Download - Balancing Acidic Redox Reactions
Balancing Acidic Redox
Reactions
Step 1: Assign oxidation numbers to all elements in the reaction.
MnO41 + SO2 Mn+2 + SO4
22 2 2+2+4+7 +6
Step 2: List the changes in oxidation numbers.
MnO41 + SO2 Mn+2 + SO4
22 2 2+2+4+7 +6
Mn +7 +2
S
change
5
+4 +6 +2
Step 3: Label the species being oxidized and reduced.
Mn +7 +2 5change
S +4 +6 +2oxidized
reduced
Step 4: Label the oxidizing and reducing agents.
Mn +7 +2 5change
S +4 +6 +2oxidized
reduced
MnO41 + SO2 Mn+2 + SO4
2
oxidizing agent
reducing agent
Step 5: Balance the change. This is done by multiplying each change by a
value to attain the least common multiple of the two numbers.
reduced Mn +7 +2 5change
oxidized S +4 +6 +2
25
=
=
10
+10
0
Step 6: Using the values chosen to balance the change, add coefficients to the particular species. Note: take into
account any subscripts present.
MnO41 + SO2 Mn+2 + SO4
25 52 2
It is necessary to multiply the manganeses by 2 and the sulfurs by 5.
Step 7: Make sure that all elements (with the exception of hydrogen and
oxygen) are balanced. Add coefficients as necessary to balance extra elements.
Note: If hydrogen or oxygen is the species being oxidized or reduced, it
must be balanced at this step.
2 MnO41 + 5 SO2 2 Mn+2 + 5 SO4
2
Step 8: Balance the charge. Part a: Multiply the coefficient by the
charge on the ion or molecule.
2 MnO41 + 5 SO2 2 Mn+2 + 5 SO4
2
(2)(1) + (5)(0) (2)(+2) + (5)(2)
(2) + (0) (+4) + (10)2 6
Step 8: Balance the charge. Part b: Add hydrogen ions to account
for the extra charges.
2 MnO41 + 5 SO2 2 Mn+2 + 5 SO4
2
2 6 + 4 H+1
2 22 MnO4
1 + 5 SO2 2 Mn+2 + 5 SO42 + 4 H+1
Step 9: Count the hydrogens and oxygens on each side of the equation.
2 MnO41 + 5 SO2 2 Mn+2 + 5 SO4
2 + 4 H+1
H HO O
0 418 20
Step 10: Balance the hydrogens and oxygens by adding water molecules.
2 MnO41 + 5 SO2 2 Mn+2 + 5 SO4
2 + 4 H+1
0 H 4 H18 O 20 O2 H2O +
Step 11: Re-write the equation and box the entire balanced reaction.
2 MnO41 + 5 SO2 + 2 H2O 2 Mn+2 + 5 SO4
2 + 4 H+1
It will be preferable for you to use the following method –
called the half-reaction method of balancing redox equations. Given your prior knowledge
and understanding, this will be much easier!
Step 1: Given the reaction to balance, separate the two half-reactions.
MnO41 Mn+2
SO2 SO42
MnO41 + SO2 Mn+2 + SO4
2
Step 2: Balance all of the atoms except H and O. For an acidic solution, next add H2O to balance the O atoms and
H+1 to balance the H atoms. In a basic solution, we would use OH-1 and H2O to balance the O and H.
Be careful: On this example the atoms except H and O are already balanced. Most of the time, they won’t already be
balanced. WATCH OUT. Do that first!!
MnO41 Mn+2
SO2 SO42
+ 4 H2O8 H+1 +
2 H2O + + 4 H+1
Step 3: Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is
accomplished by adding electrons to the reactions:
8 H+1 + MnO41 Mn+2 + 4 H2O
(+8) + (1) (+2) (+7) (+2)
2 H2O + SO2 SO42 + 4 H+1
(0) (2) + (+4) (0) (+2)
5 e1 +
+ 2 e1
Step 4: Now multiply the oxidations numbers so that the two half-reactions will have the same number of electrons
and can cancel each other out:(Remember the LCM?? Multiply the first reaction by 2
and the second reaction by 5.)
8 H+1 + MnO41 Mn+2 + 4 H2O
(+8) + (1) (+2) (+7) (+2)
2 H2O + SO2 SO42 + 4 H+1
(0) (2) + (+4) (0) (+2)
5 e1 +
+ 2 e1
10 16 2 2 8
10 5 5 20 10
Step 5: Add the two half-reactions.
10 e 1 + 16 H+1 + 2 MnO41 2 Mn+2 + 8 H2O
10 H2O + 5 SO2 5 SO4
2 + 20 H+1 + 10 e 1
16 H+1 + 2 MnO41 +10 H2O + 5 SO2
2 Mn+2 + 8 H2O + 5 SO42 + 20 H+1
Step 6: Get the overall equation by canceling out the electrons and H2O, H+1, and OH-1 that may appear on both
sides of the equation:
16 H+1 + 2 MnO41 +10 H2O + 5 SO2
2 Mn+2 + 8 H2O + 5 SO42 + 20 H+1
becomes
2 MnO41 +2 H2O + 5 SO2
2 Mn+2 + 5 SO42 + 4 H+1
Balance the following using the half-reaction method:
a. Br 1 + MnO41 Br2 + Mn+2
b. As2O3 + NO31 H3AsO4 + NO
16 H+1 + 10 Br 1 + 2 MnO41 5 Br2 + 2 Mn+2 + 8 H2O
4 H+1 + 7 H2O + 4 NO3 1 + 3 As2O3 6 H3AsO4 + 4 NO
