Each atom in a molecule has an oxidation state. We assume that each atom begins its existence alone, with the same number of electrons, protons and neutrons.
We know that when a sodium atom encounters a chlorine atom, the sodium ion will give its one valence electron to chlorine, which has seven, so that chlorine will have eight. In turn, sodium gains a full octet.
For the reaction
Zn(s) + 2H+(aq) --> Zn2+(aq) + H2(g)
A solitary atom with no charge, such as Zn(s), has an oxidation state of 0. A molecule consisting of only one type of atom with no charge, such as H2, also has an oxidation state
of zero.
The +2 superscript for the zinc ion product indicates that the zinc lost lost two electrons. Recalling "LEO goes GER", we conclude that Zn is oxidized.
For every process for which oxidation occurs, and equivalent amount of reduction should also occur (after, if something is losing electrons, then something else should be gaining
electrons.)
We notice that the hydrogens start with a +1 charge, (H+(aq), and go to H2(g), with a 0 charge.
This indicates that each hydrogen atom gained an electron,
or was reduced.
Zinc is oxidized.
Hydrogen is reduced.
We write the oxidation half-reaction as:
Zn(s) --> Zn2+(aq)
One zinc atom loses two electrons
A total of two electrons lost
and the reduction half-reaction as:
2H+(aq) --> H2(g)
Each of two hydrogen ions gains an electron
A total of two electrons gained
Thus the reaction is balanced with respect to the
gain and loss of electrons.
For the above equation, it was easy to determine the
oxidation state of each atom, because each atom was
listed by itself, or with similar atoms. If no charge
was listed (H2(g), Zn(s), etc.), the atom was neutral.
However, typical redox equations involve molecules. Given
a molecule such as MnO2, it is necessary to
use a knowledge of the periodic table to determine the
oxidation state of an atom. Some atoms, such as iron, are capable of existing in more than one oxidation state. Fortunately, for a molecule consisting of two types of
atoms, if the overall charge of the molecule, and the
oxidative state of one atom is known, the other can be
calculated.
The Rules:
oxygen is always -2.
halogens such as fluorine, chlorine, bromine, and iodine
are always -1.
hydrogen is always +1 unless the problems says otherwise.
In looking at MnO2, the rule "oxygen is always -2 is
applied, and thus, since the molecule has a zero charge,
the oxidation state of Mn must be +4.
What is the oxidation state of manganese in the ion MnO4-(aq)?
