What if there is only one reactant or product?
How do you balance the reaction?
Problems 1 - 15

Problem #1: P₄ + NaOH + H₂O ----> PH₃ + Na₂HPO₃

Comment: The NaOH tells you this is in basic solution. The equation could be written this way:

P₄ ----> PH₃ + Na₂HPO₃ [basic sol.]

This problem is also done as example #5. I decided to show you a molecular equation and then take it to net ionic and then back to full molecular.

Solution:

1) Half reactions:

P₄ ---> PH₃
P₄ ---> HPO₃²¯

2) Balance in acidic first:

12e¯ + 12H⁺ + P₄ ---> 4PH₃
12H₂O + P₄ ---> 4HPO₃²¯ + 20H⁺ + 12e¯

3) Electrons are already equal, so add:

12H₂O + 2P₄ ---> 4PH₃ + 4HPO₃²¯ + 8H⁺

This is the first point where you can reduce all the coefficients by a factor of 2. When I wrote this solution, I actually didn't notice I could reduce until the end of the problem. I decided to leave the reduction where I first noticed it.

4) Convert to basic:

8OH¯ + 12H₂O + 2P₄ ---> 4PH₃ + 4HPO₃²¯ + 8H₂O

5) Remove duplicate waters:

8OH¯ + 4H₂O + 2P₄ ---> 4PH₃ + 4HPO₃²¯

6) Divide through by a factor of 2:

4OH¯ + 2H₂O + P₄ ---> 2PH₃ + 2HPO₃²¯

7) Add sodium back in:

4NaOH + 2H₂O + P₄ ---> 2PH₃ + 2Na₂HPO₃

Problem #2a: Cl₂ ---> Cl¯ + ClO

ClO is correct. It could be ClO¯. I will do that as problem 2b.

Solution:

1) Half-reactions:

Cl₂ ---> Cl¯
Cl₂ ---> ClO

2) Balance in acidic:

2e¯ + Cl₂ ---> 2Cl¯
2H₂O + Cl₂ ---> 2ClO + 4H⁺ + 4e¯

If acid or base is not specified, pick acidic unless something like NaOH appears in the equation.

3) Equalize electrons:

4e¯ + 2Cl₂ ---> 4Cl¯
2H₂O + Cl₂ ---> 2ClO + 4H⁺ + 4e¯

4) Add:

2H₂O + 3Cl₂ ---> 4Cl¯ + 2ClO + 4H⁺

5) If so desired, you can combine:

2H₂O + 3Cl₂ ---> 4HCl + 2ClO

Problem #2b: Cl₂ ---> Cl¯ + ClO¯

Solution:

1) Half-reactions:

Cl₂ ---> Cl¯
Cl₂ ---> ClO¯

2) Balance in acidic:

2e¯ + Cl₂ ---> 2Cl¯
2H₂O + Cl₂ ---> 2ClO¯ + 4H⁺ + 2e¯

3) Electrons already equal, so add:

2H₂O + 2Cl₂ ---> 2Cl¯ + 2ClO¯ + 4H⁺

4) Combine ions (and reduce by a factor of 2) for the final answer:

H₂O + Cl₂ ---> HCl + HClO

Problem #3: CaCl(ClO) ---> Cl₂ [acid solution] (By the way, CaCl(ClO) is known as bleaching powder.)

Solution:

1) Write half-reactions:

Cl¯ ---> Cl₂
ClO¯ ---> Cl₂

Note that I eliminated the calcium ion, which I will add back it later. It's a spectator ion.

2) Balance in acid solution:

2Cl¯ ---> Cl₂ + 2e¯
2e¯ + 4H⁺ + 2ClO¯ ---> Cl₂ + 2H₂O

3) Add the two half reactions and reduce:

Cl¯ + ClO¯ + 2H⁺ ---> Cl₂ + H₂O

CaCl(ClO) + 2H⁺ ---> Cl₂ + Ca²⁺ + H₂O

Problem #4: S₈ + Na₂SO₃ + H₂O ---> Na₂S₂O₃ ⋅ 5H₂O

Solution:

1) Write half-reactions in net ionic form:

S₈ ---> S₂O₃^2-
SO₃^2- ---> S₂O₃^2-

Note that I dropped the water. I'll put it back in at the end.

2) Balance in acidic solution:

3H₂O + ¹⁄₄S₈ ---> S₂O₃^2- + 6H⁺ + 4e¯
4e¯ + 6H⁺ + 2SO₃^2- ---> S₂O₃^2- + 3H₂O

I picked acidic because it's easier to work with. Notice that the hydrogen ion and water both cancel out. Note the use of ¹⁄₄ for a coefficient. Fractions as coefficients are often used for balancing, but ¹⁄₄ doesn't show up very often.

3) Add half-reactions:

¹⁄₄S₈ + 2SO₃^2- ---> 2S₂O₃^2-

4) Convert back to molecular equation:

¹⁄₄S₈ + 2Na₂SO₃ + 10H₂O ---> 2Na₂S₂O₃ ⋅ 5H₂O

5) You might see it this way:

S₈ + 8Na₂SO₃ + 40H₂O ---> 8Na₂S₂O₃ ⋅ 5H₂O

6) Or, you might see it like this:

S + Na₂SO₃ + 5H₂O ---> Na₂S₂O₃ ⋅ 5H₂O

Problem #5: S₈ ---> S₂O₃^2- + S^2- [basic solution]

Solution:

1) Write half-reactions:

S₈ ---> S₂O₃^2-
S₈ ---> S^2-

2) Balance using "fake acid" technique for the first half-reaction:

12H₂O + S₈ ---> 4S₂O₃^2- + 24H⁺ + 16e¯
16e¯ + S₈ ---> 8S^2-

3) Add and reduce:

12H₂O + 2S₈ ---> 4S₂O₃^2- + 8S^2- + 24H⁺

6H₂O + S₈ ---> 2S₂O₃^2- + 4S^2- + 12H⁺

4) Convert to basic solution:

12OH¯ + 6H₂O + S₈ ---> 2S₂O₃^2- + 4S^2- + 12H₂O

12OH¯ + S₈ ---> 2S₂O₃^2- + 4S^2- + 6H₂O

Problem #6: NH₃ + NO₂ ---> N₂

Solution:

1) Write half-reactions:

NH₃ ---> N₂
NO₂ ---> N₂

2) Balance in acidic solution since nothing is specificed:

2NH₃ ---> N₂ + 6H⁺ + 6e¯
8e¯ + 8H⁺ + 2NO₂ ---> N₂ + 4H₂O

3) Equalize electrons:

8 [2NH₃ ---> N₂ + 6H⁺ + 6e¯]
6 [8e¯ + 8H⁺ + 2NO₂ ---> N₂ + 4H₂O]

4) Add and reduce (notice that the hydrogen ions also go away):

16NH₃ + 12NO₂ ---> 14N₂ + 24H₂O

8NH₃ + 6NO₂ ---> 7N₂ + 12H₂O

You could also play with the NH₃ and NO₂ coefficients until you got a 2:1 ratio for H and O, in order to make H₂O.

You could also decide to balance it in basic solution, adding in hydroxides to the half-reactions in step 2 of the solution. However, all the hydroxides will cancel out in the final step, just like the hydrogen ion did.

Problem #7: NO₂ ---> NO₃¯ + NO [acid solution]

Solution:

H2O + NO2 ---> NO3¯ + 2H+ + e¯

2e¯ + 2H+ + NO2 ---> NO + H+

This example is used in Balancing Redox Reactions in Acidic Solution page as problem #14, where it shows H₂O as a reactant. In problems like this, sometimes the water is shown, sometimes it is not. The reason the water is not shown is that nitrogen is the element being reduced and oxidized. By the context of the problem, the water is known to be present. So, there are times when one writer of a textbook using this problem shows the water and an author of a different text might not.

The context, by the way, is the nitrate ion. It must be in solution for the problem to work, hence the presence of water. The difficulty is that beginning students do not yet know these important little nits.

Problem #8: Se --> Se²¯ + SeO₃²¯

The ChemTeam did not write the solution below, but did think it interesting enough to reproduce.

Solution:

This is a disproportionation reaction. Se goes from state 0 to state -2 and to state +4, so for every Se oxidised, 2 are reduced.

3Se ---> 2Se²¯ + SeO₃²¯

Now to balance the oxygen, you need an oxygen-containing component on the left. In the absence of any other information, assume it to be water, and balance the hydrogen by adding H⁺ to the other side.

3Se + 3H₂O ---> 2Se²¯ + SeO₃²¯ + 6H⁺

Problem #9: K₂MnO₄ ---> MnO₂ + KMnO₄ [basic]

Solution:

1) Write half-reactions in net-ionic form:

MnO₄²¯ ---> MnO₂
MnO₄²¯ ---> MnO₄¯

2) Balance in acidic solution:

2e¯ + 4H⁺ + MnO₄²¯ ---> MnO₂ + 2H₂O
MnO₄²¯ ---> MnO₄¯ + e¯

3) Equalize electrons:

2e¯ + 4H⁺ + MnO₄²¯ ---> MnO₂ + 2H₂O
2MnO₄²¯ ---> 2MnO₄¯ + 2e¯

4) Add:

4H⁺ + 3MnO₄²¯ ---> MnO₂ + 2MnO₄¯ + 2H₂O

5) Convert to base and remove excess water:

2H₂O + 3MnO₄²¯ ---> MnO₂ + 2MnO₄¯ + 4OH¯

Problem #10: C₂N₂ ---> CN¯ + CNO¯

Solution

1) Half-reactions:

C₂N₂ ---> CN¯
C₂N₂ ---> CNO¯

2) Balance:

2e¯ + C₂N₂ ---> 2CN¯
2H₂O + C₂N₂ ---> 2CNO¯ + 4H⁺ + 2e¯

3) Electrons balance, so add:

2H₂O + 2C₂N₂ ---> 2CN¯ + 2CNO¯ + 4H⁺

4) You could make it molecular, if so desired:

2H₂O + 2C₂N₂ ---> 2HCN + 2HCNO

Problem #11: HNO₂ ---> HNO₃ + NO + H₂O

This pdf has a hand-written solution. There is a second example and it is solved via the oxidation number method for balancing redox equations.

Problem #12: ClO₂ ---> ClO₂¯ + ClO₃¯ [basic]

Solution:

1) Write half-reactions:

ClO₂ ---> ClO₂¯
ClO₂ ---> ClO₃¯

2) Balance as if in acidic solution:

e¯ + ClO₂ ---> ClO₂¯
H₂O + ClO₂ ---> ClO₃¯ + 2H⁺ + e¯

3) Add:

H₂O + 2ClO₂ ---> ClO₂¯ + ClO₃¯ + 2H⁺

4) Add two hydroxides to each side to change over to basic:

2OH¯ + 2ClO₂ ---> ClO₂¯ + ClO₃¯ + H₂O

Problem #13: (CN)₂ ---> CN¯ + OCN¯ [basic solution]

Solution:

1) Write half-reactions:

(CN)₂ ---> CN¯
(CN)₂ ---> OCN¯

2) Balance as if in acidic solution:

2e¯ + (CN)₂ ---> 2CN¯
2H₂O + (CN)₂ ---> 2OCN¯ + 4H⁺ + 2e¯

3) Add:

2H₂O + 2(CN)₂ ---> 2CN¯ + 2OCN¯ + 4H⁺

4) Reduce, then convert to basic:

H₂O + (CN)₂ ---> CN¯ + OCN¯ + 2H⁺

2OH¯ + (CN)₂ ---> CN¯ + OCN¯ + H₂O

Problem #14: P₄ ---> H₂PO₂^- + PH₃ [basic soln.]

Solution:

1) Write half-reactions:

P₄ ---> H₂PO₂^-
P₄ ---> PH₃

2) Balance as if in acidic solution:

8H₂O + P₄ ---> 4H₂PO₂^- + 8H⁺ + 4e^-
12e^- + 12H⁺ + P₄ ---> 4PH₃

3) Equalize electrons:

24H₂O + 3P₄ ---> 12H₂PO₂^- + 24H⁺ + 12e^-
12e^- + 12H⁺ + P₄ ---> 4PH₃

4) Add and eliminate like items:

24H₂O + 4P₄ ---> 12H₂PO₂^- + 4PH₃ + 12H⁺

5) Reduce:

6H₂O + P₄ ---> 3H₂PO₂^- + PH₃ + 3H⁺

6) Convert to basic:

3OH^- + 3H₂O + P₄ ---> 3H₂PO₂^- + PH₃

See another explanation here. The linked answer missed the factor of 4 in each coefficient and wound up adding 12 hydroxides to the equation I have in step #4 above. Missing a common factor in the coefficients of a balanced equation is something all of us have done.

Problem #15: Mn²⁺ + MnO₄¯ ---> MnO₂(s) [basic soln.]

Solution:

1) Separate into half-reactions:

Mn²⁺ ---> MnO₂(s)
MnO₄¯ ---> MnO₂(s)

2) Balance as if in acid:

Mn²⁺ + 2H₂O ---> MnO₂(s) + 4H⁺ + 2e¯
3e¯ + 4H⁺ + MnO₄¯ ---> MnO₂(s) + 2H₂O

3) Equalize electrons and add:

3Mn²⁺ + 6H₂O ---> 3MnO₂(s) + 12H⁺ + 6e¯
6e¯ + 8H⁺ + 2MnO₄¯ ---> 2MnO₂(s) + 4H₂O

3Mn²⁺ + 2MnO₄¯ + 2H₂O ---> 5MnO₂(s) + 4H⁺

4) Change over to basic and eliminate water:

3Mn²⁺ + 2MnO₄¯ + 4OH¯ + 2H₂O ---> 5MnO₂(s) + 4H₂O

3Mn²⁺ + 2MnO₄¯ + 4OH¯ ---> 5MnO₂(s) + 2H₂O