# Question #954ac

May 27, 2017

Chlorine gas is reduced, and carbon is oxidized.......

#### Explanation:

We can write out the stoichiometric reaction:

$C {l}_{2} \left(g\right) + C {H}_{4} \left(g\right) \rightarrow {H}_{3} C - C l \left(g\right) + H C l \left(g\right)$

And to make a meal of it, we can include some oxidation numbers to establish the FORMAL redox process........

$\stackrel{0}{C} {l}_{2} \left(g\right) + \stackrel{- I V}{C} {H}_{4} \left(g\right) \rightarrow {H}_{3} \stackrel{- I I}{C} - \stackrel{- I}{C} l \left(g\right) + H \stackrel{- I}{C} l \left(g\right)$

We generally do not assign oxidation numbers to carbon in organic chemistry. We can nevertheless do so if we follow the few simple rules that are common to redox reactions in general: i.e. for a covalent bond, the MORE electronegative atom gets the 2 electrons, i.e. ${H}_{3} C - C l \rightarrow {H}_{3} {C}^{+} + C {l}^{-}$. And the sum of the oxidation numbers must equal the charge on the resultant species (which here was ZERO!).

Remember that this is very much a formalism, an intellectual exercise; but then redox transfer in general is also such a formalism.