Transcript Balancing Oxidation-Reduction Reactions A Short Primer Oxidation-Reduction Reactions Oxidation-reduction (redox) reactions can be difficult to balance because not only must mass be balanced, so must the.

Balancing
Oxidation-Reduction
Reactions
A Short Primer
Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions can be difficult to balance
because not only must mass be balanced, so must the electrons passed
between species.
Presented here is an algorithm for balancing redox reactions quickly
using oxidation numbers. The method revolves around balancing
electron movement first, then balancing the rest of the equation without
changing the species which undergo oxidation or reduction.
It is important to apply this algorithm only to redox reactions and only
when the equation cannot be balanced by inspection alone. The
algorithm is not necessary, nor will it work for non-redox reactions.
©D.B.Green, 2000,2002
The Algorithm
1. Write the equation from words (if necessary). Ignore spectator ions (i.e., write
the net ionic equation.
2. Assign oxidation numbers to the atoms which undergo a change in oxidation
number. Connect identical atoms across the arrow with a line and place on the
line the number of electrons gained or lost by the species.
3. If necessary, adjust the stoichiometric ratios of the species. This is not a final
balance, just get the ratios correct. An illustration of this “pre-balance” is
shown in the example problems.
The following steps 4, 5, and 6 must be done in order.
©D.B.Green, 2000,2002
The Algorithm
4. Electron Balance: Balance the electrons to their least-common-multiple with
multipliers. Multiply the stoichiometric coefficients for the species connected
by the lines with the same multipliers.
5. Charge Balance: Balance the ionic charges on the left and right side of the
arrow with either H+ or OH- such that the sum of all the ionic charges on the
left and right are equal. Unless otherwise specified, usually (but not always),
hydrogen ion is used when transition metals are involved in the reaction and
hydroxide is used when nonmetals are the only species in the reaction. If there
are no ions in the reaction, this step is skipped.
6. Mass Balance: Balance the addition of H+ or OH- with H2O as needed.
7. Check the overall balance. If everything is correct, place the spectator ions
back into the equation as necessary.
©D.B.Green, 2000,2002
Review of Oxidation Numbers
 The oxidation number of an element in its most stable form at room
temperature is zero.
 The oxidation number of a monatomic ion is its ionic charge.
 Except in hydrides, the oxidation number of hydrogen in a compound
or polyatomic ion is +1.
 Except in peroxides and superoxides, the oxidation number of oxygen
in a compound or polyatomic ion is -2.
(O22- O = -1; O2- O = -1/2)
 The sum of the oxidation numbers in a molecule is zero and is the
ionic charge in a polyatomic ion.
©D.B.Green, 2000,2002
Example 1
Permanganate ion reacts with iron(II) ion to form manganese(II) ion
and iron(III) ion. Write and balance the equation.
Step 1. Write the equation from words:
MnO4- + Fe2+  Fe3+ + Mn2+
Step 2. No “pre-balance” is necessary, so….
Step 3. Assign oxidation numbers:
(+7)
(+2)
(+3)
(+2)
MnO4- + Fe2+  Fe3+ + Mn2+
©D.B.Green, 2000,2002
Example 1
(Step 3 cont’d)
And connect with a line identical atoms which undergo an oxidation state change:
+5 e(+7)
(+2)
(+3)
(+2)
MnO4- + Fe2+  Fe3+ + Mn2+
-1 e-
Step 4. Balance electrons and multiply species coefficients by the same factors:
(+5 e-) x 1
(+7)
MnO4- +
(+2)
(+3)
5 Fe2+

5 Fe3+
(+2)
+ Mn2+
(-1 e-) x 5
©D.B.Green, 2000,2002
Example 1
Step 5. Charge balance with acid (since transition metals are involved) to get
equal ionic charges on the left and right:
Ionic charge: -1 + 5(+2) = +9 (on left)
MnO4- + 5 Fe2+ +
5(+3) + +2 = +17 (on right)
8 H+
 5 Fe3+ + Mn2+
Step 6. Finally, mass balance with water on the right:
MnO4- + 5 Fe2+ + 8 H+  5 Fe3+ + Mn2+ +
4 H2O
©D.B.Green, 2000,2002
Example 1
Step 7. Perhaps the permanganate was prepared from the potassium salt, the
iron(II) from the chloride salt, and the reaction was performed in
hydrochloric acid solution:
KMnO4 + 5 FeCl2 + 8 HCl  5 FeCl3 + MnCl2 + 4 H2O + KCl
©D.B.Green, 2000,2002
Example 2
Chromium(III) ion can be converted to dichromate ion by treatment
with potassium perchlorate in acid solution. Perchlorate ion is
converted to chloride ion. Write and balance the chemical equation.
Cr3+ + ClO4-  Cr2O7-2 + Cl-
Step 1.
Step 2. “Pre-balance” the chromium:
2 Cr3+ +
ClO4-  Cr2O7-2 + Cl-
-2(3 e-) = -6 e-
Step 3.
(+3)
2
Cr3+
(+7)
(+6)2
+ ClO4  Cr2O7
-
(-1)
-2
+ Cl-
+8 e-
©D.B.Green, 2000,2002
Example 2
Step 4. Balance electrons:
(+3)
-6 e- x 4 = 24
(+7)
82 Cr3+ + 3 ClO4-
(+6)2

4 Cr2O7-2
(-1)
+
3 Cl-
+8 e- x 3 = 24
Step 5. Charge balance:
Ionic charge:
8(+3) + 3(-1) = +21 (left)
4(-2) + 3(-1) = -11 (right)
8 Cr3+ + 3 ClO4-  4 Cr2O7-2 + 3 Cl8 Cr3+ + 3 ClO4-  4 Cr2O7-2 + 3 Cl- +
?? H+
32
Step 6. Mass balance:
8 Cr3+ + 3 ClO4- + 16
?? H2O  4 Cr2O7-2 + 3 Cl- + 32 H+
©D.B.Green, 2000,2002
Example 2
Step 7. Check the balance and replace the spectators. For this example,
assume that the anionic spectator is chloride.
8 CrCl3 + 3 KClO4 + 16 H2O  4 K2Cr2O7 + 3 KCl + 32 HCl
©D.B.Green, 2000,2002
Problems
Solutions are provided on the next page. Resist the temptation to peek
at the answers until you have completed balancing the equations.
 MnO4- + SO2  Mn2+ + HSO4 Copper metal reacts in nitric acid to produce copper(II) ion and
nitrogen monoxide gas.
 C + H2SO4  CO2 + SO2
 Molecular bromine disproportionates into bromate and bromide in
basic solution.
 Bi(OH)3 + SnO22-  SnO32- + Bi (in basic solution)
©D.B.Green, 2000,2002
Solutions to Problems
 2 MnO4- + 5 SO2 + H+ + 2 H2O  2 Mn+ + 5 HSO4 3 Cu + 2 HNO3 + 6 H+  3 Cu+2 + 2 NO + 4 H2O
 C + 2 H2SO4  CO2 + 2 SO2 + 2 H2O
 6 Br2 + 12 OH-  2 BrO3- + 10 Br- + 6 H2O
 2 Bi(OH)3 + 3 SnO22-  3 SnO32- + 2 Bi + 3 H2O
©D.B.Green, 2000,2002