How do you find the oxidizing agent in a reaction?

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Answer 1 · 2017-08-19T21:23:25

The oxidizing agent is FORMALLY the species that is reduced.........

See this old answer for the formal rules of oxidation numbers. We conceive that a species, an atom LOSES electron upon oxidation.

And thus for the oxidation of methane.....

$C H_{4} (g) + O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ $CH_4(g)+O_2(g) rarr CO_2(g) + 2H_2O(l)$

We can put in oxidation numbers if we like.....

$C H_{4} (g) + O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ $CH_4(g)+O_2(g) rarr CO_2(g) + 2H_2O(l)$

$C (- I V) + O (0) \to C (+ I V) + 4 \times O (- I I)$ $C(-IV) + O(0) rarrC(+IV) + 4xxO(-II)$

A formal EIGHT electron oxidation.

We could formalize this process even further by writing the separate redox reactions:

$C H_{4} + 2 H_{2} O \to C O_{2} + 8 H^{+} + 8 e^{-}$ $CH_4 +2H_2O rarr CO_2 +8H^+ +8e^-$ $(i)$ $(i)$

$O_{2} + 2 H^{+} + 2 e^{-} \to H_{2} O$ $O_2 +2H^(+) + 2e^(-)rarr H_2O$ $(i i)$ $(ii)$

$4 \times (i i) + (i)$ $4xx(ii) + (i)$ gives the balanced combustion reaction.