How goes the redox reaction of burning methane?
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:
After cancelling, we get:
Of course, this approach would be totally impractical to do for standard combustion reactions. Even for propane,
And thus for propane,
For even-numbered alkanes, we usually require half integral coefficients. Try it out for
And thus we can avoid the tedious mucking about with oxidation numbers. Can you repeat this for the complete combustion of hexanes,
See here for another spray with regard to redox reactions in general.