BALANCING REDOX REACTIONS
IN ACIDIC AND BASIC SOLUTIONS
Redox reactions are generally carried out in solutions
containing an excess of H+ ions (acidic) or OH- ions (basic).
This is done to provide sufficient sources of excess
electrons to transfer during the gain and lose of electrons.
Balancing Redox Equations in Acidic Solutions
Given the following skeleton equation:
Cr2O7 + Fe+2
 Cr+3 + Fe+3
STEP 1: Divide the skeleton equation into half-reactions.
Except for any hydrogen and oxygen atoms present, the
same elements must appear on both sides of the halfreactions.
Cr2O7-2  Cr+3
Fe+2
 Fe+3
STEP 2: Balance the atoms other than H and O.
Cr2O7-2  2 Cr+3
Fe+2
 Fe+3
STEP 3: Balance oxygen by adding H2O to the side
needing oxygen.
Cr2O7-2  Cr+3 + 7 H2O
Fe+2
 Fe+3
STEP 4: Balance hydrogen by adding H+ to the side
needing hydrogen.
14 H+ + Cr2O7-2  Cr+3 + 7 H2O
Fe+2
 Fe+3
STEP 5: Balance the charge by finding the electrons
gained and lost by the oxidizing and reducing
agents.
First compute the net charge on each side of the reaction:
14 H+ + Cr2O7-2  2 Cr+3 + 7 H2O
(+14) + (-2) = +12
2(+3) + 0 = +6
Fe+2  Fe+3
+2
+3
Then balance using electrons:
14 H+ + Cr2O7-2 + 6 e-  2 Cr+3 + 7 H2O
Fe+2 – 1 e-  Fe+3
(+14) + (-2) + (-6) = +6
(+2)–(-1)=+3 +3
2(+3) + 0 = +6
STEP 6: Use coefficients to multiply the half-reactions
so that the number of electrons gained is equal
to the number of electrons lost.
1 x (14 H+ + Cr2O7-2 + 6 e-  2 Cr+3 + 7 H2O)
6 x (Fe+2 – 1 e-  Fe+3)
=
=
14 H+ + Cr2O7-2 + 6 e-  2 Cr+3 + 7 H2O
6 Fe+2 – 6 e-  6 Fe+3
STEP 7: Add the half-reactions together.
(Cancel anything that appears on both sides
of the equation.)
14 H+ + Cr2O7-2 + 6 e-  2 Cr+3 + 7 H2O
6 Fe+2 – 6 e-  6 Fe+3
14 H+ + Cr2O7-2 + 6 Fe+2  6 Fe+3 + 2 Cr+3 + 7 H2O
PRACTICE PROBLEMS: Balance the following redox reactions
occuring in acidic solution.
1. SO4-2 + Zn  Zn+2 + SO2
2. Cu + NO3-1  Cu+2 + NO
3. S2O3-2 + OCl-1  Cl-1 + S4O6-2
4. IO3-1 + AsO3-3  I-1 + AsO4-3
5. Mn+2 + BiO3-1  MnO4-1 + Bi+3
6. PbO2 + Cl-1  PbCl2 + Cl2
7. Ag + NO3-1  NO2 + Ag+1
8. Cr+3 + BiO-1
 Cr2O7-2 + Bi+3
9. HNO2 + I-1  I2 + NO
10. C2O4-2 + HNO2  CO2 + NO
Balancing Redox Equations in Basic Solutions
In basic solutions the H+ concentration is very small. The dominant
species are OH- and H2O. Strictly speaking, these should be used to
balance the half reactions. However, the simplest way to obtain a
balanced equation is to first pretend that the solution is acidic. We
therefore balance the equation using the seven steps described
above, and then we use a simple three-step procedure to convert the
equation to the correct form for a basic solution.
Example:
SO3-2 + MnO4-1  SO4-2 + MnO2
(Basic)
After following the seven steps above for acidic
solutions the equation would be:
2 H+1 + 3 SO3-2 + 2 MnO4-1  3 SO4-2 + 2 MnO2 + H2O
Conversion of this equation to one appropriate for a
basic solution uses the fact that H+ and OH- react in
a 1:1 ratio to give H2O. The following steps show this:
STEP 8: Add the same number of H+ as there are OHto BOTH sides of the equation.
2 OH- + 2 H+1 + 3SO3-2 + 2MnO4-1  3SO4-2 + 2MnO2 + H2O + 2 OH-
STEP 9: Combine H+ and OH- to form H2O.
2 H2O + 3 SO3-2 + 2 MnO4-1 3 SO4-2 + 2 MnO2 + H2O + 2 OHSTEP 10: Cancel any H2O that appears on both sides
of the equation.
H2O + 3 SO3-2 + 2 MnO4-1  3 SO4-2 + 2 MnO2 + 2 OH-
PRACTICE PROBLEMS: Balance the following redox reactions
occuring in basic solution.
1. CrO4-2 + S-2  S + CrO2-1
2. ClO3-1
+ N2H4  NO + Cl-1
3. NiO2 + Mn(OH)2  Mn2O3 + Ni(OH)2
4. SO3-2
+ MnO4-1  SO4-2
+ MnO2
5. CrO2-1 + S2O8-2  CrO4 + SO4-2
6. Au + CN-1 + O2  Au(CN)4 + OH-1