# 5.1 What is Oxidation and Reduction?

## Defining oxidation and reduction

Oxidation and reduction, or redox reactions, are chemical reactions in which electrons (denoted by $\ce{e-}$) are transferred from one atom to another. An element is ***oxidized*** when it loses an electron, and ***reduction*** of an element occurs when it gains an electron. Redox reactions involve the simultaneous oxidation of one element and the reduction of another.

```{dropdown} Example: Oxidation and reduction reactions 

Let's identify what's being oxidized and reduced in the reactions below:

$$
\ce{
2 Na + Cl2 &-> 2 Na+ + 2 Cl- \\
&-> 2 NaCl
} $$

>Here $\ce{Na}$ is losing an $\ce{e-}$ to become $\ce{Na+}$ and is oxidized. Also, $\ce{Cl2}$ is gaining $\ce{e-}$ to become $\ce{Cl-}$ and is reduced.

$$ 
\begin{align*} 
\ce{
2 Fe + O2 &-> 2 Fe^2+ + 2O^2- \\
&-> 2 FeO
} \end{align*} $$

>Here $\ce{Fe}$ is losing $\ce{e-}$ to become $\ce{Fe^2+}$ and is oxidized. While, $\ce{O2}$ is gaining $\ce{e-}$ to become $\ce{O^2-}$ and is reduced.
```

Representing "half" of a redox reaction is often convenient - the oxidation half or the reduction half. Reactions that only show the oxidation or reduction of an element are called half reactions. Half reactions help balance redox reactions.

```{dropdown} Example: Oxidation and reduction half-reactions 

Below are examples of oxidation reactions where an element loses $\ce{e-}$.

>$$ \begin{align*} 
\ce{
Fe(s) &-> Fe^2+ + 2 e- \\
&-> Fe^3+ + 3 e- \\
C(s) &-> C^4+ + 4 e-
} \end{align*} $$

Below are reduction reactions where an element gains $\ce{e-}$.

>$$ \begin{align*} 
\ce{
O + 2 e- &-> O^2- \\
N + 3 e- &-> N^3-
} 
\end{align*} $$
```

The above examples are shown in the form of a ***half-reaction***, where oxidation or reduction of an element is clearly shown. No free $\ce{e-}$ transfer is possible. Every oxidation reaction must be accompanied by a reduction reaction(s), where the $\ce{e-}$ are consumed. The above sets of half-reactions can be combined so that the final redox reaction will show no free $\ce{e-}$.

```{dropdown} Example: Balancing half-reactions 

Let's combine the reactions to write a balanced redox reaction. 

>Since each half-reaction produces 2  $\ce{e-}$ while half-reaction requires 4 $\ce{e-}$, multiply half-reaction by 2. Now combine the half-reactions and cancel the $\ce{e-}$.
>
>$$ \begin{align*} 
\ce{
2 Fe(s) &-> 2 Fe^2+ + 4e- \\
O2(g) + 4 e- &-> 2 O^2- \\
\\
2 Fe(s) + O2(g) &-> 2 Fe^2+ + 2 O^2-
} 
\end{align*} $$
>
>Note: The $\ce{e-}$ from the half-reactions appear on opposite sides to be able to cancel out.
```
 

Since any reduced element can release $\ce{e-}$ that can combine with (i.e., reduce) another element, reduced compounds are sometimes called reducing agents or reductants. Similarly, oxidized elements can gain $\ce{e-}$ from other elements; in the process, the (initially) oxidized element becomes reduced and acts as an oxidizing agent or oxidant.

```{dropdown} Example: Oxidizing and reducing agents 

Let's identify the oxidant and reductant in the following reaction:

$$\ce{
2 Fe + O2 -> 2 FeO
}$$

>On left-side of reaction, $\ce{Fe}$ is "helping" $\ce{O2}$ reduce is, hence, a reducing agent. Likewise, $\ce{O2}$ is helping $\ce{Fe}$ oxidize and is an oxidizing agent.
```


## Oxidation State

The oxidation state (OS) describes the degree of oxidation of an atom in a chemical compound. Oxidation states allow us to balance electrons in a chemical equation. It is the charge an atom would have if electrons were transferred completely. Conceptually, the oxidation state is the hypothetical charge that an atom would have if all bonds to atoms of different elements were $\pu{100 \%}$ ionic, with no covalent component. Oxidation states are typically represented by integers, which may be positive, zero, or negative. In some cases, the average oxidation state of an element is a fraction (in complex organic molecules).

Oxidation numbers enable us to identify oxidized and reduced elements at a glance. The elements that show an increase in oxidation state are being oxidized, whereas the elements that show a decrease in oxidation state are being reduced.

```{admonition} Rules for Determining Oxidation State (OS)

1. OS of a free element $=0$. E.g., $\ce{Na}$, $\ce{Be}$, $\ce{K}$, $\ce{Pb}$, $\ce{H2}$, $\ce{O2}$, $=0$
2. For monoatomic ions $=$ net charge on ion. E.g., $\ce{Na+} = +1$,
$\ce{Fe^3+} = +3$, $\ce{O^2-} = -2$
3. OS of $\ce{O} = -2$ in most compounds. Exception: OS $= -1$ in $\ce{H2O2}$
4. OS of $\ce{H}$ $= +1$. Exception: OS $= -1$ in metal hydrides (e.g., $\ce{LiH}$)
5. Group I metals $= +1$
6. Group II metals $= +2$
7. Halides $=$ negative
8. Algebraic sum of OS in neutral molecules $= 0$
9. Algebraic sum of OS in ionic molecules $=$ charge on an ion
10. OS can be a fraction
```

```{dropdown} Example: Determining oxidation states of nitrogen 

Determine the OS of $\ce{N}$ in $\ce{NH3}$, $\ce{NO3-}$, $\ce{NO2-}$, and $\ce{N2}$. 

>Per the OS rules, OS for $\ce{H}$ and $\ce{O}$ in these compounds are $+1$ and $-2$, respectively.
>
>We can use a simple algebra trick to determine the OS for an unknown. Let's start with $\ce{NH3}$. Let's call OS of $\ce{N}$ as $x$. Therefore, based on Rule #8, $x+3(+1) = 0$. Or, $x = -3$, or OS for $\ce{N}$ in $\ce{NH3}$ is $3$.
>
>Similarly, let's write the algebraic notation for $\ce{NO3-}$ using Rule #9 as $x+3(-2) = -1$, Therefore, $x = +5$ or OS for $\ce{N}$ in $\ce{NO3-}$ is $5$.
>
>Therefore, the OS of all $\ce{N}$ compounds are:
>1. $\ce{NH3}$ : $-3$
>2. $\ce{NO3-}$: $+5$
>3. $\ce{NO2-}$: $+3$
>4. $\ce{N2}$: $0$
>
>We can rank these $\ce{N}$ compounds from the most oxidized to most reduced forms as follows: $\ce{NO3-}$, $\ce{NO2-}$, $\ce{N2}$, and $\ce{NH3}$.
```