""^(+IV))Cstackrel(""^(-II))O_2(g) + 2stackrel(""^(+I))H_2stackrel(""^(-II))O#

Alternatively, we could write the individual redox reactions...as we would for an inorganic redox process....

#"Reduction of dioxygen..(i)"#

#1/2O_2(g)+2H^+ + 2e^(-) rarr H_2O(l)#

#"Oxidation of methane..(ii)"#

#CH_4(g) +2H_2O rarr CO_2(g)+8H^+ +8e^(-)#

And in the usual way...we add #4xx(i)+(ii)# to eliminate the electrons....

#CH_4(g) +cancel(2H_2O) +2O_2(g)+cancel(8H^+ + 8e^(-))rarr CO_2(g)+cancel(8H^+) +cancel(4)2H_2O + cancel(8e^(-))#

...to give...

#CH_4(g) +2O_2(g)rarr CO_2(g)+ 2H_2O +Delta#

And energy is released, because we make STRONG #C=O# and #H-O# bonds....

Of course it is easier to balance these combustion reactions the old-fashioned and straightforward way...balance the carbons as carbon dioxide, balance the hydrogens as water, and then add appropriate dioxygen gas..