Redox reactions differ from the other reaction types by the changing the configuration of the electrons in the outer shell. Often it is a matter of removing or getting electrons, but it may also be a rearrangement of the electrons in the orbitals. In a few select cases it is a matter of breaking a specific type of bonds, e.g. the single bond between two oxygens, O-O, providing the redox reaction, but for the most part, redox reactions is about changing electron configurations. In redox reactions, this is called changing the oxidation state. Oxidation state, or oxidation number, is a purely mathematical tool for doing calculations on redox reactions, which can make it look a bit odd, e.g. you can have oxidation states that are fractions, but in real life you cannot move only a fraction of an electron.
In our normal way of looking at the outer electrons, we strive for having the outer shell filled up. This can be done by removing or absorbing electrons. Quite a few will have learned in school, that this means removing or absorbing electrons to obtain eight electrons in the outer shell, also known as the octet rule. This is not correct, but a misleading simplification of the real world! That would mean the elements could only have two oxidation states, 0 when no electrons had been removed or absorbed, or the charge related to the number of electrons absorbed or removed. Quite a few elements only have two oxidation steps, but that is the result of the atom's orbitals, not the total number of electrons in the outer shell.
Hydrogen will, as a single atom or in its natural form H2 have the oxidation state 0. All elements in their free/pure form have the oxidation state 0. Hydrogen comes with one electron in its outer (and only) shell, and there is room for two in total. H now has the option of giving off an electron and become H+, this is the most common form, or absorb an electron end become H−. The oxidation state is now either +1 or -1, like the charge. -1 for hydrogen only occurs for the chemical compounds called hydrides. In this case, the oxidation state reflects the number of electrons absorbed or removed, compared to the free element. The oxidation state is normally written above the specific element, like this:
0
+1
+2
+1 -2
H2
H+
Ca2+
H2O
The oxidation state can also be written under the element, if you prefer. There is no rule dictating where to write the oxidation state, and no rule about using Roman numerals or ordinary (Arabic) numbers. It is a matter of personal preferences and irrelevant, as long as you are consistent and the oxidation states are used correctly. From a practical point of view, Roman numerals may be a good idea for handwriting, as you easily get a lot of information in a small area. Having a typographical difference is a big help in avoiding that the informations get mixed up. Note that there is a + in front of the positive values. Normally + is implicit, when writing positive numbers, but for oxidation states you use an explicit + for the positive values.
For a redox reaction the oxidation states looks like this:
0
+2
+2
0
Zn(s) +
Cu2+(aq)
Zn2+(aq) +
Cu(s)
Two electrons have heen transferred from zinc to copper. Zinc has been oxidized (the oxidation state has increased) and copper is reduced (the oxidation stated has decreased).
In the educational system many students have been raised thinking of the l shell and outwards as one shell containing 8 electrons. For educational material designed for this kind of simplified thinking, it is an excellent model for educational purposes, but this is not how it really works. In the l shell there is room for 1 s orbital, which can contain 2 electrons, and 3 p orbitals, which can contain 2 electrons each. Each type of orbital has a different energy level, so instead of removing all the outer electrons or fill up to 8, you have several solutions in between. On top of this, you have the transition metals and the rare earth metals, where you have far more than 8 electrons in the outer shell.
Phosphorous has 5 electrons in the outer shell, which is the m shell. These are distributed as 2 s and 3 p electrons. The first two options are to remove 5 electrons or absorb 3, to have the outer shell filled. That would be oxidation states +5 and -3. As s and p orbitals are two different types of orbitals with different energy levels, phosphorous can limit the removal of electrons to the 3 p electrons, so the outer shell contains 2 s electrons. We are now in a situation where the outer shell is not filled up, but the outer orbital is (this is why it is important to be acutely aware of the difference between shells and orbitals), so for the phosphorous atom, the oxidation state +3 is also fine.
If we look at oxygen, it is way too difficult to remove 6 electrons, compared to absorbing 2. Therefore oxygen can be found having the oxidation states 0 and -2. In special cases, oxygen is found as peroxides and super oxides. The most common of these is hydrogen peroxide, H2O2. Here the oxygen atoms are linked as H-O-O-H and oxygen has the oxidation state -1. The electrons are not arranged any different from the way they are arranged in the water molecule, H-O-H, but hydrogen is limited to being +1 when it is not a hydride, and the total oxidation state is 0. Having 2 hydrogen atoms, each contributing with +1, each of the 2 oxygens are forced to contribute with -1 to have the total charge 0. Contrary to the previous examples, the oxidation state here is a purely mathematical phenomenon, reflecting that peroxide is relatively easy to divide, and the energy from this division can drive a redox reaction. In use, it looks like this:
0
+1 -1
+1
0
+1 -2
2 I−(aq) +
H2O2(aq) +
2 H+(aq)
I2(aq) +
H2O(l)
As seen, iodide is oxidized to iodine, while oxygen is reduced from peroxide to oxide.
The last way to change the oxidation state is hybridization of the orbitals, which is especially important for carbon and carbon chemistry. In its ground state carbon has 2 s and 2 p electrons, i.e. you have 2 unpaired electrons (Px and Py) which means 2 covalent bonds and a lone pair. However, s and p orbitals can be mixed (this includes the unused Pz orbital), resulting in the formation of 4 separate orbitals, that are a mixture of s and p orbitals. These are called sp3 orbitals and carbon can now have 4 covalent bonds. Along with oxygen, carbon can thus form both CO and CO2, where the oxidation states are +2 and +4, respectively. In this case, carbon has not changed the number of electrons, the way they move around the nucleus has just changed.
Therefore, oxidation states should only be seen as a mathematical way of describing states and changes in the atom's electron configuration during chemical reactions, and should only be used as a tool for balancing reaction equations.
Calculation of oxidation states
The free elements always have oxidation state 0. The state of the free element is irrelevant:
0
0
0
0
0
I2(s)
S8(s)
Fe(s)
Hg(l)
Ne(g)
For elements in their ionic form, the oxidation state is the same as the charge:
+1
+2
-1
Cu+
Cu2+
I−
For chemical compounds, the sum of the oxidation states is the same as the charge of the chemical compound.
If we look at water, the formula is H2O. The total charge is 0, so the sum of the oxidation states for H and O must be 0. H is limited to +1:
+1
H2O
As there is 2 hydrogens, giver this gives a total contribution of 2·(+1) = +2. With the total charge being 0, oxygen must have the oxidation state -2 for the oxidation states to balance.
+1 -2
H2O
Same procedure for the nitrate ion, NO3−. The total charge is -1, so the total oxidation state is -1. Oxygen we set to -2 (as it is not a peroxide):
-2
NO3−
Having an oxygen content of 3, the total contribution from oxygen is -6 (3·(-2) from each oxygen). To have a total oxidation state of -1, nitrogen must be +5 (+5-6 = -1):
+5 -2
NO3−
One of the places where the difference between the math and the chemistry becomes noticeable, in regards to oxidation states, is when the oxidation state is a fraction. Potassium can be found as a superoxide, under special conditions. Here oxygen appears with the oxidation state -½, equivalent to only having received half an electron:
+1 -½
KO2
In principle it might as well have been potassium having oxidation state +4, as it is only math, but here math has to bow to the chemical reality, that potassium can only have oxidation states 0 and +1, due to the placement in the periodical system, i.e. the electron configuration. When balancing reaction equation containing oxidation states that are fractions, you just ignore that in real life fractions are nonsense, and do the calculations using the fractions.
Balancing redox reactions
Oxidation steps are used for balancing redox reaction. For simple reactions they may appear redundant, but when it comes to complex reactions, they become a rather good tool.
Considering that nothing just appears or disappears, it has to be that you cannot have oxidation in one place without having a reduction somewhere else, and vice versa. You cannot have one without the other, and they have to have the same size, otherwise something would be appearing or disappearing.
If we take a look at the reaction between metallic zinc and silver ions:
0
+1
+2
0
Zn(s) +
Ag+(aq)
Zn2+(aq) +
Ag(s)
Zn increases by 2, from 0 to +2, and Ag decreases with 1, from +1 to 0. This is written like this:
Zn
↑
2
Ag
↓
1
In some places (schools) they want it written explicitly with words. Like this:
Oxidation:
Zn
↑
2
Reduction:
Ag
↓
1
The information is redundant, as you cannot have an increase in oxidation state without the element being oxidized, and vice versa, but for some educational institutions this is the only correct notation.
As oxidation and reduction needs to be of equal size, you need 2 silver ions, as 2·1 = 2. Graphically it can be written like this:
Zn
↑
2
1
Ag
↓
1
2
From a mathematical point of view, the switch is equivalent to finding a common denominator for fractions by multiplying the denominators.
In essence, you need 1 Zn per 2 Ag for the reaction. The increase and decrease in oxidation state becomes the inverse ratio for the reactants in the reaction equation.
The reaction now looks like this:
0
+1
+2
0
Zn(s) +
2 Ag+(aq)
Zn2+(aq) +
2 Ag(s)
As seen, you now have the same number of Ag and Zn on both sides of the reaction equation, and the charge is the same on both sides, the the reaction equation is balanced.
A variation of the shown notation is this one:
0
+1
+2
0
Zn(s) +
2 Ag+(aq)
Zn2+(aq) +
2 Ag(s)
There are no fixed rules for how the notation should look. From the reaction equation and the notation, you must be able to plainly see the oxidation states and what is oxidized and reduced, and how much. It has to be easy for other readers to read and understand the reaction, but otherwise the notation is a matter of personal preferences. Some are visual in their writing, others prefer the mathematical approach.
Some reactions only occur under acidic or alkaline conditions. Permanganate e.g. will react with iodine, forming manganese(III) ions, but only under acidic conditions. Balancing such an equation is done in multiple steps. The first part is writing the basic reaction:
MnO4−(aq) + I−(aq) Mn3+(aq) + I2(aq)
We know that the conditions are acidic, so we are going to need some H+ somewhere in the equation, but we don not know where to put them yet. The next step is adding oxidation states:
+7 -2
-1
+3
0
MnO4−(aq) +
I−(aq)
Mn3+(aq) +
I2(aq)
Mn has decreased with 4 and I has increased with 1, i.e. we need 4 iodine atoms per 1 manganese atom:
I
↑
1
4
Mn
↓
4
1
Starting on the left side:
+7 -2
-1
+3
0
MnO4−(aq) +
4 I−(aq)
Mn3+(aq) +
I2(aq)
To have 4 iodine atoms on the right side, we need 2 I2.
+7 -2
-1
+3
0
MnO4−(aq) +
4 I−(aq)
Mn3+(aq) +
2 I2(aq)
We now have the same number of manganese and iodine atoms on both sides of the equation. On the left side, the total charge is -5 (-1 from MnO4− and 4·(-1) from the 4 I−). On the right side, the total charge +3. As we know the conditions are acidic, the difference in charge must be evened by adding H+. They need to be added on the left side, to get from -5 to +3, and you need 8 of them:
+7 -2
-1
+1
+3
0
MnO4−(aq) +
4 I−(aq) +
8 H+(aq)
Mn3+(aq) +
2 I2(aq)
The chance of ever putting H+ or OH− on the right side of an equation are slim to none. You can have an acid catalyzed reaction, and those are quite common in organic chemistry, in which case the charge is the same on both sides of the equation, and you ad the H+ above the arrow like this:
H+ (aq)
. You could have an acid catalyzed cumulative reaction forming H+, in which case the H+ would go on the right side, so you cannot rule out the chance of having to have the H+ (or OH− for the alkaline reactions) on the right side. It is just highly unlikely.
Now the charges are balanced and we have a surplus of 4 oxygen and 8 hydrogen. Unless something else is specified, it is reasonable to assume that hydrogen and oxygen becomes water. In this case, 4 water molecules are formed from the excess H and O:
+7 -2
-1
+1
+3
0
MnO4−(aq) +
4 I−(aq) +
8 H+(aq)
Mn3+(aq) +
2 I2(aq) +
4 H2O(aq)
In its finalized form, balanced and with explanatory text regarding oxidation/reduction, the reaction looks something like this (depending on the preferred notation):
