What if you do not have a net ionic equation?
Problems #1 - 10

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If you don't have a net ionic equation to balance, that means you have what is usually called a molecular equation. These are harder to balance because the elements being oxidized and reduced are mixed in with substances that are are not being reduced or oxidized.

The solution is to, as much as possible, remove the "extraneous" substances: the ones that are neither oxidized or reduced. This gives you an unbalanced net ionic equation, which you proceed to balance. You then add back in the substances you had removed.

The problem? There is no explicit set of steps for how to do this. So, I will just let the technique evolve through some examples.

Comment: one thing that frustrates students learning the technique discussed in this tutorial is knowing how to decide what to eliminate and what not to. The easy answer to that is to advise you to eliminate everything but that which gets oxidized and reduced. For example, this equation:

CuCl₂ + KI --> CuI + I₂ + KCl

You must be able to see that the Cu is reduced from +2 to +1, which the I goes from -1 to zero. That means that the chloride and the potassium ion are spectator ions and we arrive at this:

Cu²⁺ + I¯ ---> Cu⁺ + I₂

2Cu²⁺ + 2I¯ ---> 2Cu⁺ + I₂

That is why it must seem to a student that the teacher just magically knows what to do and the proper order of steps to follow. This is the experience factor. In the ChemTeam's case, high school chemistry was taken in the late 60's, well before many of you were born. Much experience has transpired since then!

September 30, 2011: I replaced example #8 with a different problem and, while constructing the answer, I solved it in a way I literally had never done before. The old dog did learn a new trick! (BTW, I was listening to BWV 831 (played by Glenn Gould) at the time of the solution. I find that listening to Bach does make you smarter.

Here are all the problems:

Example #1: SeCl2 ---> H2SeO3 + Se + HCl	Example #6: Ti + K2CrO4 + H2SO4 ---> Ti(SO4)2 + K2SO4 + Cr2(SO4)3 + H2O
Example #2: H2SO3 + KMnO4 ---> MnSO4 + H2SO4 + K2SO4 + H2O	Example #7: K2Cr2O7 + SnCl2 + HCl ---> CrCl3 + SnCl4 + KCl + H2O
Example #3: Zn + H3AsO4 + HNO3 ---> AsH3 + Zn(NO3)2 + H2O	Example #8: K2Cr2O7 + FeSO4 + H2SO4 ---> K2SO4 + Fe2(SO4)3 + Cr2(SO4)3 + H2O
Example #4a: HCl + KMnO4 ---> H2O + KCl + MnCl2 + Cl2	Example #9: C12H22O11 + H2SO4 + K2Cr2O7 ---> CO2 + Cr2(SO4)3 + K2SO4
Example #5: Fe(s) + NO3¯(aq) ---> Fe3+(aq) + NO2(g)	Example #10: Na2HAsO3 + KBrO3 + HCl ---> NaCl + KBr + H3AsO4

Example #1: SeCl₂ ---> H₂SeO₃ + Se + HCl

1) Here is what can be removed:

a) the SeCl₂ can be changed to Se²⁺
b) the H₂SeO₃ can be changed to SeO₃²¯
c) the HCl.

I dropped the chlorine and hydrogen because neither was oxidized or reduced. The prime skill needed in changing a molecular equation to a net ionic is recognizing which elements have had their oxidation states changed.

I will discuss the SeO₃²¯ a bit more below. Also, the HCl tells me that this solution is to be balanced in acidic solution.

2) We now have this unbalanced net ionic equation:

Se²⁺ ---> SeO₃²¯ + Se

3) Here are the two half-reactions:

Se²⁺ ---> SeO₃²¯
Se²⁺ ---> Se

4) Balance the two half-reactions (in acidic solution):

3H₂O + Se²⁺ ---> SeO₃²¯ + 6H⁺ + 2e¯
Se²⁺ + 2e¯ ---> Se

5) Add the two half-reactions and eliminate electrons:

3H₂O + 2Se²⁺ ---> 2Se + SeO₃²¯ + 6H⁺

6) Restore items that were removed (shown in two steps). This gives the final answer:

a) 3H₂O + 2SeCl₂ ---> 2Se + SeO₃²¯ + 4HCl + 2H⁺
b) 3H₂O + 2SeCl₂ ---> 2Se + H₂SeO₃ + 4HCl

Why not drop the oxygen and get this half-reaction?

Se²⁺ ---> Se⁴⁺ + 2e¯

It does work because you get this balanced net ionic equation as an answer:

2Se²⁺ ---> Se⁴⁺ + Se

However, it seems to me to be a bit of a hassle to get back to the final answer. You might wish to try that on your own. Also, H₂SeO₃ is a weak acid. You might want to keep the full formula in its half-reaction and then balance. You can also get to the correct answer that way.

Example #2: H₂SO₃ + KMnO₄ ---> MnSO₄ + H₂SO₄ + K₂SO₄ + H₂O

You must be able to see that it is the Mn that gets reduced (from +7 to +2) and that the S gets oxidized (from +4 to +6).

1) Here is what I will remove:

a) the H₂O (it will get back in via both half-reactions)
b) the H in H₂SO₃ and H₂SO₄
c) the potassium ion

Here is the unbalanced net-ionic equation that results:

SO₃²¯ + MnO₄¯ ---> MnSO₄

In order to make the half-reactions, I going to allow the MnSO₄ to ionize. Why?

I know that the sulfate will play no role in balancing the reduction half-reaction, so I don't need it. How did I know? I know because the sulfur in the sulfate is being oxidized. I can't have an oxidation occuring in a reduction half-reaction. I will recombine the Mn²⁺ and the sulfate when I recover the balanced, molecular equation.

Also, suppose I keep the MnSO₄ unionized. That means I would have to include the permanganate in the equation, to keep Mn present on both sides. That means I have to balance the net-ionic equation. Ah, no, I won't do that!

2) Here are the half-reactions that result:

SO₃²¯ ---> SO₄²¯
MnO₄¯ ---> Mn²⁺

3) Here are the balanced half-reactions (note that it is acidic solution and that I have equalized the electrons):

5 [H₂O + SO₃²¯ ---> SO₄²¯ + 2H⁺ + 2e¯]
2 [5e¯ + 8H⁺ + MnO₄¯ ---> Mn²⁺ + 4H₂O]

4) The final, balanced net ionic equation (extra water and hydrogen ion have been removed):

5SO₃²¯ + 2MnO₄¯ + 6H⁺ ---> 5SO₄²¯ + 2Mn²⁺ + 3H₂O

5) This link goes to a step-by-step discussion about how to change from the net ionic back to the molecular form of the equation.

Example #3: Zn + H₃AsO₄ + HNO₃ ---> AsH₃ + Zn(NO₃)₂ + H₂O

This is a very difficult one because it involves three half-reactions. Click this sentence for a discussion of the answer to this problem.

By the way, here are several other examples that require three half-reactions to balance.

Example #4a: HCl + KMnO₄ ---> H₂O + KCl + MnCl₂ + Cl₂

Example #4b: KMnO₄ + KCl + H₂SO₄ ---> K₂SO₄ + MnSO₄ + Cl₂ + H₂O

(What follows is a discussion of 4a. Please attempt 4b on your own after study of 4a. A link to an answer for 4b is provided at the end of the discussion of 4a.)

We can eliminate the following:

H⁺ from the HCl
H₂O
K⁺
Cl¯, but only from the right-hand side

Why only from the right-hand side? It is because the Cl¯ will be oxidized. I know this because there is Cl₂ on the right-hand side as well. This tells me the chlorine is going from a -1 oxidation state to zero. It is being oxidized. So, I need Cl¯ on the left-hand side, but not the right-hand side.

Here is the net-ionic equation:

Cl¯ + MnO₄¯ ---> Mn²⁺ + Cl₂

From this, we get the half-reactions:

MnO₄¯ ---> Mn²⁺
Cl¯ ---> Cl₂

The balanced half-reactions:

2 [5e¯ + 8H⁺ + MnO₄¯ ---> Mn²⁺ + 4H₂O]
5 [2Cl¯ ---> Cl₂ + 2e¯]

Note the addition of the factors 2 and 5, which will balance the electrons.

The balanced equation, in net-ionic form:

16H⁺ + 2MnO₄¯ + 10Cl¯ ---> 2Mn²⁺ + 5Cl₂ + 8H₂O

Now, we must re-create the molecular equation, but in balanced form. First, make some HCl:

6H⁺ + 2MnO₄¯ + 10HCl ---> 2Mn²⁺ + 5Cl₂ + 8H₂O

Next, add potassium ion back in:

6H⁺ + 2KMnO₄ + 10HCl ---> 2Mn²⁺ + 2K⁺ + 5Cl₂ + 8H₂O

To both sides, add in six chlorine ions:

2KMnO₄ + 16HCl ---> 2MnCl₂ + 2KCl + 5Cl₂ + 8H₂O

Where did they go? On the left-hand side, they hooked up with the hydrogen ion to make 6 HCl, giving a total of 16HCl. On the right-hand side, two went with the potassium ion to make KCl and the other 4 made 2MnCl₂.

I can add Cl¯ back in since we know, from the starting equation, that it was always present. Some of the Cl¯ reacted, to form Cl₂, and some remained as spectator ions.

That last equation is your answer.

Link to solution for Example 4b

Example #5: Fe(s) + NO₃¯(aq) ---> Fe³⁺(aq) + NO₂(g)

Notice that that reaction is already in net ionic form. I propose to balance it and then put it into molecular form.

Balance by half-reactions in acidic solution:

Fe(s) ---> Fe³⁺(aq) + 3e¯
e¯ + 2H⁺ + NO₃¯(aq) ---> NO₂(g) + H₂O

Fe(s) ---> Fe³⁺(aq) + 3e¯
3e¯ + 6H⁺ + 3NO₃¯(aq) ---> 3NO₂(g) + 3H₂O

6H⁺ + Fe(s) + 3NO₃¯(aq) ---> Fe³⁺(aq) + 3NO₂(g) + 3H₂O

I knew to balance in acidic solution because of the nitrate. I needed positive ions coupled with the nitrate to give me a compound. Hydroxide did not fit my needs. Also, note that the iron(III) ion is in solution. In the presence of hydroxide, Fe(OH)₃ would precipitate.

To make a molecular equation, join up some hydrogen ion with some nitrate ion:

3H⁺ + Fe(s) + 3HNO₃(aq) ---> Fe³⁺(aq) + 3NO₂(g) + 3H₂O

We need more nitrate ion to combine with the rest of the hydrogen ion AND the ferric ion:

Fe(s) + 6HNO₃(aq) ---> Fe(NO₃)₃(aq) + 3NO₂(g) + 3H₂O

It's now a fully-balanced molecular equation. Note that iron(III) nitrate is soluble, making the three nitrate ions added in this last step as the only spectator ions in the reaction.

Example #6: In the reaction of titanium metal with excess potassium chromate and sulfuric acid, the products are titanium (IV) sulfate, potassium sulfate, chromium (III) sulfate, and water. Write and balance the equation and calculate the coefficient for the potassium sulfate in the balanced equation.

1) Write the equation (with correct formulas!):

Ti + K₂CrO₄ + H₂SO₄ ---> Ti(SO₄)₂ + K₂SO₄ + Cr₂(SO₄)₃ + H₂O

2) Convert it to net ionic:

Ti + CrO₄^2- ---> Ti⁴⁺ + Cr³⁺

3) Write and balance half-reactions:

Ti ---> Ti⁴⁺
CrO₄^2- ---> Cr³⁺

Ti ---> Ti⁴⁺ + 4e¯
3e¯ + 8H⁺ + CrO₄^2- ---> Cr³⁺ + 4H₂O

4) Balance net ionic equation:

32H⁺ + 3Ti + 4CrO₄^2- ---> 3Ti⁴⁺ + 4Cr³⁺ + 16H₂O

5) Make chromium(III) sulfate:

20H⁺ + 6H₂SO₄ + 3Ti + 4CrO₄^2- ---> 3Ti⁴⁺ + 2Cr₂(SO₄)₃ + 16H₂O

Note how six sulfuric acid has been added in as well. This balances the six sulfates added.

6) Make titanium(IV) sulfate:

8H⁺ + 12H₂SO₄ + 3Ti + 4CrO₄^2- ---> 3Ti(SO₄)₂ + 2Cr₂(SO₄)₃ + 16H₂O

Note how another six sulfuric acid has been added in as well. This balances the six sulfates added. Also, see the hydrogen ion deceasing by 12 each time, for a total of 24 so far.

7) Make potassium chromate:

8H⁺ + 12H₂SO₄ + 3Ti + 4K₂CrO₄ ---> 3Ti(SO₄)₂ + 2Cr₂(SO₄)₃ + 16H₂O + 4K₂SO₄

Notice how 4 potassium sulfate showed up on the right. That's to balance the potassium. Now, I'll fix the extra 4 sulfates that currently exist on the right-hand side of the equation.

8) Introduce 4 sulfates to the left-hand side of the equation:

16H₂SO₄ + 3Ti + 4K₂CrO₄ ---> 3Ti(SO₄)₂ + 2Cr₂(SO₄)₃ + 16H₂O + 4K₂SO₄

Did you see what I did? I simply introduced 4 more sulfuric acids. I took the 8 hydrogen ion already there and combined them with 4 sulfate ions I needed to add to the left-hand side. Notice that the coefficient of the sulfuric acid went from 12 to 16.

It's balanced and the answer the problem wants is 4.

Example #7: K₂Cr₂O₇ + SnCl₂ + HCl ---> CrCl₃ + SnCl₄ + KCl + H₂O

1) Write the net ionic equation:

Cr₂O₇^2- + Sn²⁺ ---> Cr³⁺ + Sn⁴⁺

2) Break into half-reactions and balance:

Cr₂O₇^2- ---> Cr³⁺
Sn²⁺ ---> Sn⁴⁺

6e¯ + 14H⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O
Sn²⁺ ---> Sn⁴⁺ + 2e¯

14H⁺ + 3Sn²⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

3) Add chloride to the Cr(III) and Sn(IV) ions:

2H⁺ + 12 HCl + 3SnCl₂ + Cr₂O₇^2- ---> 2CrCl₃ + 3SnCl₄ + 7H₂O

4) Add potassium ion:

2H⁺ + 12HCl + 3SnCl₂ + K₂Cr₂O₇ ---> 2CrCl₃ + 3SnCl₄ + 7H₂O + 2KCl

Notice that there are now two extra chlorides on the right-hand side.

5) Add two chlorides to the left-hand side:

14HCl + 3SnCl₂ + K₂Cr₂O₇ ---> 2CrCl₃ + 3SnCl₄ + 7H₂O + 2KCl

Example #8: K₂Cr₂O₇ + FeSO₄ + H₂SO₄ ---> K₂SO₄ + Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + H₂O

1) Write the net ionic equation:

Cr₂O₇²¯ + Fe²⁺ ---> Fe₂⁶⁺ + Cr₂⁶⁺

Notice how I preserved the subscripts for the Fe(III) ion and the Cr(III) ion on the right-hand side. In the next example, I do not employ this little trick. In several other problem I use the above trick. One way will be effective for some and the other way will be effective for others.

2) Write the half-reactions and balance them:

Cr₂O₇²¯ ---> Cr₂⁶⁺
Fe²⁺ ---> Fe₂⁶⁺

6e¯ + 14H⁺ + Cr₂O₇²¯ ---> Cr₂⁶⁺ + 7H₂O
2Fe²⁺ ---> Fe₂⁶⁺ + 2e¯

3) The balanced net ionic equation:

6Fe²⁺ + 14H⁺ + Cr₂O₇²¯ ---> Cr₂⁶⁺ + 7H₂O + 3Fe₂⁶⁺

4) I'm going to add everything to the left-hand side and nothing to the right-hand side:

6FeSO₄ + 7H₂SO₄ + K₂Cr₂O₇ ---> Cr₂⁶⁺ + 7H₂O + 3Fe₂⁶⁺

I added thirteen sulfates and two potassium ions. Now, I have to do the same to the right-hand side.

5) Add sulfate to the chromium(III) ion:

6FeSO₄ + 7H₂SO₄ + K₂Cr₂O₇ ---> Cr₂(SO₄)₃ + 7H₂O + 3Fe₂⁶⁺

That's three (of thirteen).

6) Add sulfate to the iron(III) ion:

6FeSO₄ + 7H₂SO₄ + K₂Cr₂O₇ ---> Cr₂(SO₄)₃ + 7H₂O + 3Fe₂(SO₄)₃

That's nine more (remember the coefficient of three in front of the iron(III) on the right-hand side), for a total of twelve. One more sulfate to go.

7) Pick up the last sulfate with the two potassium ions still unbalanced:

6FeSO₄ + 7H₂SO₄ + K₂Cr₂O₇ ---> Cr₂(SO₄)₃ + 7H₂O + 3Fe₂(SO₄)₃ + K₂SO₄

Example #9: C₁₂H₂₂O₁₁ + H₂SO₄ + K₂Cr₂O₇ ---> CO₂ + Cr₂(SO₄)₃ + K₂SO₄

1) Write the net ionic equation:

C₁₂H₂₂O₁₁ + Cr₂O₇^2- ---> CO₂ + Cr³⁺

2) Write the half-reactions and balance them:

C₁₂H₂₂O₁₁ ---> CO₂
Cr₂O₇^2- ---> Cr³⁺

13H₂O + C₁₂H₂₂O₁₁ ---> 12CO₂ + 48H⁺ + 48e¯
6e¯ + 14H⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O

3) The balanced net ionic equation:

64H⁺ + C₁₂H₂₂O₁₁ + 8Cr₂O₇^2- ---> 12CO₂ + 16Cr³⁺ + 43H₂O

16H⁺ + 24H₂SO₄ + C₁₂H₂₂O₁₁ + 8Cr₂O₇^2- ---> 12CO₂ + 8Cr₂(SO₄)₃ + 43H₂O

Notice how the sulfates are balanced with sulfuric acid. Notice also that this consumes 48 of the 64 hydrogen ions.

16H⁺ + 24H₂SO₄ + C₁₂H₂₂O₁₁ + 8K₂Cr₂O₇ ---> 12CO₂ + 8Cr₂(SO₄)₃ + 43H₂O + 8K₂SO₄

Notice how the potassium is balanced with potassium sulfate. This introduces eight extra sulfate on the right-hand side.

32H₂SO₄ + C₁₂H₂₂O₁₁ + 8K₂Cr₂O₇ ---> 12CO₂ + 8Cr₂(SO₄)₃ + 43H₂O + 8K₂SO₄

Notice how the 16 remaining hydrogen ions are included in the sulfuric acid.

Example #10: Na₂HAsO₃ + KBrO₃ + HCl ---> NaCl + KBr + H₃AsO₄

1) Write the net ionic equation:

HAsO₃^2- + BrO₃¯ ---> AsO₄^3- + Br¯

2) Write the half-reactions and balance:

HAsO₃^2- ---> AsO₄^3-
BrO₃¯ ---> Br¯

H₂O + HAsO₃^2- ---> AsO₄^3- + 3H⁺ + 2e¯
6e¯ + 6H⁺ + BrO₃¯ ---> Br¯ + 3H₂O

3HAsO₃^2- + BrO₃¯ + 6H⁺ ---> 3H₃AsO₄ + Br¯

3) Add one potassium to each side:

3HAsO₃^2- + KBrO₃ + 6H⁺ ---> 3H₃AsO₄ + KBr

4) Add six sodium and six chloride to the left-hand side:

3Na₂HAsO₃ + KBrO₃ + 6HCl ---> 3H₃AsO₄ + KBr

5) Add six sodum chloride to the right-hand side:

3Na₂HAsO₃ + KBrO₃ + 6HCl ---> 3H₃AsO₄ + KBr + 6NaCl

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