Editing Half-reaction (section)

==Half-reaction balancing method==
Consider the reaction below:

:<chem>Cl2 + 2Fe^2+ -> 2Cl- + 2Fe^3+</chem>

The two elements involved, [[iron]] and [[chlorine]], each change oxidation state; iron from +2 to +3, chlorine from 0 to &minus;1. There are then effectively two ''half'' reactions occurring. These changes can be represented in formulas by inserting appropriate [[electron]]s into each half reaction:

:<math chem>\begin{align}
& \ce{Fe^2+ -> Fe^3+ + e-} \\
& \ce{Cl2 + 2e- -> 2Cl-}
\end{align}</math>

Given two half reactions it is possible, with knowledge of appropriate electrode potentials, to arrive at the complete (original) reaction the same way. The decomposition of a reaction into half reactions is key to understanding a variety of chemical processes. For example, in the above reaction, it can be shown that this is a [[redox reaction]] in which Fe is oxidised, and Cl is reduced. Note the transfer of electrons from Fe to Cl. Decomposition is also a way to simplify the balancing of a [[chemical equation]]. A chemist can atom balance and charge balance one piece of an equation at a time.

For example:
* {{chem2|Fe(2+) -> Fe(3+) + e-}} becomes {{chem2|2Fe(2+) -> 2Fe(3+) + 2e-}}
* is added to {{chem2|Cl2 + 2e- -> 2Cl-}}
* and finally becomes {{chem2|Cl2 + 2Fe(2+) -> 2Cl- + 2Fe(3+)}}

It is also possible and sometimes necessary to consider a half reaction in either basic or acidic conditions, as there may be an acidic or basic [[electrolyte]] in the [[redox reaction]]. Due to this electrolyte it may be more difficult to satisfy the balance of both the atoms and charges. This is done by adding {{chem2|H2O, OH-, e-}}, and/or {{chem2|H+}} to either side of the reaction until both atoms and charges are balanced.

Consider the half reaction below:

: <chem>PbO2 -> PbO</chem>

{{chem2|OH-, H2O}}, and {{chem2|e-}} can be used to balance the charges and atoms in basic conditions, as long as it is assumed that the reaction is in water.

: <chem>2e- + H2O + PbO2 -> PbO + 2OH-</chem>

Again consider the half reaction below:

: <chem>PbO2 -> PbO</chem>

{{chem2|H+, H2O}}, and {{chem2|e-}} can be used to balance the charges and atoms in acidic conditions, as long as it is assumed that the reaction is in water.

: <chem>2e- + 2H+ + PbO2 -> PbO + H2O</chem>

Notice that both sides are both charge balanced and atom balanced.

Often there will be both {{chem2|H+}} and {{chem2|OH-}} present in acidic and basic conditions but that the resulting reaction of the two ions will yield water, {{chem2|H2O}} (shown below):

: <chem>H+ + OH- -> H2O</chem>