Balancing Redox Equations - Some Worked Samples
T1:
magnesium reduces lead ions to lead metal
Oxidation:
Mg  Mg2+ + 2e–
Reduction:
Pb2+ + 2e–
 Pb
-------------------------------------------------------------------------------------------------Total Redox:
Mg +
Pb2+  Mg2+ + Pb
Note: Mg’s oxidation # changes from a 0 to a 2+ in the oxidation half reaction… this is a 2 electron loss
Pb’s oxidation # changes from a 2+ to a 0 in the reduction half reaction… this is a 2 electron gain
Solution provided by Alex, Caleb, Aaron, and Zeb
T2:
sulfur dioxide is oxidized to sulfate, iodine is reduced to iodide ions
(assume acidic solution)
Oxidation:
SO2 + 2 H2O
 SO42– + 2e– + 2 H+
Reduction:
I2 + 2e– +  2 I–
-------------------------------------------------------------------------------------------------Total Redox:
SO2 + 2 H2O + I2  SO42– + 2 H+ + 2 I–
Note: S’s oxidation # changes from a 4+ to a 6+ in the oxidation half reaction… this is a 2 electron loss
I’s oxidation # changes from a 0 to a 1– in the reduction half reaction… this is a 1 electron gain
per iodine atom
Solution provided by Lisa, Julia B, Bree, and Anna
T3:
hydrogen peroxide oxidizes iron(II) to iron(III) in acidic solution
Oxidation:
2(Fe2+
 Fe3+ + 1e–)
Reduction:
H2O2 + 2e– + 2 H+
 2 H2O
-------------------------------------------------------------------------------------------------Total Redox:
2 Fe2+ + H2O2 + 2 H+  2 Fe3+ + 2 H2O
Note: Fe’s oxidation # changes from a 2+ to a 3+ in the oxidation half reaction… this is a 1 electron loss
O’s oxidation # changes from a 1– to a 2– in the reduction half reaction… this is a 1 electron gain
per oxygen atom
Solution provided by Tessa, Lindsey, and Julia R
T4:
zinc reduces acidified dichromate ions to chromium(III)
Oxidation:
3(Zn  Zn2+ + 2e–)
Reduction:
Cr2O72– + 6e– + 14 H+  2 Cr3+ + 7 H2O
-------------------------------------------------------------------------------------------------Total Redox:
3 Zn + Cr2O72– + 14 H+  3 Zn2+ + 2 Cr3+ + 7 H2O
Note: Zn’s oxidation # changes from a 0 to a 2+ in the oxidation half reaction… this is a 2 electron loss
Cr’s oxidation # changes from a 6+ to a 3+ in the reduction half reaction… this is a 3 electron gain
per atom of chromium
Solution provided by Robby, Chris, Ariel, Stephanie, and Alex
T5:
acidified permanganate ions oxidize methanol to carbon dioxide and
water
Oxidation:
5(CH3OH + H2O  CO2 + 6 H+ + 6e–)
Reduction:
6(MnO4– + 5e– + 8 H+  Mn2+ + 4 H2O )
-------------------------------------------------------------------------------------------------Total Redox: 5 CH3OH + 6 MnO4– + 18 H+  5CO2 + 6 Mn2+ + 19 H2O
Note: C’s oxidation # changes from a 2– to a 4+ in the oxidation half reaction… this is a 6 electron loss
Mn’s oxidation # changes from a 7+ to a 2- in the reduction half reaction… this is a 5 electron gain
Solution provided by Tudor, Max, Christienne, and Sarah
T6:
chlorate ion (ClO3-) is oxidized to the perchlorate ion (ClO4-) and is
reduced to the chloride ion
Oxidation:
3(ClO3–

ClO4– + 2e–)
Reduction:
ClO3– + 6e–

Cl–
-------------------------------------------------------------------------------------------------Total Redox:
4 ClO3–

Cl– + 3 ClO4–
Note: Cl’s oxidation # changes from a 5+ to a 7+ in the oxidation half reaction… this is a 2 electron loss
Cl’s oxidation # changes from a 5+ to a 1- in the reduction half reaction… this is a 6 electron gain
Solution provided by Dan, Dominique, and Mike