Download Balancing Redox Reactions

A. Romero 2017
Balancing Redox Reactions
CHEM 1A/B
Steps for balancing redox reactions with the ½ reaction method:
Be sure the reaction is redox
Look at the oxidation numbers for the atoms in the reaction. The oxidation numbers of some
elements must increase, and others must decrease as reactants go to products.
Write the two redox ½ reactions
Identify what is being oxidized (losing electrons) and what is being reduced (gaining electrons).
Split the reaction into an oxidation ½ reaction and a reduction ½ reaction.
Balance both ½ reactions
For each ½ reaction, balance all elements other than O and H as you would in a normal chemical
equation.
Balance the O by adding the appropriate number of H2O molecules to the other side of the
equation.
Balance H by adding the appropriate number of protons (H+) to the other side of the equation.
If the reaction is occurring in a basic solution, add enough hydroxide ions (OH–) to both sides of
the ½ reaction to neutralize any H+. The side that formerly had the H+ now has water molecules
(H+ + OH–  H2O), and the other side of the equation has excess OH– anions.
Balance any difference in charge by adding electrons (e–) to the more positive side of the ½
reaction.
Look at the number of electrons in the balanced oxidation and reduction ½ reactions. If the
numbers of electrons are not the same, multiply each equation by the smallest whole number that
will make the number of electrons in the two ½ reactions equal. Be sure to multiply all the
coefficients in the equation by this number, not just the electrons.
Add the ½ reactions together and cancel
Add the oxidation ½ reaction and reduction ½ reaction together, and cancel out anything that
appears on both sides of the equation.
Check
Make sure that the number of each type of atom and the total charge on each side are balanced.
COLOR CODING KEY:
Balancing Redox Reactions using the ½ Reaction Method
Steps:
For each ½ reaction:
 Split overall equation into two ½ reactions
 Balance elements other than O & H
 Balance O with H2O
 Balance H with H+
 If in basic solution, add enough OH− to both sides to neutralize the H+
 Balance charge with e–
 Multiply through to make # of e− in the two ½ reactions match
 Add ½ reactions back together
 Cancel out species that appear on both sides
 Check both sides of the equation for same # atoms and overall charge
Example #1:
In the following unbalanced redox equation, the oxidation number of each atom has been
labeled, the oxidizing and reducing agents have been identified, and the reaction has been
balanced using the ½ reaction method.
ox
red
As2O3(s) + NO3–(aq)  H3AsO4(aq) + NO(g)
(+3)2 (−2)3
+5 (−2)3
(+1)3 +5 (−2)4
+2 −2
−
N
Element being reduced: ______________
NO3
Oxidizing Agent: ______________
As
Element being oxidized: ______________
As2O3
Reducing Agent: ______________
Reduction ½ Reaction:
Oxidation ½ Reaction:
 unbalanced 
As2O3  2 H3AsO4
NO3−  NO
+ 5 H2 O
+ 2 H2 O
+4H
+
+ 4 e−
+4H
+
+ 3 e−
x3
3 As2O3 + 15 H2O  6 H3AsO4 + 12 H+ + 12 e−
x4
4 NO3− + 16 H+ + 12 e−  4 NO + 8 H2O
Add & Cancel
7
4
−
3 As2O3 + 15 H2O + 4 NO3 + 16 H+ + 12 e−  6 H3AsO4 + 12 H+ + 12 e− + 4 NO + 8 H2O
Balanced Redox Equation:
3 As2O3 + 7 H2O + 4 NO3− + 4 H+  6 H3AsO4 + 4 NO
Check:
As 6
O 28
H 18
N 4
charge 0





Example #2: Occurring in basic solution
In the following unbalanced redox equation, the oxidation number of each atom has been
labeled, the oxidizing and reducing agents have been identified, and the reaction has been
balanced using the ½ reaction method.
red
ox
NO2–(aq) + Al(s)  NH3(g) + AlO2–(aq)
+3 (−2)2
0
−3 (+1)3
+3 (−2)2
−
N
Element being reduced: ______________
NO2
Oxidizing Agent: ______________
Al
Element being oxidized: ______________
Al
Reducing Agent: ______________
Oxidation ½ Reaction:
Reduction ½ Reaction:
Al  AlO2−
NO2−  NH3
+ 2 H2 O
+ 2 H2 O
+
+ 4 OH
−
+
+4H
+ 4 OH−
+7H
+ 7 OH−
5
7 H2O
2
4 H2O
+ 3 e−
x2
2 Al + 8 OH−  2 AlO2− + 4 H2O + 6 e−
+ 7 OH−
+ 6 e−
x1
NO2− + 5 H2O + 6 e−  NH3 + 7 OH−
1
1
2 Al + 8 OH− + NO2− + 5 H2O + 6 e−  2 AlO2− + 4 H2O + 6 e− + NH3 + 7 OH−
Balanced Redox Equation:
2 Al + OH− + NO2− + H2O  2 AlO2− + NH3
Check:
Al 2
O 4
H 3
N 1
charge 2−




