What happens to an oxidizing agent during a redox reactions?

1 Answer
Jun 16, 2017

The oxidizing reagent is FORMALLY reduced.......

Explanation:

Redox reactions are conceived to occur on the basis of CONCEPTUAL electron transfer, and typically we write out separate redox processes.........

For oxidation of methane we could write...........

${\stackrel{- I V}{\text{ CH"_4(g) +2H_2O rarrstackrel(+IV)"CO"_2(g)+"4H}}}^{+} + 8 {e}^{-}$ $\left(i\right)$

And for every oxidation, there must be a corresponding reduction; here of dioxygen gas to water..........

${\stackrel{0}{O}}_{2} \left(g\right) + 4 {H}^{+} + 4 {e}^{-} \rightarrow 2 {H}_{2} \stackrel{- I I}{O}$ $\left(i i\right)$

And this is the FORMAL reduction reaction. Oxygen decreases in oxidation number, and it has FORMALLY GAINED electrons.....

stackrel(0)O_2+4e^(-) rarr 2stackrel(-II)O""^(2-)

And add the two half equation together.......so that we eliminate the electrons: $\left(i\right) + 2 \times \left(i i\right)$

$\stackrel{0}{C} {H}_{4} \left(g\right) + 2 {\stackrel{0}{O}}_{2} \left(g\right) \rightarrow {\stackrel{+ I V}{\text{CO}}}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

Of course, I am making a meal of it here; I could simply follow the usual rigmarole and (i) balance the carbons, (ii) balance the hydrogens as water, and (iii) balance the oxygens.