What if you do not have a net ionic equation?
What if it is a complete molecular equation?
Fifteen Examples

If you don't have a net ionic equation to balance, that means you have what is usually called a molecular equation (or a complete 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 and that 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

and eventually this:

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!

Example #1: NaIO₃ + H₂SO₃ ---> NaI + H₂SO₄

Solution:

1) The sodium ion never changes its oxidation state. It's a +1 in NaIO₃ and a +1 in NaI. We can eliminate it on our way to the net ionic equation:

IO₃¯ + H₂SO₃ ---> I¯ + H₂SO₄

2) The hydrogen never changes its oxidation state. I will choose to remove it from the equation:

IO₃¯ + SO₃²¯ ---> I¯ + SO₄²¯

I actually could have left the hydrogen ion and still got to the final answer. My personal opinion is to take them out.

3) I have run across teachers that will say to drop the oxygens from the sulfite and the sulfate. My preference is to keep them. The reason is that sulfate and sulfate ions exist as discrete species in solution, so I choose to keep them that way. That leaves the equation in step 2 as the net ionic.

4) Separate into half reactions:

IO₃¯ ---> I¯
SO₃²¯ ---> SO₄²¯

5) Balance:

6e¯ + 6H⁺ + IO₃¯ ---> I¯ + 3H₂O
H₂O + SO₃²¯ ---> SO₄²¯ + 2H⁺ + 2e¯

6) Equalize electrons:

6e¯ + 6H⁺ + IO₃¯ ---> I¯ + 3H₂O
3H₂O + 3SO₃²¯ ---> 3SO₄²¯ + 6H⁺ + 6e¯

7) Add:

IO₃¯ + 3SO₃²¯ ---> I¯ + 3SO₄²¯

8) Put spectator ions back in:

NaIO₃ + 3H₂SO₃ ---> NaI + 3H₂SO₄

I added one Na⁺ and six H⁺ back in.

Example #2: SeCl₂ ---> H₂SeO₃ + Se + HCl

Solution:

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⁺ ---> Se + SeO₃²¯ + 6H⁺

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

a) 3H₂O + 2SeCl₂ ---> Se + SeO₃²¯ + 4HCl + 2H⁺
b) 3H₂O + 2SeCl₂ ---> Se + 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 #3: H₂SO₃ + KMnO₄ ---> MnSO₄ + H₂SO₄ + K₂SO₄ + H₂O

Solution:

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) Make some MnSO₄:

5SO₃²¯ + 2MnO₄¯ + 6H⁺ ---> 3SO₄²¯ + 2MnSO₄ + 3H₂O

6) Make some potassium permanganate:

5SO₃²¯ + 2KMnO₄ + 6H⁺ ---> 2SO₄²¯ + 2MnSO₄ + 3H₂O + K₂SO₄

Notice how I made some potassium sulfate at the same time.

7) Make some H₂SO₃:

2SO₃²¯ + 2KMnO₄ + 3H₂SO₃ ---> 2SO₄²¯ + 2MnSO₄ + 3H₂O + K₂SO₄

8) Now, I will add 4 hydrogens on each side:

2KMnO₄ + 5H₂SO₃ ---> 2H₂SO₄ + 2MnSO₄ + 3H₂O + K₂SO₄

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

Solution:

1) 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.

2) Here is the net-ionic equation:

Cl¯ + MnO₄¯ ---> Mn²⁺ + Cl₂

3) From this, we get the half-reactions:

MnO₄¯ ---> Mn²⁺
Cl¯ ---> Cl₂

4) 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.

5) The balanced equation, in net-ionic form:

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

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

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

7) Next, add potassium ion back in:

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

8) 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.

This equation is your answer.

Example #5: KMnO₄ + KCl + H₂SO₄ ---> K₂SO₄ + MnSO₄ + Cl₂ + H₂O

Solution #1:

1) Eliminate all spectator ions (this means anything that isn't reduced or oxidized). The list is:

K⁺ (from the KMnO₄ and the KCl)
H₂SO₄
K₂SO₄
SO₄²¯ (from the MnSO₄)
H₂O

That leaves this:

MnO₄¯ + Cl¯ ---> Mn²⁺ + Cl₂

Why did I keep those? It's because they are the exact items oxidized (the Cl) or reduced (the Mn).

2) Balance the net ionic reaction (in acidic solution because of the sulfuric acid):

MnO₄¯ ---> Mn²⁺
Cl¯ ---> Cl₂

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

16H⁺ + 2MnO₄¯ + 10Cl¯ ---> 2Mn²⁺ + 5Cl₂ + 8H₂O (this is the balanced, net ionic equation)

a) add back in 8 sulfates:

8H₂SO₄ + 2MnO₄¯ + 10Cl¯ ---> 2MnSO₄ + 5Cl₂ + 8H₂O + 6K₂SO₄

b) This added 12 potassium ions to the right side, so add 12 K⁺ to the left:

8H₂SO₄ + 2KMnO₄ + 10KCl ---> 2MnSO₄ + 5Cl₂ + 8H₂O + 6K₂SO₄

The last equation is the balanced, molecular equation.

Solution #2:

1) Identify the atoms being oxidized and reduced:

Mn goes from +7 to +2; it is reduced
Cl goes from -1 to zero; it is oxidized

2) Write the two half-reactions, showing ONLY the items reduced/oxidized:

Mn⁷⁺ + 5e¯ ---> Mn²⁺
2Cl¯ ---> Cl₂ + 2e¯

3) Multiply the first reaction by 2 and the second by 5, and add them to get:

2Mn⁷⁺ + 10Cl¯ ---> 2Mn²⁺ + 5Cl₂

4) Since this a complete redox reaction and the number of electrons are equal on both sides, we can now use these coefficients in the molecular equation to balance it:

2KMnO₄ + 10KCl + H₂SO₄ ---> K₂SO₄ + 2MnSO₄ + 5Cl₂ + H₂O

This is left to the reader.

Example #6: CH₃CH₂OH + K₂Cr₂O₇ + H₂SO₄ ---> HC₂H₃O₂ + Cr₂(SO₄)₃ + K₂SO₄ + H₂O

1) Write the net ionic equation:

CH₃CH₂OH + Cr₂O₇²¯ ---> Cr³⁺ + HC₂H₃O₂

2) Half-reactions and balance:

CH₃CH₂OH ---> HC₂H₃O₂
Cr₂O₇²¯ ---> Cr³⁺

H₂O + CH₃CH₂OH ---> HC₂H₃O₂ + 4H⁺ + 4e¯
6e¯ + 14H⁺ + Cr₂O₇²¯ ---> 2Cr³⁺ + 7H₂O

Top half-reaction by 3, bottom by 2; then add and eliminate water and hydrogen ion.

16H⁺ + 3CH₃CH₂OH + 2Cr₂O₇²¯ ---> 4Cr³⁺ + 3HC₂H₃O₂ + 11H₂O

3) Add 4 K⁺ to the left, then the right. Add two sulfate to the right (thus making the potassium sulfate), then to the left:

12H⁺ + 2H₂SO₄ + 3CH₃CH₂OH + 2K₂Cr₂O₇ ---> 4Cr³⁺ + 3HC₂H₃O₂ + 11H₂O + 2K₂SO₄

4) Add six sulfates to the Cr(III) ion, then make the rest of the sulfuric acid. Result:

8H₂SO₄ + 3CH₃CH₂OH + 2K₂Cr₂O₇ ---> 2Cr₂(SO₄)₃ + 3HC₂H₃O₂ + 11H₂O + 2K₂SO₄

Example #7: PbCrO₄ + HCl + FeSO₄ ---> PbCl₂ + Cr₂(SO₄)₃ + FeCl₃ + H₂O + Fe₂(SO₄)₃

Solution:

1) Half-reactions:

CrO₄²¯ ---> Cr³⁺
Fe²⁺ ---> Fe³⁺

2) Balance in acidic solution:

3e¯ + 8H⁺ + CrO₄²¯ ---> Cr³⁺ + 4H₃O
Fe²⁺ ---> Fe³⁺ + e¯

3) Equalize electrons:

3e¯ + 8H⁺ + CrO₄²¯ ---> Cr³⁺ + 4H₃O
3Fe²⁺ ---> 3Fe³⁺ + 3e¯

4) Add:

8H⁺ + CrO₄²¯ + 3Fe²⁺ ---> Cr³⁺ + 3Fe³⁺ + 4H₃O

5) Inspection of Cr₂(SO₄)₃ leads us to double the balanced equation before turning it into a molecular equation:

16H⁺ + 2CrO₄²¯ + 6Fe²⁺ ---> 2Cr³⁺ + 6Fe³⁺ + 8H₃O

6) Restore the full formulas on the reactant side:

16HCl + 2PbCrO₄ + 6FeSO₄ ---> 2Cr³⁺ + 6Fe³⁺ + 8H₃O

16 chloride ions, two Pb(II) ions and six sulfate ions were added.

7) Add sulfates to the product side:

16HCl + 2PbCrO₄ + 6FeSO₄ ---> Cr₂(SO₄)₃ + 4Fe³⁺ + Fe₂(SO₄)₃ + 8H₃O

8) Add Pb(II) to the product side:

16HCl + 2PbCrO₄ + 6FeSO₄ ---> 2PbCl₂ + Cr₂(SO₄)₃ + 4Fe³⁺ + Fe₂(SO₄)₃ + 8H₃O

Note how four chorides were also added.

9) Add twelve more chlorides to finish:

16HCl + 2PbCrO₄ + 6FeSO₄ ---> 2PbCl₂ + Cr₂(SO₄)₃ + 4FeCl₃ + Fe₂(SO₄)₃ + 8H₃O

Example #8: C₆H₁₂O₆ + KMnO₄ + H₂SO₄ ---> CO₂ + K₂SO₄ + MnSO₄ + H₂O

Solution:

1) Half-reactions:

C₆H₁₂O₆ ---> CO₂
MnO₄¯ ---> Mn²⁺

2) Balance:

6H₂O + C₆H₁₂O₆ ---> 6CO₂ + 24H⁺ + 24e¯
5e¯ + 8H⁺ + MnO₄¯ ---> Mn²⁺ + 4H₂O

3) Equalize electrons:

30H₂O + 5C₆H₁₂O₆ ---> 30CO₂ + 120H⁺ + 120e¯
120e¯ + 192H⁺ + 24MnO₄¯ ---> 24Mn²⁺ + 96H₂O

4) Add:

72H⁺ + 5C₆H₁₂O₆ + 24MnO₄¯ ---> 30CO₂ + 24Mn²⁺ + 66H₂O

5) Complete the reactant side only:

36H₂SO₄ + 5C₆H₁₂O₆ + 24KMnO₄ ---> 30CO₂ + 24Mn²⁺ + 66H₂O

36 sulfates and 24 potassium ions were added.

6) Add 24 sulfates to the product side:

36H₂SO₄ + 5C₆H₁₂O₆ + 24KMnO₄ ---> 30CO₂ + 24MnSO₄ + 66H₂O

This leaves 24 potassium ions and 12 sulfate ions left over. Imagine that!

7) Finish:

36H₂SO₄ + 5C₆H₁₂O₆ + 24KMnO₄ ---> 30CO₂ + 24MnSO₄ + 12K₂SO₄ + 66H₂O

Example #9: KMnO₄ + KOH + FeSO₄ ---> K₂MnO₄ + Fe(OH)₃ + K₂SO₄

Solution:

1) Net ionic:

MnO₄¯ + OH¯ + Fe²⁺ ---> MnO₄²¯ + Fe(OH)₃

2) Half reactions:

e¯ + MnO₄¯ ---> MnO₄²¯
3OH¯ + Fe²⁺ ---> Fe(OH)₃ + e¯

3) Add (since electrons are already equalized):

MnO₄¯ + 3OH¯ + Fe²⁺ ---> MnO₄²¯ + Fe(OH)₃

4) Add four potassium ion and one sulfate to the left:

KMnO₄ + 3KOH + FeSO₄ ---> MnO₄²¯ + Fe(OH)₃

5) Add the same to the right side:

KMnO₄ + 3KOH + FeSO₄ ---> K₂MnO₄ + Fe(OH)₃ + K₂SO₄

6) I wrote Fe(OH)₃ because I know, from experience, that it is an insoluble precipitate. Let us suppose you did not know that and wrote this:

MnO₄¯ + OH¯ + Fe²⁺ ---> MnO₄²¯ + Fe³⁺

Could this be balanced correctly? The answer, of course is yes.

7) Here are the revised half-reactions:

e¯ + MnO₄¯ ---> MnO₄²¯
Fe²⁺ ---> Fe³⁺ + e¯

8) Add (since electrons are already equalized):

MnO₄¯ + Fe²⁺ ---> MnO₄²¯ + Fe³⁺

9) Four potassiums and one sulfate, just as above. The only other thing is to add three hydroxides to each side.

Example #10: K₂Cr₂O₇ + KI + H₂SO₄ ---> Cr₂(SO₄)₃ + K₂SO₄ + I₂ + H₂O

Solution:

1) Identify the net ionic:

Cr₂O₇²¯ + I¯ ---> Cr³⁺ + I₂

2) Balanced half-reactions:

6e¯ + 14H⁺ + Cr₂O₇²¯ ---> 2Cr³⁺ + 7H₂O
2I¯ ---> I₂ + 2e¯

3) Equalize electrons and add:

14H⁺ + Cr₂O₇²¯ + 6I¯ ---> 2Cr³⁺ + 3I₂ + 7H₂O

4) Add seven sulfates and eight potassium ions to the left-hand side:

7H₂SO₄ + K₂Cr₂O₇ + 6KI ---> 2Cr³⁺ + 3I₂ + 7H₂O

5) Add three sulfates to the right-hand side:

7H₂SO₄ + K₂Cr₂O₇ + 6KI ---> Cr₂(SO₄)₃ + 3I₂ + 7H₂O

6) There are eight potassium ions and four sulfates. Whatever shall we do with them?

7H₂SO₄ + K₂Cr₂O₇ + 6KI ---> Cr₂(SO₄)₃ + 3I₂ + 7H₂O + 4K₂SO₄

Example #11: C₆H₄(CH₃)₂ + K₂Cr₂O₇ + H₂SO₄ ---> C₆H₄(COOH)₂ + Cr₂(SO₄)₃ + H₂O

Solution:

Comment: this is the reaction to change xylene [C₆H₄(CH₃)₂] to terephthalic acid [C₆H₄(COOH)₂]. I won't bother with the fact that each compound name actually has three different isomers since that does not affect the redox balancing.

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

C₆H₄(CH₃)₂ ---> C₆H₄(COOH)₂
Cr₂O₇²¯ ---> Cr₂⁶⁺

2) Balance:

4H₂O + C₆H₄(CH₃)₂ ---> C₆H₄(COOH)₂ + 12H⁺ + 12e¯
6e¯ + 14H⁺ + Cr₂O₇²¯ ---> Cr₂⁶⁺ + 7H₂O

Comment: when I balanced this equation for posting on the website, I initially put a 2 in front of the H₂O in the first half-reaction. I finally saw the mistake but, even so, even veterans can still make mistakes (and miss them for a few moments).

3) Equalize electrons:

4H₂O + C₆H₄(CH₃)₂ ---> C₆H₄(COOH)₂ + 12H⁺ + 12e¯
12e¯ + 28H⁺ + 2Cr₂O₇²¯ ---> 2Cr₂⁶⁺ + 14H₂O

4) Add:

16H⁺ + C₆H₄(CH₃)₂ + 2Cr₂O₇²¯ ---> C₆H₄(COOH)₂ + 2Cr₂⁶⁺ + 10H₂O

5) Add eight sulfates and 4 potassium ion to the left-hand side:

8H₂SO₄ + C₆H₄(CH₃)₂ + 2K₂Cr₂O₇ ---> C₆H₄(COOH)₂ + 2Cr₂⁶⁺ + 10H₂O

6) Only six sulfates are used to restore chromium(III) sulfate:

8H₂SO₄ + C₆H₄(CH₃)₂ + 2K₂Cr₂O₇ ---> C₆H₄(COOH)₂ + 2Cr₂(SO₄)₃ + 10H₂O

7) Add two potassium sulfate to the right-hand side to finish the balancing:

8H₂SO₄ + C₆H₄(CH₃)₂ + 2K₂Cr₂O₇ ---> C₆H₄(COOH)₂ + 2Cr₂(SO₄)₃ + 2K₂SO₄ + 10H₂O

8) You could write the first half-reaction like this:

(CH₃)₂²¯ ---> (COOH)₂²¯

and add the C₆H₄ back in at the end. You are welcome to try that on your own.

Example #12: H₂SeO₃ + HBr ---> Se + Br₂ + H₂O

Solution:

1) I'm going to not write a net ionic equation. Here are the half-reactions:

H₂SeO₃ ---> Se + H₂O
HBr ---> Br₂

2) Do the balancing!

4e¯ + 4H⁺ + H₂SeO₃ ---> Se + 3H₂O
2HBr ---> Br₂ + 2H⁺ + 2e¯

3) Equalize electrons:

4e¯ + 4H⁺ + H₂SeO₃ ---> Se + 3H₂O
4HBr ---> 2Br₂ + 4H⁺ + 4e¯

4) Add:

H₂SeO₃ + 4HBr ---> Se + 2Br₂ + 3H₂O

Example #13: KI + CoSO₄ + KIO₃ + H₂O ---> Co(OH)₂ + K₂SO₄ + I₂

Solution:

1) Examine the equation for what gets oxidized and reduced and we find this:

I¯ ---> I₂
IO₃¯ ---> I₂

There are a number of rections in which cobalt is either oxidized or reduced. Not in this example.

2) Balance in acidic solution:

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

3) Equalize electrons:

10I¯ ---> 5I₂ + 10e¯
10e¯ + 12H⁺ + 2IO₃¯ ---> I₂ + 6H₂O

4) Add:

12H⁺ + 10I¯ + 2IO₃¯ ---> 6I₂ + 6H₂O

5) Divide through by 2, then convert to basic:

6H⁺ + 5I¯ + IO₃¯ ---> 3I₂ + 3H₂O

6H₂O + 5I¯ + IO₃¯ ---> 3I₂ + 3H₂O + 6OH¯

3H₂O + 5I¯ + IO₃¯ ---> 3I₂ + 6OH¯

6) Add three cobalt(II) ion back into the equation:

3H₂O + 3Co²⁺ + 5I¯ + IO₃¯ ---> 3I₂ + 3Co(OH)₂

I added three because I knew I had to use six hydroxides.

7) Add six potassium ion and three sulfate to the left-hand side:

3H₂O + 3CoSO₄ + 5KI + KIO₃ ---> 3I₂ + 3Co(OH)₂

7) Add six potassium ion and three sulfate to the right-hand side:

3H₂O + 3CoSO₄ + 5KI + KIO₃ ---> 3I₂ + 3Co(OH)₂ + 3K₂SO₄

Done!

Example #14: NaCl + H₂SO₄ + MnO₂ ---> Na₂SO₄ + MnCl₂ + H₂O + Cl₂

Solution:

1) What to remove?

NaCl: sodium ion is a classic spectator ion and it will be removed.
H₂SO₄; notice how only sulfate is present on the right-hand side. It will be removed.
MnO₂: this stays since Mn is reduced in the reaction.
Cl₂: it stays since the chloride from the left-hand side is oxidized.

Here is what remains: Cl¯ + H⁺ + MnO₂ ---> Mn²⁺ + H₂O + Cl₂

That's the net ionic.

Notice I kept the MnO₂ intact.

2) Now for the half-reactions:

Cl¯ ---> Cl₂
MnO₂ ---> Mn²⁺

3) Balance them:

2Cl¯ ---> Cl₂ + 2e¯
2e¯ + 4H⁺ + MnO₂ ---> Mn²⁺ + 2H₂O

4) Since the electrons are already balanced, we add:

2Cl¯ + 4H⁺ + MnO₂ ---> Cl₂ + Mn²⁺ + 2H₂O

5) We add in two more chloride to make MnCl₂:

4Cl¯ + 4H⁺ + MnO₂ ---> Cl₂ + MnCl₂ + 2H₂O

6) We add in 4 sodium ion to make NaCl:

4NaCl + 4H⁺ + MnO₂ ---> 4Na⁺ + Cl₂ + MnCl₂ + 2H₂O

7) Add in four sulfate ions:

4NaCl + 2H₂SO₄ + MnO₂ ---> 2Na₂SO₄ + Cl₂ + MnCl₂ + 2H₂O

Example #15: Fe(s) + NO₃¯(aq) ---> Fe³⁺(aq) + NO₂(g)

Solution:

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

1) 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.

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

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

3) 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.

Bonus Example: Mo₂₄O₃₇ + KMnO₄ + H₂SO₄ ---> MoO₃ + MnSO₄ + K₂SO₄ + H₂O

Solution:

1) Separate into half-reactions (note that potassium ion and sulfate ion have been eliminated):

Mo₂₄O₃₇ ---> MoO₃
MnO₄¯ ---> Mn²⁺

2) Balance the half-reactions in acidic solution. Steps for the first one are given, then the answer for the second half-reaction is given.

Mo₂₄O₃₇ ---> 24MoO₃

Mo₂₄O₃₇ + 35H₂O ---> 24MoO₃ (balance oxygen)

Mo₂₄O₃₇ + 35H₂O ---> 24MoO₃ + 70H⁺ (balance hydrogen)

Mo₂₄O₃₇ + 35H₂O ---> 24MoO₃ + 70H⁺ + 70e¯ (balance charge)

MnO₄¯ + 8H⁺ + 5e¯ ---> Mn²⁺ + 4H₂O (the second half-reaction)

3) Equalize elections:

Mo₂₄O₃₇ + 35H₂O ---> 24MoO₃ + 70H⁺ + 70e¯
14MnO₄¯ + 112H⁺ + 70e¯ ---> 14Mn²⁺ + 56H₂O

4) Add (note that only electrons have been eliminated):

Mo₂₄O₃₇ + 14MnO₄¯ + 35H₂O + 112H⁺ ---> 24MoO₃ + 14Mn²⁺ + 56H₂O + 70H⁺

5) Eliminate excess hydrogen ion and water:

Mo₂₄O₃₇ + 14MnO₄¯ + 42H⁺ ---> 24MoO₃ + 14Mn²⁺ + 21H₂O

6) Add 21 sulfate ions on each side:

Mo₂₄O₃₇ + 14MnO₄¯ + 21H₂SO₄ ---> 24MoO₃ + 14MnSO₄ + 21H₂O + 7SO₄²¯

7) Add 14 potassium ions to each side:

Mo₂₄O₃₇ + 14KMnO₄ + 21H₂SO₄ ---> 24MoO₃ + 14MnSO₄ + 21H₂O + 7H₂SO₄

And we are done.