Transcript 20120815105143

Balancing
Acidic Redox
Reactions
Step 1: Assign oxidation numbers
to all elements in the reaction.
+7 2
+4 2
+2
MnO41 + SO2  Mn+2 +
+6 2
SO42
Step 2: List the changes in
oxidation numbers.
+7 2
+4 2
+6 2
+2
MnO41 + SO2  Mn+2 + SO42
Mn
+7  +2
S
+4  +6
change
5
+2
Step 3: Label the species being
oxidized and reduced.
reduced
oxidized
Mn +7  +2
S
+4  +6
change
5
+2
Step 4: Label the oxidizing and
reducing agents.
MnO4 + SO2 
1
oxidizing
agent
+2
Mn +
2
SO4
reducing
agent
change
reduced
Mn +7  +2
5
oxidized
S
+4  +6
+2
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.
change
reduced Mn +7  +2 5 2 = 10
oxidized S +4  +6 +2 5 = +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.
It is necessary to multiply the
manganeses by 2 and the sulfurs by 5.
1 +
+2 +
2
MnO
SO

Mn
SO
2
5 2
2
5
4
4
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 MnO41 + 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 MnO41 + 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 MnO41 + 5 SO2  2 Mn+2 + 5 SO42
2  6
2  2
+ 4 H+1
2 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1
Step 9: Count the hydrogens and
oxygens on each side of the equation.
2 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1
0 H
18 O
 4 H
 20 O
Step 10: Balance the hydrogens and
oxygens by adding water molecules.
2 MnO41 + 5 SO2  2 Mn+2 + 5 SO42 + 4 H+1
0 H
2 H2O + 18 O
 4 H
 20 O
Step 11: Re-write the equation and
box the entire balanced reaction.
2 MnO41 + 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.
MnO41 + SO2  Mn+2 + SO42
MnO41  Mn+2
SO2  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!!
8
+1
H
+ MnO41  Mn+2 + 4 H2O
2
+1
SO

SO
2 H2O +
+4H
2
4
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:
5 e1 + 8 H+1 + MnO41  Mn+2 + 4 H2O
(+8) + (1)  (+2)
(+7)  (+2)
2 H2O + SO2  SO42 + 4 H+1 + 2 e1
(0)  (2) + (+4)
(0)  (+2)
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.)
105 e1 + 16
8 H+1 + 2 MnO41  2 Mn+2 + 48 H2O
(+8) + (1)  (+2)
(+7)  (+2)
2 + 20
+1
1
5
5
2
H
O
+
SO

SO
4
H
10 2
+
10
2
e
2
4
(0)  (2) + (+4)
(0)  (+2)
Step 5: Add the two half-reactions.
10 e 1 + 16 H+1 + 2 MnO41  2 Mn+2 + 8 H2O
10 H2O + 5 SO2  5 SO42 + 20 H+1 + 10 e 1
16 H+1 + 2 MnO41 +10 H2O + 5 SO2 
2 Mn+2 + 8 H2O + 5 SO42 + 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 MnO41 +10 H2O + 5 SO2 
2 Mn+2 + 8 H2O + 5 SO42 + 20 H+1
becomes
2 MnO41 +2 H2O + 5 SO2 
2 Mn+2 + 5 SO42 + 4 H+1
Balance the following using the half-reaction method:
a. Br 1 + MnO41  Br2 + Mn+2
16 H+1 + 10 Br 1 + 2 MnO41  5 Br2 + 2 Mn+2 + 8 H2O
b. As2O3 + NO31  H3AsO4 + NO
4 H+1 + 7 H2O + 4 NO3 1 + 3 As2O3  6 H3AsO4 + 4 NO