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

Problem #1: HBr + NaMnO4 ---> NaBr + Br2 + MnBr2 + H2O

1) Half-reactions:

Br¯ ---> Br2
MnO4¯ ---> Mn2+

2) Balance them:

2Br¯ ---> Br2 + 2e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Equalize electrons, then combine half-reactions:

16H+ + 10Br¯ + 2MnO4¯ ---> 5Br2 + 2Mn2+ + 8H2O

4) Add two sodium ions and two bromide ions to each side:

16H+ + 12Br¯ + 2NaMnO4 ---> 2NaBr + 5Br2 + 2Mn2+ + 8H2O

5) Add four bromide ions to each side:

16HBr + 2NaMnO4 ---> 2NaBr + 5Br2 + 2MnBr2 + 8H2O

Note that, on the left-hand side, I also combined the 16 hydrogen ions with the 16 bromide ions.

Problem #2: HNO3 + KI ---> KNO3 + I2 + NO + H2O

1) Half-reactions:

NO3¯ ---> NO
I¯ ---> I2

2) Balance them:

3e¯ + 4H+ + NO3¯ ---> NO + 2H2O
2I¯ ---> I2 + 2e¯

3) Equalize electrons, then combine half-reactions:

8H+ + 2NO3¯ + 6I¯ ---> 3I2 + 2NO + 4H2O

4) Add in six potassium ions and 6 nitrate ions:

8HNO3 + 6KI ---> 6KNO3 + 3I2 + 2NO + 4H2O

Note that, on the left-hand side, I also combined the 8 hydrogen ions with the 8 nitrate ions.

Problem #3: KOH + Zn + H2O + KNO3 ---> K2Zn(OH)4 + NH3

1) Half-reactions:

Zn ---> Zn(OH)42¯
NO3¯ ---> NH3

2) Balance:

4OH¯ + Zn ---> Zn(OH)42¯ + 2e¯
8e¯ + 6H2O + NO3¯ ---> NH3 + 9OH¯

I first balanced the second half-reaction in acidic solution, then changed it over to basic.

3) Equalize electrons, then combine half-reactions:

7OH¯ + 4Zn + 6H2O + NO3¯ ---> 4Zn(OH)42¯ + NH3

4) Add eight potassium ions to each side:

7KOH + 4Zn + 6H2O + KNO3 ---> 4K2Zn(OH)42¯ + NH3

Problem #4: KNO2 + KMnO4 + H2SO4 ---> KNO3 + MnSO4 + K2SO4 + H2O

1) Write the net ionic equation:

NO2¯ + MnO4¯ ---> NO3¯ + Mn2+

2) Half-reactions and balance:

NO2¯ ---> NO3¯
MnO4¯ ---> Mn2+

H2O + NO2¯ ---> NO3¯ + 2H+ + 2e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

Top half-reaction by 5, bottom by 2; then add and eliminate water and hydrogen ion.

6H+ + 5NO2¯ + 2MnO4¯ ---> 5NO3¯ + 2Mn2+ + 3H2O

3) Add sulfate to the Mn2+:

2H+ + 5NO2¯ + 2MnO4¯ + 2H2SO4 ---> 5NO3¯ + 2MnSO4 + 3H2O

2H+ + 5KNO2 + 2KMnO4 + 2H2SO4 ---> 5KNO3 + 2MnSO4 + 3H2O + 2K+

5KNO2 + 2KMnO4 + 3H2SO4 ---> 5KNO3 + 2MnSO4 + 3H2O + K2SO4

Problem #5: K2Cr2O7 + K2C2O4 + H2C2O4 ---> K3Cr(C2O4)3 + CO2 + H2O

1) Write the net ionic equation:

Cr2O72¯ + C2O42¯ ---> Cr3+ + CO2

The potassium ion is one of the usual suspects when it comes to spectator ions, so that's easy. Be sure to notice, however, that the oxalate is both a spectator ion and the species oxidized (the C going from +3 to +4). That means that only some of the oxalate in the final answer gets oxidized.

2) Half-reactions and balance:

Cr2O72¯ ---> Cr3+
C2O42¯ ---> CO2

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
C2O42¯ ---> 2CO2 + 2e¯

14H+ + Cr2O72¯ + 3C2O42¯ ---> 2Cr3+ + 6CO2 + 7H2O

4) Add six oxalate ions to each side:

14H+ + Cr2O72¯ + 9C2O42¯ ---> 2Cr(C2O4)33¯ + 6CO2 + 7H2O

5) Make seven oxalic acids:

Cr2O72¯ + 2C2O42¯ + 7H2C2O4 ---> 2Cr(C2O4)33¯ + 6CO2 + 7H2O

K2Cr2O7 + K2C2O4 + 7H2C2O4 ---> 2K3Cr(C2O4)3 + 6CO2 + 7H2O

Problem #6: K2MnF6 + SbF5 ---> KSbF6 + MnF3 + F2

Personal note: I rather enjoyed figuring this one out.

1) Net ionic, half-reactions and balanced net ionic:

Mn4+ + F¯ ---> Mn3+ + F2

e¯ + Mn4+ ---> Mn3+
2F¯ ---> F2 + 2e¯

2Mn4+ + 2F¯ ---> 2Mn3+ + F2

2) Add 12 F¯ and one Sb5+ to each side:

add 12 F to left side: 2MnF62¯ + 2F¯ ---> 2Mn3+ + F2

add 12 F and one Sb5+ to the right side: 2MnF62¯ + 2F¯ ---> SbF6¯ + 2MnF3 + F2

add one Sb5+ to the left: 2MnF62¯ + Sb5+ + 2F¯ ---> SbF6¯ + 2MnF3 + F2

3) Add 4 potassium ions to the left:

2K2MnF6 + Sb5+ + 2F¯ ---> SbF6¯ + 2MnF3 + F2

4) This means we MUST add 4 potassium to the right side:

2K2MnF6 + Sb5+ + 2F¯ ---> 4KSbF6 + 2MnF3 + F2

5) This means we MUST add 3 Sb and 18 F to the left side. I'll do it as ions first:

2K2MnF6 + 4Sb5+ + 20F¯ ---> 4KSbF6 + 2MnF3 + F2

6) Combine ions for the final answer:

2K2MnF6 + 4SbF5 ---> 4KSbF6 + 2MnF3 + F2

Problem #7: Ca(OH)2 + NaOH + ClO2 + C ---> NaClO2 + CaCO3 + H2O

1) Half-reactions:

C ---> CO32¯
ClO2 ---> ClO2¯

2) Balance using acid solution & write net-ionic in basic solution:

3H2O + C ---> CO32¯ + 6H+ + 4e¯
e¯ + ClO2 ---> ClO2¯

3H2O + C + 4ClO2 ---> CO32¯ + 4ClO2¯ + 6H+

6OH¯ + C + 4ClO2 ---> CO32¯ + 4ClO2¯ + 3H2O

3) Add one calcium ion back in:

Ca(OH)2 + 4OH¯ + C + 4ClO2 ---> CaCO3 + 4ClO2¯ + 3H2O

Ca(OH)2 + 4NaOH + C + 4ClO2 ---> CaCO3 + 4NaClO2 + 3H2O

Problem #8: K2S + K2Cr2O7 + H2SO4 ---> S8 + K2SO4 + Cr2(SO4)3 + H2O

1) Write the net ionic equation:

S2- + Cr2O72- ---> S8 + Cr3+

Note that the only chemical species in the net ionic equation are the ions (which includes polyatomic ions) which were oxidized or reduced.

Also, the chromium(III) sulfate is a precipitate, yet I deleted the sulfate ion. Why? Because I know the sulfates are not involved in the production of elemental sulfur. How do I know that? Below, I already have a half-reaction producing elemental sulfur. To introduce a third half-reaction of sulfate to elemental sulfur is unnecessary. At the end, I'll just reform chromium(III) sulfate when I remake the balanced molecular equation.

2) Write the half-reactions and balance:

S2- ---> S8
Cr2O72- ---> Cr3+

8S2- ---> S8 + 16e¯
6e¯ + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O

Use a factor of three for the first half-reaction and eight for the second half-reaction.

112H+ + 8Cr2O72- + 24S2- ---> 3S8 + 16Cr3+ + 56H2O

3) Add this to the left-hand side: 56 sulfate + 16 K+ + 48 K+

56H2SO4 + 8K2Cr2O7 + 24K2S ---> 3S8 + 16Cr3+ + 56H2O

4) Add 24 sulfates to the right-hand side:

56H2SO4 + 8K2Cr2O7 + 24K2S ---> 3S8 + 8Cr2(SO4)3 + 56H2O

Note how the coefficient went from 16 to 8.

5) Pick up the remaining 32 sulfates and 64 potassium ions:

56H2SO4 + 8K2Cr2O7 + 24K2S ---> 3S8 + 8Cr2(SO4)3 + 56H2O + 32K2SO4

In chemical equations, it is common to use S rather than S8. Here's an example:

K2Cr2O7 + 3H2S + 4H2SO4 ---> K2SO4 + 3S + Cr2(SO4)3 + 7H2O

I'll start you with the net-ionic:

Cr2O72- + S2- ---> S + Cr3+

You can take it from there, if so desired.

Problem #9: K2Cr2O7 + H2S ---> Cr3+ + H2SO4 + K+

Solution: I propose to balance it in net-ionic form and then convert it to molecular form.

1) The net-ionic equation and half-reactions:

Cr2O72¯ + S2¯ ---> Cr3+ + SO42¯

Cr2O72¯ ---> Cr3+
S2¯ ---> SO42¯

2) Balance in acidic:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯

3) The least common multiple between 6 and 8 is 24. Equalize the electrons:

24e¯ + 56H+ + 4Cr2O72¯ ---> 8Cr3+ + 28H2O
12H2O + 3S2¯ ---> 3SO42¯ + 24H+ + 24e¯

32H+ + 4Cr2O72¯ + 3S2¯ ---> 8Cr3+ + 3SO42¯ + 16H2O

5) Convert back to molecular, step one:

26H+ + 4K2Cr2O7 + 3H2S ---> 8Cr3+ + 3SO42¯ + 16H2O + 8K+

Added eight potassium ions to each side and used six hydrogen ions to make three H2S.

5) Convert back to molecular, step two:

13H2SO4 + 4K2Cr2O7 + 3H2S ---> 8Cr3+ + 16SO42¯ + 16H2O + 8K+

Added 13 sulfate to each side.

6) Make formulas on the right-hand side:

13H2SO4 + 4K2Cr2O7 + 3H2S ---> 4Cr2(SO4)3 + 16H2O + 4K2SO4

Notice that no sulfuric acid winds up on the right-hand side. I wonder if it was a misdirection attempt by the person who wrote the equation. They were looking to see if you were confident enough to overcome the misdirection.

Problem #10: KMnO4 + FeSO4 + H2SO4 ---> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O

1) Write the net ionic equation:

MnO4- + Fe2+ ---> Mn2+ + Fe3+

2) Half-reactions and balance in net ionic form:

MnO4¯ ---> Mn2+
Fe2+ ---> Fe3+

8H+ + MnO4¯ + 5Fe2+ ---> Mn2+ + 5Fe3+ + 4H2O

However, there will be a major problem, so much so that the balanced net ionic equaton will be insufficient for what lies ahead.

The problem has to do with adding the potassium back in to make the molecular equation. One the left-hand side we have potassium permanganate (with only one potassium per formula unit) and on the right-hand side, the only place to add potassium is with potassium sulfate (with two potassium per formula unit). I can never balance the potassium with the net ionic as written above. (Yes, I could put a 2 in front of KMnO4. Try it and see what happens. It will probably be much harder to figure out than what I'm going to do below.)

So, how to solve?

Notice that I said the balanced net ionic is insufficient, not incorrect. Perhaps something can be done to it to make it sufficient? And, suprisingly enough, the answer is yes. (I'll bet you weren't suprised!)

In the net ionic above, I used factors of 1 and 5 to multiply through so that, when I added the two half-reactions, the electrons cancel out. To solve my potassium problem, I will use factors of 2 and 10 to get this:

16H+ + 2MnO4¯ + 10Fe2+ ---> 2Mn2+ + 10Fe3+ + 8H2O

Notice that I will now use 2KMnO4 (and one K2SO4) to balance the potassium. So, let's do that:

3) Balance the potassium:

16H+ + 2KMnO4 + 10Fe2+ ---> 2Mn2+ + 10Fe3+ + 8H2O + K2SO4

I balance the sulfate introduced on the right like this:

H2SO4 + 14H+ + 2KMnO4 + 10Fe2+ ---> 2Mn2+ + 10Fe3+ + 8H2O + K2SO4

4) Add sulfates to the iron(II) and iron(III) ions (follow this carefully):

H2SO4 + 14H+ + 2KMnO4 + 10FeSO4 ---> 2Mn2+ + 5Fe2(SO4)3 + 8H2O + K2SO4

The problem is that I added 10 sulfates to the left and 15 to the right. So add 5 sulfates to the left:

6H2SO4 + 4H+ + 2KMnO4 + 10FeSO4 ---> 2Mn2+ + 5Fe2(SO4)3 + 8H2O + K2SO4

Notice how the sulfuric acid goes from 1 to 6 and the hydrogen ion goes down by 10.