# How do I balance this equation using the "half reaction" method: CH_4+O_2+H_2O=CO+CO_2+H_2 ?

Apr 21, 2017

You separate out the oxidation reaction from the reduction process....I don't think dihydrogen figures as a product here.

#### Explanation:

Methane is oxidized to give $\text{(i) carbon monoxide}$, and $\text{(ii) carbon dioxide...........}$

$C {H}_{4} \left(g\right) + {H}_{2} O \rightarrow C O \left(g\right) + 6 {H}^{+} + 6 {e}^{-}$

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

And overall, this gives..........

$2 C {H}_{4} \left(g\right) + 3 {H}_{2} O \rightarrow C {O}_{2} \left(g\right) + C O \left(g\right) + 14 {H}^{+} + 14 {e}^{-}$ $\left(i\right)$

But for oxidation there must be reduction.........and clearly dioxygen is the oxidizing agent (that is of course reduced..........to ${O}^{2 -}$, a 2 electron reduction).

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

And we balance this in the usual way; i.e. eliminate electrons, so $7 \times \left(i i\right) + 2 \times \left(i\right)$, to give (finally):

$4 C {H}_{4} \left(g\right) + 7 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 2 C O \left(g\right) + 8 {H}_{2} O \left(l\right)$

Which (I think) is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical, i.e. physical, reality.

Perhaps you should try this out for yourself, using the standard redox equation method that have been taught to you. The oxidation number of carbon in methane is ${C}^{- I V}$, of $C O$ and $C {O}_{2}$, we have ${C}^{- + I I}$, and ${C}^{+ I V}$.

To assign oxidations numbers in higher alkanes, we recall that the definition of oxidation number, the charge left on the central atom, when all the bonds are broken, with the charge distributed to the most electronegative atom. For element-element bonds, the carbons have equal electronegativity, and thus for ethane, we have $C \left(- I I I\right)$, and for propane, we have ${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I I}{C} {H}_{2} - C {H}_{3}$, etc.