# Question #9d2da

Dec 15, 2014

What you are looking for is acetylen's combustion reaction, which can also be viewed as an oxidation-reduction reaction.

$2 {C}_{2} {H}_{2} + 5 {O}_{2} \to 4 C {O}_{2} + 2 {H}_{2} O$

Let's assign oxidation numbers for all atoms involved in the reaction

$2 {C}_{2}^{- 1} {H}_{2}^{+ 1} + 5 {O}_{2}^{0} \to 4 {C}^{+ 4} {O}_{2}^{- 2} + 2 {H}_{2}^{+ 1} {O}^{- 2}$

We can see that $C$'s oxidation number goes from (-1) on the reactants' side, to (+4) on the products' side

${C}^{- 1} \to {C}^{+ 4} + 5 {e}^{_}$ (for each individual $C$ atom);

$O$'s oxidation number went from (0) an the reactants' side, to (-2) for both products

${O}^{0} + 2 {e}^{-} \to {O}^{- 2}$ (for each individual $O$ atom);

The total number of electrons lost by $C$ is $4 \cdot 5 {e}^{-} = 20 {e}^{-}$, while the total number of electrons gained by $O$ is $8 \cdot 2 {e}^{-} + 2 \cdot 2 {e}^{-} = 20 {e}^{-}$ -> determined from the balanced REDOX reaction.

Therefore, $C$ is being oxidized - acts as a reducing agent, while $O$ is being reduced - acts as an oxidation agent.