Redox Reactions

Key Questions

• 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.

• Redox reactions involve the transfer of electrons (${e}^{-}$)from one compound to another. These reactions differ from acid-base reactions because we are no longer dealing with protons (${H}^{+}$), but instead with electrons.

The substance that lost electrons (the substance that was oxidized) is called a reducing agent, while the substance that gained electrons (the substance that was reduced) is called an oxidation agent.

We use oxidation numbers to keep track of the electrons transferred in a redox reaction. The rules for assigning oxidation numbers to each atom belonging to a compound involved in a redox reaction are pretty straightforward, as you can see here

http://www.occc.edu/kmbailey/chem1115tutorials/oxidation_numbers.htm

Therefore, a reduction is a chemical reaction in which a compound gains electrons. Here's an example to better illustrate this:

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

Let's assign oxidation numbers (ON) for every atom (I will not go into the rules for doing this here)

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

As you can see, $C$ went from ON = $- 4$ on the reactsnts' side, to ON = $+ 4$ on the products' side - it LOST $8 {e}^{-}$

$O$, on the other hand, went from ON = $0$ on the reactsnts' side, to ON = $- 8$ on the products' side - it GAINED $8 {e}^{-}$.

Therefore, the oxygen was reduced by the carbon, while the carbon was oxidized by the oxygen.

• A redox reaction occurs when one substance involved in the reaction is reduced, and another substance is oxidized.

Thanks, but that doesn't help me too much!

Q: What does it mean for something to be reduced?
A: That means the charge of that substance has been reduced due to the addition of negatively charged electrons.

Q: So what does it mean when something is oxidized?
A: When something is oxidized, it means it has lost electrons and gains a positive charge (subtraction of a negative)

Q: What is an example of a redox reaction?
A: Here is an example...

2$A g N {O}_{3}$ + Cu -> $C u {\left(N {O}_{3}\right)}_{2}$ + 2Ag

In this reaction the nitrate ion ($N {O}_{3}$) keeps the same -1 charge from the reactant side to the product side.

Cu begins with a charge of 0 as a reactant (pure elements are neutral) to a +2 charge when paired with 2 copies of the nitrate ion. This means the copper was oxidized.

The silver begins with a +1 charge (you know this because it's paired in a 1:1 ratio with the -1 nitrate) to having a charge of 0 when in its pure elemental form after the reaction. So the silver was reduced.

Here is a video discussing this topic:

Hope this helps!
Noel P.