### 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¯

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¯

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¯

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¯

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
4) Add back in the calcium and recreate the bleaching powder formula:
CaCl(ClO) + 2H+ ---> Cl2 + Ca2+ + H2O

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

Solution:

1) Write half-reactions in net ionic form:

S8 ---> S2O32-
SO32- ---> S2O32-

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

2) Balance in acidic solution:

3H2O + 14S8 ---> S2O32- + 6H+ + 4e¯
4e¯ + 6H+ + 2SO32- ---> S2O32- + 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 14 for a coefficient. Fractions as coefficients are often used for balancing, but 14 doesn't show up very often.

14S8 + 2SO32- ---> 2S2O32-

4) Convert back to molecular equation:

14S8 + 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 ---> S2O32- + S2- [basic solution]

Solution:

1) Write half-reactions:

S8 ---> S2O32-
S8 ---> S2-

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

12H2O + S8 ---> 4S2O32- + 24H+ + 16e¯
16e¯ + S8 ---> 8S2-

12H2O + 2S8 ---> 4S2O32- + 8S2- + 24H+

6H2O + S8 ---> 2S2O32- + 4S2- + 12H+

4) Convert to basic solution:

12OH¯ + 6H2O + S8 ---> 2S2O32- + 4S2- + 12H2O

12OH¯ + S8 ---> 2S2O32- + 4S2- + 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¯

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¯

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¯

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¯

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