Balance the following REDOX reactions in an acidic solution:
1.
2.
3.
4.
5.
ClO3¯ + SO2 ---> SO42¯ + Cl¯
H2S + NO3¯ ---> S8 + NO
MnO4¯ + H2S ---> Mn2+ + S8
Cu + SO42¯ ---> Cu2+ + SO2
MnO4¯ + CH3OH ---> HCOOH + Mn2+
Balance the following REDOX reactions in a basic solution:
1.
2.
3.
4.
5.
NH3 + ClO¯ ---> N2H4 + Cl¯
Au + O2 + CN¯ ---> Au(CN)2¯ + H2O2
Br¯ + MnO4¯ ---> MnO2 + BrO3¯
AlH4¯ + H2CO ---> Al3+ + CH3OH
Se + Cr(OH)3 ---> Cr + SeO32¯
Acidic Solution
Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯
Solution:
1) Split into unbalanced half-reactions:
ClO3¯ ---> Cl¯
SO2 ---> SO42¯
2) Balance the half-reactions:
6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯
3) Make the number of electrons equal:
6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 3
4) Add the two half-reactions for the final answer:
ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+
Note that items duplicated on each side were cancelled out. The duplicates are 6e¯,
3H2O, and 6H+
Example #2: H2S + NO3¯ ---> S8 + NO
Solution:
1) The unbalanced half-reactions:
H2S ---> S8
NO3¯ ---> NO
2) balance each half-reaction:
8H2S ---> S8 + 16H+ + 16e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O
3) Make the number of electrons equal:
24H2S ---> 3S8 + 48H+ + 48e¯ <--- multiplied by a factor of 3
48e¯ + 64H+ + 16NO3¯ ---> 16NO + 32H2O <--- multiplied by a factor of 16
Note that 16 and 3 have no common factors except 1, so both 16 and 3 had to be used
to obtain the lowest common multiple of 48 for the number of electrons.
4) Add:
24H2S + 16H+ + 16NO3¯ ---> 3S8 + 16NO + 32H2O
Comment: removing a factor of 8 does look tempting, doesn't it? However, the three
in front of the S8 (or the five in the next example) makes it impossible. Also, note that
duplicates of 48 electrons and 48 hydrogen ions were removed.
Example #3: MnO4¯ + H2S ---> Mn2+ + S8
Solution:
1) Half-reactions:
H2S ---> S8
MnO4¯ ---> Mn2+
2) Balance:
8H2S ---> S8 + 16H+ + 16e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
3) Make the number of electrons equal (note that there are no common factors
between 5 and 16 except 1):
40H2S ---> 5S8 + 80H+ + 80e¯ <--- factor of 5
80e¯ + 128H+ + 16MnO4¯ ---> 16Mn2+ + 64H2O <--- factor of 16
4) The final answer:
40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O
Another possibility of removing a factor of 8 destroyed by an odd number, in this
case, the 5 in front of the S8. Curses, foiled again!
Example #4: Cu + SO42¯ ---> Cu2+ + SO2
1) The unbalanced half-reactions:
Cu ---> Cu2+
SO42¯ ---> SO2
2) The balanced half-reactions:
Cu ---> Cu2+ + 2e¯
2e¯ + 4H+ + SO42¯ ---> SO2 + 2H2O
3) The final answer:
Cu + 4H+ + SO42¯ ---> Cu2+ + SO2 + 2H2O
No need to equalize electrons since it turns out that, in the course of balancing the
half-reactions, the electrons are equal in amount. Note how easy it was to balance the
copper half-reaction. All you needed were the two electrons.
Example #5: MnO4¯ + CH3OH ---> HCOOH + Mn2+
1) The balanced half-reactions:
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + CH3OH ---> HCOOH + 4H+ + 4e¯
2) Equalize electrons:
20e¯ + 32H+ + 4MnO4¯ ---> 4Mn2+ + 16H2O <--- factor of 4
5H2O + 5CH3OH ---> 5HCOOH + 20H+ + 20e¯ <--- factor of 5
3) The final answer:
12H+ + 5CH3OH + 4MnO4¯ ---> 5HCOOH + 4Mn2+ + 11H2O
Basic Solution
Example #1: NH3 + ClO¯ ---> N2H4 + Cl¯
Solution:
1) The two half-reactions, balanced as if in acidic solution:
2NH3 ---> N2H4 + 2H+ + 2e¯
2e¯ + 2H+ + ClO¯ ---> Cl¯ + H2O
2) Electrons already equal, convert to basic solution:
2OH¯ + 2NH3 ---> N2H4 + 2H2O + 2e¯
2e¯ + 2H2O + ClO¯ ---> Cl¯ + H2O + 2OH¯
Comment: that's 2 OH¯, not 20 H¯. Misreading the O in OH as a zero is a common
mistake.
3) The final answer:
2HN3 + ClO¯ ---> N2H4 + Cl¯ + H2O
Notice that no hydroxide appears in the final answer. That means this is a basecatalyzed reaction. For the reaction to occur, the solution must be basic and hydroxide
IS consumed. It is just regenerated in the exact same amount, so it cancels out in the
final answer.
Example #2: Au + O2 + CN¯ ---> Au(CN)2¯ + H2O2
Solution:
1) the two half-reactions, balanced as if in acidic solution:
2CN¯ + Au ---> Au(CN)2¯ + e¯
2e¯ + 2H+ + O2 ---> H2O2
2) Make electrons equal, convert to basic solution:
4CN¯ + 2Au ---> 2Au(CN)2¯ + 2e¯ <--- multiplied by a factor of 2
2e¯ + 2H2O + O2 ---> H2O2 + 2OH¯
3) The final answer:
4CN¯ + 2Au + 2H2O + O2 ---> 2Au(CN)2¯ + H2O2 + 2OH¯
Comment: the CN¯ is neither reduced nor oxidized, but it is necessary for the
reaction. For example, you might see this way of writing the problem:
Au + O2 ---> Au(CN)2¯ + H2O2
Notice that CN¯ does not appear on the left side, but does so on the right. Since you
MUST balance the equation, that means you are allowed to use CN¯ in your
balancing. An important point here is that you know the cyanide polyatomic ion has a
negative one charge.
Example #3: Br¯ + MnO4¯ ---> MnO2 + BrO3¯
Solution:
1) The two half-reactions, balanced as if in acidic solution:
3H2O + Br¯ ---> BrO3¯ + 6H+ + 6e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
2) Make the number of electrons equal:
3H2O + Br¯ ---> BrO3¯ + 6H+ + 6e¯
6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O <--- multiplied by a factor of 2
3) Convert to basic solution, by adding 6OH¯ to the first half-reaction and 8OH¯ to
the second:
6OH¯ + Br¯ ---> BrO3¯ + 3H2O + 6e¯
6e¯ + 4H2O + 2MnO4¯ ---> 2MnO2 + 8OH¯
4) The final answer:
H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + 2OH¯
5) What happens if you add the two half-reactions without converting them to basic?
You get this:
2H+ + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + H2O
Then, add 2OH¯ to each side:
2H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + H2O + 2OH¯
Eliminate one water for the final answer:
H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + 2OH¯
The answer to the question? Nothing happens. You get the right answer if convert
before adding the half-reactions or after. There will even be cases where balancing
one half-reaction using hydroxide can easily be done while the other half-reaction gets
balanced in acidic solution before converting. You can add the two half-reactions
while one is basic and one is acidic, then convert after the adding (see below for an
example of this).
Example #4: AlH4¯ + H2CO ---> Al3+ + CH3OH
Solution:
1) The two half-reactions, balanced as if in acidic solution:
AlH4¯ ---> Al3+ + 4H+ + 8e¯
2e¯ + 2H+ + H2CO ---> CH3OH
2) Converted to basic by addition of hydroxide, second half-reaction multiplied by 4
(note that the hydrogen is oxidized from -1 to +1):
4OH¯ + AlH4¯ ---> Al3+ + 4H2O + 8e¯
8e¯ + 8H2O + 4H2CO ---> 4CH3OH + 8OH¯
3) The final answer:
AlH4¯ + 4H2O + 4H2CO ---> Al3+ + 4CH3OH + 4OH¯
Example #5: Se + Cr(OH)3 ---> Cr + SeO32¯
Solution:
1) The unbalanced half-reactions:
Se ---> SeO32¯
Cr(OH)3 ---> Cr
2) Note that only the first half-reaction is balanced using the balance-first-in-acid
technique, the second is balanced using hydroxide:
Se + 3H2O ---> SeO32¯ + 6H+ + 4e¯
3e¯ + Cr(OH)3 ---> Cr + 3OH¯
3) Convert the first half-reaction by adding 6 hydroxide to each side, eliminate
duplicate waters, then make the electrons equal (factor of 3 for the first half-reaction
and a factor of 4 for the second). The final answer:
6OH¯ + 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO32¯ + 9H2O
4) What would happen if we didn't make the first half-reaction basic and just added
them?
first, make the electrons equal:
3Se + 9H2O ---> 3SeO32¯ + 18H+ + 12e¯
12e¯ + 4Cr(OH)3 ---> 4Cr + 12OH¯
then, add:
3Se + 4Cr(OH)3 + 9H2O ---> 4Cr + 3SeO32¯ + 18H+ + 12OH¯
combine hydrogen ion and hydroxide ion on the right-hand side:
3Se + 4Cr(OH)3 + 9H2O ---> 4Cr + 3SeO32¯ + 6H+ + 12H2O
eliminate water:
3Se + 4Cr(OH)3 ---> 4Cr + 3SeO32¯ + 6H+ + 3H2O
add six hydroxides:
6OH¯ + 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO32¯ + 9H2O
Note that I combined the H+ and the OH¯ to make six waters and then added it to the
three waters that were already there.