User:Wallace.miller/sandbox

Original Text:

Consider the reaction below:

Cl₂ + 2Fe²⁺ → 2Cl⁻ + 2Fe³⁺

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

Fe²⁺ → Fe³⁺ + e⁻

Cl₂ + 2e⁻ → 2Cl⁻

In the same way given two half-reactions it is possible, with knowledge of appropriate electrode potentials, to arrive at the full (original) reaction.[1]

Suggested Addition:

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 H₂O, OH^-, e^-, and or H⁺ to either side of the reaction until both atoms and charges are balanced.

Consider the half-reaction below:

PbO₂ → PbO

OH^-, H₂O, and e^- can be used to balance the charges and atoms in basic conditions.

2e^- + H₂O + PbO₂ → PbO + 2OH^-

Again Consider the half-reaction below:

PbO₂ → PbO

H⁺, H₂O, and e^- can be used to balance the charges and atoms in acidic conditions.

2e^- + 2H⁺ + PbO₂ → PbO + H₂O

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

It is important to know that often times there will be both H⁺ and OH^- present in acidic and basic conditions but that the resulting reaction of the two ions will yield water H₂O (shown below):

H⁺ + OH^- → H₂O