http://www.ck12.org Chapter 22. Oxidation-Reduction Reactions
better than the oxidation-number method when the substances in the reaction are dissolved in water. The aqueous
solution is typically either acidic or basic, so hydrogen ions or hydroxide ions can be added if necessary to balance
the atoms and charge.
In general, the half-reactions are first balanced separately by looking just at the number of each atom. Then, electrons
are added to each half-reaction in order to balance the charges. Then, all of the species in each half-reaction are
multiplied by some factor so that the number of electrons lost in the oxidation is equal to the number of electrons
gained in the reduction. Finally, the two half-reactions are added back together. The following example shows the
oxidation of Fe^2 +ions to Fe^3 +ions by dichromate (Cr 2 O 72 −) in acidic solution. The dichromate ions (Figure
below) are reduced to Cr^3 +ions.
FIGURE 22.7
Potassium dichromate (K 2 Cr 2 O 7 ) is a
striking orange color. The dichromate ion
is a good oxidizing agent because of the
+6 oxidation state of the chromium.Step 1:Write the unbalanced ionic equation.
Fe^2 +(aq)+Cr 2 O^27 −(aq)→Fe^3 +(aq)+Cr^3 +(aq)Notice that the equation is far from balanced, as there are no oxygen atoms on the right side. This will be resolved
by the balancing method.
Step 2: Write separate half-reactions for the oxidation and the reduction processes. Determine the oxidation
numbers first, if necessary.
Oxidation: Fe^2 +(aq)→Fe^3 +(aq)
Reduction:+ 6
Cr 2 O^27 −(aq)→Cr^3 +(aq)Step 3:Balance the atoms in each half-reaction, but ignore hydrogen and oxygen. In the oxidation half-reaction
above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced as follows:
Cr 2 O^27 −(aq)→2Cr^3 +(aq)Step 4:Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction
half-reaction above, seven H 2 O molecules will be added to the product side.
Cr 2 O^27 −(aq)→2Cr^3 +(aq)+7H 2 O(l)