Balancing redox reactions in basic solution
Problems 1 - 10

Examples #1-5 A list of the examples and problems (without any solutions)
Problems 11 - 30 Balancing redox reactions in acidic solution Return to Redox menu

Problem #1: OCN¯ + OCl¯ ---> CO32¯ + N2 + Cl¯

Solution:

On initial inspection, this problem seems like it might require three half-reactions. However, it only needs two half-reactions. This is because the oxidation number on the C does not change. The C is +4 in OCN¯ and also +4 in the carbonate.

1) Here are the two half-reactions:

OCN¯ ---> CO32¯ + N2
OCl¯ ---> Cl¯

2) Balance them as if in acid solution:

4H2O + 2OCN¯ ---> 2CO32¯ + N2 + 8H+ + 6e¯
2e¯ + 2H+ + OCl¯ ---> Cl¯ + H2O

3) Equalize the electrons:

4H2O + 2OCN¯ ---> 2CO32¯ + N2 + 8H+ + 6e¯
6e¯ + 6H+ + 3OCl¯ ---> 3Cl¯ + 3H2O

4) Add:

H2O + 2OCN¯ + 3OCl¯ ---> 2CO32¯ + N2 + 3Cl¯ + 2H+

5) Convert to basic:

2OH¯ + 2OCN¯ + 3OCl¯ ---> 2CO32¯ + N2 + 3Cl¯ + H2O

Problem #2: Dentrification in soils and oceans occurs when the nitrate ion is reduced to nitrous oxide by anaerobic bacteria in the presence of water. Oxygen and the hydroxyl ion are also produced during this process. Write a balanced net-ionic equation for this reaction.

Solution:

1) Reaction from the information in the question:

NO3¯ + H2O ---> N2O + O2 + OH¯

2) We balance by half-reactions (in basic solution):

NO3¯ ---> N2O
H2O ---> O2
8e¯ + 5H2O + 2NO3¯ ---> N2O + 10OH¯
4OH¯ ---> O2 + 2H2O + 4e¯

3) Equalize electrons:

8e¯ + 5H2O + 2NO3¯ ---> N2O + 10OH¯
8OH¯ ---> 2O2 + 4H2O + 8e¯

4) Add and eliminate electrons, water and hydroxide:

H2O + 2NO3¯ ---> N2O + 2O2 + 2OH¯

Note: I balanced the second half-reaction via the "balance in acid first" method:

H2O ---> O2

2H2O ---> O2

2H2O ---> O2 + 4H+ + 4e¯

Then, I added hydroxide to both sides:

4OH¯ + 2H2O ---> O2 + 4H2O + 4e¯

Then, I dropped the excess water:

4OH¯ ---> O2 + 2H2O + 4e¯


Problem #3: Al(s) + NO2¯ (aq) ---> AlO2¯ (aq) + NH3(aq)

Comment: ammonia is a base, consequently we balance in basic solution.

This equation is balanced as part of a presentation here.

1) Half reactions:

NO2¯ (aq) ---> NH3(aq)
Al(s) ---> AlO2¯ (aq)

2) Balance:

6e¯ + 7H+ + NO2¯ (aq) ---> NH3(aq) + 2H2O
2H2O + Al(s) ---> AlO2¯ (aq) + 4H+ + 3e¯

We could change to basic right now. I propose to simply leave it in acid and change over at the end.

3) Equalize the electrons:

6e¯ + 7H+ + NO2¯ (aq) ---> NH3(aq) + 2H2O
4H2O + 2Al(s) ---> 2AlO2¯ (aq) + 8H+ + 6e¯

4) Now add the half-reactions and eliminate excess items:

4H2O + 2Al(s) + 7H+ + NO2¯ (aq) ---> NH3(aq) + 2H2O + 2AlO2¯ (aq) + 8H+

2H2O + 2Al(s) + NO2¯ (aq) ---> NH3(aq) + 2AlO2¯ (aq) + H+

5) Change over to basic by adding one hydroxide, then eliminate one water:

OH¯ + H2O + 2Al(s) + NO2¯ (aq) ---> NH3(aq) + 2AlO2¯ (aq)

Problem #4: Cr(OH)3 + ClO3¯ ---> CrO42¯ + Cl¯

Solution:

Cr(OH)3 ---> CrO42¯
ClO3¯ ---> Cl¯

H2O + Cr(OH)3 ---> CrO42¯ + 5H+ + 3e¯
6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O

2H2O + 2Cr(OH)3 ---> 2CrO42¯ + 10H+ + 6e¯
6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O

2Cr(OH)3 + ClO3¯ ---> 2CrO42¯ + Cl¯ + 4H+ + H2O

Change over to basic solution by adding four hydroxides to each side:

2Cr(OH)3 + ClO3¯ + 4OH¯ ---> 2CrO42¯ + Cl¯ + 5H2O


Problem #5: Bi(OH)3 + SnO22- ----> SnO32- + Bi

What happens if one half-reaction is balanced in basic and one in acid, then the half-reactions are added before the one balanced in acid is converted to basic solution? The answer, of course, is nothing. You get the correct answer regardless of when you convert from acid to base.

Solution:

1) Half-reactions:

Bi(OH)3 ----> Bi
SnO22- ----> SnO32-

2) Balance them:

3e- + Bi(OH)3 ----> Bi + 3OH-
H2O + SnO22- ----> SnO32- + 2H+ + 2e-

Note that one is basic and one is acidic.

3) Equalize electrons:

6e- + 2Bi(OH)3 ----> 2Bi + 6OH-
3H2O + 3SnO22- ----> 3SnO32- + 6H+ + 6e-

4) Add:

3H2O + 3SnO22- + 2Bi(OH)3 ----> 2Bi + 6OH- + 3SnO32- + 6H+

5) Change to basic:

3H2O + 3SnO22- + 2Bi(OH)3 ----> 2Bi + 3SnO32- + 6H2O

6) Remove water for the final answer:

3SnO22- + 2Bi(OH)3 ----> 2Bi + 3SnO32- + 3H2O

7) If I had converted the second half-reaction in basic before adding, I would have this:

3e- + Bi(OH)3 ----> Bi + 3OH-
2OH- + SnO22- ----> SnO32- + H2O + 2e-

When you equalize the electrons, then add, there will be six hydroxide on each side and three water on the right. You remove the six hydroxide from each side and you will recover the balanced equation I have just above in step 6.


Problem #6: Ni(OH)2 + N2H4 ---> Ni + N2

Solution:

1) Separate into half-reactions:

Ni(OH)2 ---> Ni
N2H4 ---> N2

2) I propose to balance the first half-reaction in basic and the second in acid:

2e¯ + Ni(OH)2 ---> Ni + 2OH¯
N2H4 ---> N2 + 4H+ + 4e¯

3) Equalize electrons:

4e¯ + 2Ni(OH)2 ---> 2Ni + 4OH¯
N2H4 ---> N2 + 4H+ + 4e¯

4) Add:

2Ni(OH)2 + N2H4 ---> 2Ni + N2 + 4H2O

Note that I formed the four hydrogen ions and the four hydroxide ions into four waters.


Problem #7: Fe(OH)2 + H2O2 ---> Fe(OH)3

Solution:

Comment: this equation can be balanced by sight. Hint: think of the peroxide as two OH groups. You might want to try before going through the redox balancing below.

1) Half-reactions:

Fe(OH)2 ---> Fe(OH)3
H2O2 ---> H2O

2) Balance (one in basic, one in acidic):

OH¯ + Fe(OH)2 ---> Fe(OH)3 + e¯
2e¯ + 2H+ + H2O2 ---> 2H2O

3) Equalize electrons:

2OH¯ + 2Fe(OH)2 ---> 2Fe(OH)3 + 2e¯
2e¯ + 2H+ + H2O2 ---> 2H2O

4) Add:

2H+ + 2OH¯ + 2Fe(OH)2 + H2O2 ---> 2Fe(OH)3 + 2H2O

Note that the two hydrogen ions and the two hydroxide ion on the left-hand side form two waters.

5) Eliminate two waters from each side:

2Fe(OH)2 + H2O2 ---> 2Fe(OH)3

Problem #8: ClO2 + H2O + KOH ---> H2O + KClO2 + O2

Solution #1 (pick the oxygen in water to be reduced):

1) Half-reactions:

ClO2 ---> ClO2¯
H2O ---> O2

2) Balance as if in acid:

e¯ + ClO2 ---> ClO2¯
2H2O ---> O2 + 4H+ + 4e¯

3) Equalize electrons:

4e¯ + 4ClO2 ---> 4ClO2¯
2H2O ---> O2 + 4H+ + 4e¯

4) Add:

4ClO2 + 2H2O ---> 4ClO2¯ + O2 + 4H+

5) Change to basic:

4ClO2 + 4OH¯ ---> 4ClO2¯ + O2 + 2H2O

Solution #2 (pick the oxygen in hydroxide to be reduced):

1) Half-reactions:

ClO2 ---> ClO2¯
OH¯ ---> O2

2) Balance as if in acid:

e¯ + ClO2 ---> ClO2¯
2OH¯ ---> O2 + 2H+ + 4e¯

3) Equalize electrons:

4e¯ + 4ClO2 ---> 4ClO2¯
2OH¯ ---> O2 + 2H+ + 4e¯

4) Add:

4ClO2 + 2OH¯ ---> 4ClO2¯ + O2 + 2H+

5) Change to basic:

4ClO2 + 4OH¯ ---> 4ClO2¯ + O2 + 2H2O

The change back to molecular consists of adding four potassium ions:

4ClO2 + 4KOH ---> 4KClO2 + O2 + 2H2O

Problem #9: Cr3+ + H2O2 + OH¯ --> CrO42¯ + H2O

Solution:

1) Half-reactions:

Cr3+ ---> CrO42¯
H2O2 ---> H2O

2) I will balance as if in acidic solution, then change over to basic at the end.

4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯
2e¯ + 2H+ + H2O2 ---> 2H2O

3) Equalize electrons:

8H2O + 2Cr3+ ---> 2CrO42¯ + 16H+ + 6e¯
6e¯ + 6H+ + 3H2O2 ---> 6H2O

4) Add:

2H2O + 3H2O2 + 2Cr3+ ---> 2CrO42¯ + 10H+

5) Change to basic:

10OH¯ + 3H2O2 + 2Cr3+ ---> 2CrO42¯ + 8H2O

Problem #10: MnO4¯ + H2O + NO2¯ ---> MnO2(s) + NO3¯ + OH¯

Solution:

1) Half-reactions:

MnO4¯ ---> MnO2
NO2¯ ---> NO3¯

2) Balance in acidic solution (change to basic at the end):

3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
H2O + NO2¯ ---> NO3¯ + 2H+ + 2e¯

Equalize electrons:

6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O
3H2O + 3NO2¯ ---> 3NO3¯ + 6H+ + 6e¯

4) Add:

2H+ + 2MnO4¯ + 3NO2¯ ---> 2MnO2 + 3NO3¯ + H2O

5) Change to basic by two hydroxides to each side:

H2O + 2MnO4¯ + 3NO2¯ ---> 2MnO2 + 3NO3¯ + 2OH¯

Examples #1-5 A list of the examples and problems (without any solutions)
Problems 11 - 30 Balancing redox reactions in acidic solution Return to Redox menu