# 18.03: Balancing Oxidation-Reduction Reactions Using the Half Reaction Method

Some redox reactions are not easily balanced by inspection. Consider this redox reaction:

Al + Ag+ → Al3+ + Ag

At first glance, this equation seems balanced: there is one Ag atom on both sides and one Al atom on both sides. However, if you look at the total charge on each side, there is a charge imbalance: the reactant side has a total charge of 1+, while the product side has a total charge of 3+. Something is amiss with this chemical equation; despite the equal number of atoms on each side, it is not balanced.

A fundamental point about redox reactions that has not arisen previously is that the total number of electrons being lost must equal the total number of electrons being gained for a redox reaction to be balanced. This is not the case for the aluminum and silver reaction: the Al atom loses three electrons to become the Al3+ ion, while the Ag+ ion gains only one electron to become elemental silver.

To balance this, we will write each oxidation and reduction reaction separately, listing the number of electrons explicitly in each. Individually, the oxidation and reduction reactions are called half reactions. We will then take multiples of each reaction until the number of electrons on each side cancels completely and combine the half reactions into an overall reaction, which should then be balanced. This method of balancing redox reactions is called the half reaction method.

The oxidation half reaction involves aluminum, which is being oxidized:

Al → Al3+

This half reaction is not completely balanced because the overall charges on each side are not equal. When an Al atom is oxidized to Al3+, it loses three electrons. We can write these electrons explicitly as products:

Al → Al3+ + 3e

Now this half reaction is balanced—in terms of both atoms and charges.

The reduction half reaction involves silver:

Ag+ → Ag

The overall charge is not balanced on both sides. But we can fix this by adding one electron to the reactant side because the Ag+ ion must accept one electron to become the neutral Ag atom:

Ag+ + e → Ag

This half reaction is now also balanced.

When combining the two half reactions into a balanced chemical equation, the key is that the total number of electrons must cancel, so the number of electrons lost by atoms are equal to the number of electrons gained by other atoms. This may require we multiply one or both half reaction(s) by an integer to make the number of electrons on each side equal. With three electrons as products and one as reactant, the least common multiple of these two numbers is three: we can use a single aluminum reaction but must take three times the silver reaction:

Al → Al3+ + 3e3 × [Ag+ + e → Ag]

The 3 on the second reaction is distributed to all species in the reaction:

Al → Al3+ + 3e3Ag+ + 3e → 3Ag

Now the two half reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. The same species on opposite sides of the arrow can be canceled:

$Al+3Ag^{+}+3e^{-}\rightarrow Al^{3+}+3Ag+3e^{-}$

The net balanced redox reaction is as follows:

Al + 3Ag+ → Al3+ + 3Ag

There is still only one Al atom on each side of the chemical equation, but there are now three Ag atoms, and the total charge on each side of the equation is the same (3+ for both sides). This redox reaction is balanced. It took more effort to use the half reaction method than by inspection, but the correct balanced redox reaction was obtained.

Example $$\PageIndex{1}$$:

Balance this redox reaction by using the half reaction method.

Fe2+ + Cr → Fe + Cr3+

Solution

We start by writing the two half reactions. Chromium is being oxidized, and iron is being reduced:

Cr → Cr3+ oxidationFe2+ → Fe reduction

Then we include the appropriate number of electrons on the proper side to balance the charges for each reaction:

Cr → Cr3+ + 3eFe2+ + 2e → Fe

The first reaction involves three electrons, while the second reaction involves two electrons. The least common multiple of these two numbers is six, so to get six electrons in each reaction we need to double the first reaction and triple the second one:

2 × [Cr → Cr3+ + 3e] = 2Cr → 2Cr3+ + 6e3 × [Fe2+ + 2e → Fe] = 3Fe2+ + 6e → 3Fe

We can combine the two final reactions, noting that the electrons cancel:

$2Cr+3Fe^{2+}+6e^{-}\rightarrow 2Cr^{3+}+3Fe+6e^{-}$

The overall, balanced redox reaction is

2Cr + 3Fe2+ → 2Cr3+ + 3Fe

Exercise $$\PageIndex{1}$$

Balance this redox reaction by using the half reaction method.

O2− + F2 → O2 + F