Step 1: Write the unbalanced ionic equation.

Fe2+(aq)+Cr2O2−7(aq)→Fe3+(aq)+Cr3+(aq)(22.10.1) 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:Fe2+(aq)→Fe3+(aq)Reduction:Cr2+6O2−7(aq)→Cr3+(aq)(22.10.2)(22.10.3) Step 3: Balance the atoms in the half-reactions other than hydrogen and oxygen. In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced with the chromium atoms.

Cr2O2−7(aq)→2Cr3+(aq)(22.10.4) Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction half-reaction above, seven H2O molecules will be added to the product side.

Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)(22.10.5) Now the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance. In this example, fourteen H+ ions will be added to the reactant side.

14H+(aq)+Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)(22.10.6) Step 5: Balance the charges by adding electrons to each half-reaction. For the oxidation half-reaction, the electrons will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant side. By adding one electron to the product side of the oxidation half-reaction, there is a 2+ total charge on both sides.

Fe2+(aq)→Fe3+(aq)+e−(22.10.7) There is a total charge of 12+ on the reactant side of the reduction half-reaction (14−2) . The product side has a total charge of 6+ due to the two chromium ions (2×3) . To balance the charge, six electrons need to be added to the reactant side.

6e−+14H+(aq)+Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)(22.10.8) Now equalize the electrons by multiplying everything in one or both equations by a coefficient. In this example, the oxidation half-reaction will be multiplied by six.

6Fe2+(aq)→6Fe3+(aq)+6e−(22.10.9) Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by inspection. If necessary, cancel out H2O or H+ that appear on both sides.

6Fe2+(aq)6e−+14H+(aq)+Cr2O2−7(aq)14H+(aq)+6Fe2+(aq)+Cr2O2−7(aq)→6Fe3+(aq)+6e−→2Cr3+(aq)+7H2O(l)→6Fe3+(aq)+2Cr3+(aq)+7H2O(l)(22.10.10)(22.10.11)(22.10.12) Step 7: Check the balancing. In the above equation, there are 14H , 6Fe , 2Cr , and 7O on both sides. The net charge is 24+ on both sides. The equation is balanced.