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

Jun 21, 2018

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...

$\stackrel{{\text{^(-IV))Cstackrel(""^(+I))H_4+2stackrel(""^0)O_2(g) rarr stackrel(""^(+IV))Cstackrel(""^(-II))O_2(g) + 2stackrel(""^(+I))H_2stackrel(}}^{- I I}}{O}$

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

$\text{Reduction of dioxygen..(i)}$

$\frac{1}{2} {O}_{2} \left(g\right) + 2 {H}^{+} + 2 {e}^{-} \rightarrow {H}_{2} O \left(l\right)$

$\text{Oxidation of methane..(ii)}$

$C {H}_{4} \left(g\right) + 2 {H}_{2} O \rightarrow C {O}_{2} \left(g\right) + 8 {H}^{+} + 8 {e}^{-}$

And in the usual way...we add $4 \times \left(i\right) + \left(i i\right)$ to eliminate the electrons....

$C {H}_{4} \left(g\right) + \cancel{2 {H}_{2} O} + 2 {O}_{2} \left(g\right) + \cancel{8 {H}^{+} + 8 {e}^{-}} \rightarrow C {O}_{2} \left(g\right) + \cancel{8 {H}^{+}} + \cancel{4} 2 {H}_{2} O + \cancel{8 {e}^{-}}$

...to give...

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O + \Delta$

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..