As we discussed in this previous post, redox reactions are reactions where reduction and oxidation occur simultaneously.
Reduction is the gain of electrons while oxidation is the loss of electrons.
In redox reactions, there can be a total transfer of electrons (from metal to non-metal) or a partial transfer of electrons (from non-metal to non-metal).
Total transfer of electrons
We already know that metals, when forming ionic compounds, will become (positively charged) cations by losing their valence electron(s). This means that the metal is oxidized, since it loses electrons.
So, we can say that a magnesium atom is oxidized when it loses both of its valence electrons:
Mg → Mg²⁺ + 2e⁻
Non-metal ions can also be oxidized to atoms and compounds:
2Cl⁻ → Cl₂ + 2e⁻
Here, both chlorine ions lose that electron that gives them the single negative charge, forming chlorine atoms. The atoms then pair up to form the chlorine molecule.
We also already know that non-metals will gain electrons from a metal when forming an ionic compound, becoming negatively charged anions. This means that the non-metal is reduced, as it gains electrons.
So, when a chlorine atom gains an electron, it is said to be reduced to the chlorine ion:
Cl₂ + 2e⁻ → 2Cl⁻
Each chlorine atom gains an electron, forming chlorine ions.
Metal ions may also be reduced- Magnesium ions can be reduced to magnesium atoms by gaining electrons:
Mg²⁺ + 2e⁻ → Mg
So, as you can see, when electron loss occurs, there must be another atom or ion to gain those electrons. And, for electron gain to occur, there must be an atom that gives those electrons:
In the reaction shown above between sodium and chlorine, sodium loses an electron and sodium gains that electron to form the ionic compound sodium chloride. So, sodium is oxidized to the sodium ion and, at the same time, chlorine is reduced to the chlorine ion.
You can recognize whether reduction or oxidation has occurred based on oxidation numbers. For ions, the oxidation number is the number of electrons lost or gained when the ion is formed from its element. For atoms in covalent compounds, it is the number of electrons that an atom partially loses or gains in a molecule. The oxidation number is positive if electrons are gained and negative if electrons are lost.
Please note that oxidation number and charge are completely different and should not be used interchangeably.
Oxidation number is written with the sign before the number. Charge is written with the sign after then number. So the oxidation number of the oxide ion is -2 but the charge is 2-.
How do you find the oxidation number?
There are certain rules that allow you determine oxidation number:
Elements in their elemental or free states have an oxidation number of 0. For example, Al in Al (its uncombined free state) would have an oxidation number of zero. The same applies for diatomic elements- so N in N₂ and O in O₂ both have oxidation numbers of 0.
Monoatomic ions have an oxidation number equal to the charge of the ion. So, the Al³⁺ ion has an oxidation number of +3.
The oxidation numbers of the atoms in polyatomic ions add up to the charge on the polyatomic ion. So, in the nitrate ion (NO₃⁻), the oxidation numbers of 3 oxygen ions (3 × -2) and one nitrogen ion (+5) add up to -1.
Group I metals in their compounds have an oxidation number of +1. Na in NaCl is +1, K in KOH is +1...
Group II metals in their compounds have an oxidation number of +2. Mg in MgCl2 is +2, Ca in CaO is +2...
Hydrogen in its compounds usually has an oxidation number of +1, except for when it is the hydride ion in some compounds, where it has an oxidation number of -1. H in H20 is +1, H in NH3 is +1, but H in NaH is (-1).
Oxygen usually has an oxidation number of -2 in its compounds- except for in peroxides and superoxides, and of course except in its free state O2, where its oxidation state is 0. O in Na2O is -2, O in CO2 is -2,but O in H2O2 (hydrogen peroxide) is -1.
In compounds between non-metals, the more electronegative part will have a negative oxidation number while the less electronegative part will have a positive oxidation number. For example, in hydrofluoric acid, HF (an incredibly corrosive acid), the fluorine (being the most electronegative element) has an oxidation number that is negative (-1) and the hydrogen has an oxidation number of +1.
The oxidation numbers in a neutral compound (i.e. not an ion) add up to 0. In NaCl Na is +1 and Cl is -1. The two numbers add up to 0.
We can use this information to determine the oxidation number of a certain element in a compound.
Example: Find the oxidation number of N in N₂H₄.
We already know that H has an oxidation number of +1. This is a neutral compound, so the oxidation number must add up to 0. We can write this as an expression:
(2 × oxidation number of N) + (4 × oxidation number of H) = 0
(2 × oxidation number of N) + (4 × +1) = 0
(2 × oxidation number of N) + 4 = 0
2 × oxidation number of N = -4
oxidation number of N = -4/2
oxidation number of N = -2
The oxidation number of N in N₂H₄ is -2.
So, how do oxidation numbers relate to reduction and oxidation?
Look at the following reaction:
2Na(s) + Cl₂(g) → 2NaCl(aq)
This can be split into two separate equations known as half equations.
The first is with sodium. In the reactants, sodium is in its free state. But in the products, sodium is an ion bonded to the chlorine ion. So, sodium becomes the sodium ion by losing an electron to chlorine (oxidation). We can show this like this:
2Na(s) → 2Na⁺(aq) + 2e⁻
Na(s) → Na⁺(aq) + e⁻
So, the oxidation number of sodium increases from 0 in its free state to +1 as the sodium ion. Therefore, oxidation causes an increase in oxidation number.
The second half equation is with chlorine. In the reactants, chlorine is in its free state. But in the products, chlorine is an ion bonded to the sodium ion. So, chlorine becomes the chlorine ion by gaining an electron from sodium (reduction). We can show this like this:
Cl₂(g) + 2e⁻ → 2Cl⁻(aq)
So, the oxidation number decreases from 0 in its free state as the chlorine molecule to -1 as the chlorine ion. Therefore, reduction reduces the oxidation number.
When we combine these half equations:
2Na(s) → 2Na⁺(aq) + 2e⁻
+ Cl₂(g) + 2e⁻ → 2Cl⁻(aq)
2Na(s) + Cl₂(g) → 2NaCl(aq)
We get back our original redox reaction.
Oxidation numbers also help us to distinguish between compounds with the same elements in different ratios due to differences in oxidation number.
Look at these compounds:
They are all called sodium chlorate.
But how? That seems weird, right? But in each compound, chlorine has a different oxidation number.
We can name each compound using the oxidation state to help distinguish among them.
NaClO - Chlorine's oxidation number is +1 here (in the chlorate ion), so this compound can be called sodium chlorate (I)
NaClO₂ - Chlorine's oxidation number is +3 here, so this compound can be called sodium chlorate (III)
NaClO₃ - Chlorine's oxidation number is +5 here, so this compound can be called sodium chlorate (V)
NaClO₄ - Chlorine's oxidation number is +7 here, so this compound can be called sodium chlorate (VII)
Transition metals are particularly pesky with their variable oxidation numbers. So, in compounds containing them, we make sure to write the oxidation number of the transition metal. For example, in MnO₂, the transition metal, manganese, has an oxidation number of +4, so we call this compound manganese (IV) oxide.