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

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The key point to remember is fairly easy: repeat the single substance when you write the two half-reactions needed to balance the reaction. Then balance the half-reactions as you normally would. When you add the two half-reactions together at the end, remember to add items that are the same (for example, the P₄ in problem three).

Here are five examples:

1) ClO¯ ---> Cl¯ + ClO₃¯ [basic solution]

2) IO₃¯ + I¯ ---> I₂ [acidic solution]

3) P₄ ---> HPO₃²¯ + PH₃ [basic solution]

4) NO₂ ---> NO₃¯ + NO [acid solution]

5) Cl₂ ---> Cl¯ + ClO₃¯ [basic solution]

This last could also use Br instead of Cl. Another equation is: Cl₂ ---> Cl¯ + ClO

What follows below are not complete solutions to each and every problem. Some have been left incomplete. After example five, there are more examples, which are completely solved.

Partial Solution to Example One:

1) Half-reactions:

ClO¯ ---> Cl¯
ClO¯ ---> ClO₃¯

2) Balance:

2e¯ + H₂O + ClO¯ ---> Cl¯ + 2OH¯
4OH¯ + ClO¯ ---> ClO₃¯ + 2H₂O + 4e¯

Partial Solution to Example Two:

1) Half-reactions:

IO₃¯ ---> I₂
I¯ ---> I₂

2) Balance:

10e¯ + 12H⁺ + 2IO₃¯ ---> I₂ + 6H₂O
2I¯ ---> I₂ + 2e¯

Solution to Example Three:

P₄ ---> HPO₃²¯
P₄ ---> PH₃

The P in P₄ is reduced in the first half-reaction (0 to -3) and is oxidized in the second (0 to +3).

Solve first half-reaction by the "fake acid" method:

12e¯ + 12H⁺ + P₄ ---> 4PH₃

Add 12 hydroxide ions to each side (remember, hydrogen ion plus hydroxide ion makes water):

12e¯ + 12H₂O + P₄ ---> 4PH₃ + 12OH¯

Now, solve the second half-reaction by the same method (balancing it first in acid, then adding hydroxide to convert it to basic solution.

The rest of the solution to example 3. This problem is also #16 in the Balancing Redox Reactions in Basic Solution worksheet.

Solution to Example Four:

H2O + NO2 ---> NO3¯ + 2H+ + e¯

2e¯ + 2H+ + NO2 ---> NO + H+

This example is used in Balancing Redox Reactions in Acidic Solution page as problem #19, 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.

Solution to Example Five:

A solution to example five is provided on video.

Here's Example Five done with bromine:

Br₂ ---> BrO₃¯ + Br¯ [basic solution]

Solution:

1) Half-reactions:

Br₂ ---> BrO₃¯
Br₂ ---> Br¯

2) Balance the half-reactions using the "fake acid" technique:

6H₂O + Br₂ ---> 2BrO₃¯ + 12H⁺ + 10e¯
2e¯ + Br₂ ---> 2Br¯

3) Equalize the electrons (multiply second half-reaction by 5):

6H₂O + Br₂ ---> 2BrO₃¯ + 12H⁺ + 10e¯
10e¯ + 5Br₂ ---> 10Br¯

The rest of the balancing is left to you, dear reader.

Example #6: 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

Example #7: 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 + (1/4) S₈ ---> 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 used, but 1/4 doesn't show up very often.

3) Add half-reactions:

(1/4) S₈ + 2SO₃^2- ---> 2S₂O₃^2-

4) Convert back to molecular equation:

(1/4) S₈ + 2 Na₂SO₃ + 10 H₂O ---> 2 Na₂S₂O₃ ⋅ 5H₂O

5) You might see it this way:

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

6) Or, you might see it like this:

S + Na₂SO₃ + 5 H₂O ---> Na₂S₂O₃ ⋅ 5H₂O

Example #8: 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:

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

3) Add and reduce:

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

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

4) Convert to basic solution:

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

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

Example #9: NH₃ + NO₂ ---> N₂ + H₂O

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.

Example #10: I₃¯ ---> I¯ + IO₃¯ [acidic sol.]

Solution:

1) Half-reactions:

I₃¯ ---> I¯
I₃¯ ---> IO₃¯

2) Balance half-reactions:

2e¯ + I₃¯ ---> 3I¯
9H₂O + I₃¯ ---> 3IO₃¯ + 18H⁺ + 16e¯

3) Equalize electrons and add:

8 [2e¯ + I₃¯ ---> 3I¯]
9H₂O + I₃¯ ---> 3IO₃¯ + 18H⁺ + 16e¯

9H₂O + 9I₃¯ ---> 3IO₃¯ + 24 I¯ + 18H⁺

3H₂O + 3I₃¯ ---> IO₃¯ + 8I¯ + 6H⁺

Example #11: 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⁺

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