# How goes the redox reaction of burning methane?

Feb 11, 2017

$\text{Carbon undergoes a formal 8 electron oxidation..........}$

$C \left(- I V\right) \rightarrow C \left(+ I V\right)$

#### Explanation:

$\text{Oxidation:}$

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

$\text{Reduction:}$

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

In both instances, mass and charge have been conserved (and they must be in order to reflect reality).

And thus we cross multiply so that electrons do not appear in the final redox equation: $\left(i\right) + 2 \times \left(i i\right) :$

$C {H}_{4} + \cancel{2 {H}_{2} O} + 2 {O}_{2} + \cancel{8 {H}^{+} + 8 {e}^{-}} \rightarrow C {O}_{2} + \cancel{8 {H}^{+} + 8 {e}^{-}} + \cancel{4} 2 {H}_{2} O$

After cancelling, we get:

$C {H}_{4} + 2 {O}_{2} \rightarrow C {O}_{2} + 2 {H}_{2} O$

Of course, this approach would be totally impractical to do for standard combustion reactions. Even for propane, ${H}_{3} C - C {H}_{2} - C {H}_{3}$, we have formal oxidation numbers of ${C}_{\text{terminal}} = - I I I$ and ${C}_{\text{central}} = - I I$. The sequence that is generally adopted is:

$\text{(i)}$ $\text{Balance the carbons as carbon dioxide}$

$\text{(ii) Balance the hydrogens as water}$

$\text{(iii) Balance the oxygens as dioxygens}$

And thus for propane, ${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$

For even-numbered alkanes, we usually require half integral coefficients. Try it out for $\text{ethane}$.

And thus we can avoid the tedious mucking about with oxidation numbers. Can you repeat this for the complete combustion of hexanes, ${C}_{6} {H}_{14}$?

See here for another spray with regard to redox reactions in general.