# Can oxidation happen without reduction in a chemical reaction?

Jan 26, 2016

Nope. Oxidation occurs because an agent that causes the oxidation---the oxidizing agent---must itself get reduced.

HALF-REACTIONS

You may be confused as to what it means to have a half-reaction; in those, one atom or compound gets reduced or oxidized, but nothing else gets oxidized or reduced.

In a half-reaction, there is a theoretical transfer of electrons out from or into a compound or atom, changing its oxidation state. For example:

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

Here, two electrons are removed from iron solid, hence the production of two lone ${e}^{-}$'s and an iron(II) cation of a more positive (less negative) charge. We've written out the reaction for the oxidation of iron.

The charges add up to neutrality on either side; $0 = 2 + \left(- 2\right)$.

The thing is, where do the electrons go? They don't just escape the iron solid and go out into the world; the probability of an electron existing outside of orbitals is 0%. It has to be donated to another compound or atom.

"RELEASED" ELECTRONS MUST GO TO AN ELECTRON ACCEPTOR

We say that iron must be "active" enough to donate this electron; so a less "active" metal that can accept this electron is silver, according to the activity series, as ${\text{Ag}}^{+}$.

So, we can write another theoretical half-reaction:

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

But one equivalent of silver(I) only wants one electron, so we need two equivalents of silver(I) to accept iron's two electrons.

THE FULL REDOX (REDUCTION-OXIDATION) REACTION

Now we can sum up the two reactions to get the full redox reaction:

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

Because the two electrons were transferred from iron to silver, no electrons are "out in limbo" (as they were in the theoretical half-reactions), and we can cancel out the donated electrons on the products and reactants sides.

In this full redox reaction:

• Because silver decreased in positive charge (or, increased in negative charge), silver got reduced. ${\text{Ag"^(+1) -> "Ag}}^{0}$
• Because iron increased in positive charge (or, decreased in negative charge), iron got oxidized. ${\text{Fe"^(0) -> "Fe}}^{+ 2}$

As for the roles:

• Iron is the reducing agent. It donated the electrons to silver, reducing it.
• Silver is the oxidizing agent. Silver(I) cation accepted the electrons from iron, oxidizing it.

As an aside, because iron is more active than silver, you can't reverse this reaction to get:

${\text{Fe"^(2+)(aq) + 2"Ag"(s) color(red)(cancel(color(black)(->))) "Fe"(s) + 2"Ag}}^{+} \left(a q\right)$

and it doesn't occur.