What is oxidation?

Dec 3, 2015

Oxidation is generally defined in two ways:

• organic oxidation by addition of oxygen or removal of hydrogen
• inorganic oxidation by the removal of electrons, thereby increasing the oxidation state (i.e. making it more positive) by $1$ for each electron (${e}^{-}$).

For example...

ORGANIC OXIDATION

$\text{CH"_3"C"color(red)("H"_2)"OH" stackrel("H"_2"CrO"_4)(->) "CH"_3("C"="O")"H" stackrel("cont.")(->) "CH"_3("C"="O")color(green)"O""H}$

You can see the loss of two protons on ethanol, and the addition of one oxygen onto ethanaldehyde as a result of these oxidations. This is a type of reaction you would learn in organic chemistry later on.

INORGANIC OXIDATION

${\text{Fe"(s) -> "Fe}}^{2 +} \left(a q\right) + 2 {e}^{-}$

Iron in its elemental form has an oxidation state of $0$, while it clearly has a $+ 2$ oxidation state (and evidently, charge) after removing two electrons. Thus, iron has been oxidized in this scenario.

Note, however, that this is only the oxidation half-reaction. This tends to be paired with a reduction half-reaction. If we want the full context, let us consider a reduction half-reaction as well, which is just the opposite of an oxidation half-reaction.

$\text{Mg"^(2+)(aq) + 2e^(-) -> "Mg} \left(s\right)$

So, when we combine these, we get:

${\text{Fe"(s) -> "Fe}}^{2 +} \left(a q\right) + \cancel{2 {e}^{-}}$ $\left({E}_{\text{red"^@ = +"0.45 V}}\right)$
$\text{Mg"^(2+)(aq) + cancel(2e^(-)) -> "Mg} \left(s\right)$ $\left({E}_{\text{red"^@ = -"2.37 V}}\right)$
$\text{-----------------------------------}$
$\text{Fe"(s) + "Mg"^(2+)(aq)-> "Fe"^(2+)(aq) + "Mg} \left(s\right)$

${E}_{\text{cell"^@ stackrel(?)(=) -"1.92 V}}$

And just to verify whether or not this is right as-written, recall how the more active a metal is, the better the reducing agent it is. In other words, the more readily it donates electrons (thus becoming oxidized), the more active it is.

That means for this to work, the element being oxidized must be higher up in the activity series than the element being reduced. Magnesium is more active than iron, so this reaction is backwards. Actually, the much more likely reaction would be:

$\textcolor{b l u e}{{\text{Fe"^(2+)(aq) + "Mg"(s) -> "Fe"(s) + "Mg}}^{2 +} \left(a q\right)}$

$\textcolor{b l u e}{{E}_{\text{cell"^@ = +"1.92 V}}}$

We can see that this is spontaneous because of the following (admittedly obscure) equation:

$\Delta G = - n F {E}_{\text{cell}}^{\circ}$
where $n$ is the number of electrons transferred and $F$ is Faraday's constant (~"96500 C/V").

Both $n$ and $F$ must be positive, so with a positive ${E}_{\text{cell}}^{\circ}$, $\Delta G$ is negative and thus the redox reaction is spontaneous.