### Balancing redox equations when three half-reactions are requiredProblems #1 - 15

Problem #1: HNO3 + H3AsO4 + Zn ---> AsH3 + Zn(NO3)2

Solution

1) Split into half-reactions:

H33+ ---> H33¯
AsO43¯ ---> As3+
Zn ---> Zn2+

2) Balance:

6e¯ + H33+ ---> H33¯
2e¯ + 8H+ + AsO43¯ ---> As3+ + 4H2O
Zn ---> Zn2+ + 2e¯

3) Combine the first two half-reactions:

8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O
Zn ---> Zn2+ + 2e¯

This was done to get the H3AsO4 back.

4) Equalize electrons:

8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O
4 [Zn ---> Zn2+ + 2e¯]

8H+ + H3AsO4 + 4Zn ---> AsH3 + 4Zn2+ + 4H2O

6) Take it back to the full molecular equation:

8HNO3 + H3AsO4 + 4Zn ---> AsH3 + 4Zn(NO3)2 + 4H2O

In problem #14 below, I solve the exact same problem, just with a different explanation. It turned out I did the same problem in two different files so, when I discovered that, I decided to put them in the same file.

Problem #2: CNS¯ + MnO4¯ -----> CO2 + NO + SO2 + Mn2+

Solution

1) Examine the oxidation numbers:

a) thiocyanate ion: C = +4, N = −3, S = −2
b) carbon dioxide: C = +4
c) nitrogen monoxide: N = +2
d) sulfr dioxide: S = +4
e) permanganate ion: Mn = +7 (and it goes to a +2)

Comment: notice that the carbon does not change in its oxidation state.

2) Write the half-reactions:

N3¯ ---> NO
S2¯ ---> SO2
MnO4¯ ---> Mn2+

3) Balance them in acidic solution:

H2O + N3¯ ---> NO + 2H+ + 5e¯
2H2O + S2¯ ---> SO2 + 4H+ + 6e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

4) Recreate the thiocyanate ion:

a) add the first two half-reactions:
3H2O + NS5¯ ---> NO + SO2 + 6H+ + 11e¯

3H2O + CNS¯ ---> CO2 + NO + SO2 + 6H+ + 11e¯

c) Balance it:

5H2O + CNS¯ ---> CO2 + NO + SO2 + 10H+ + 11e¯
Comment 1: in balancing the half-reaction just above, adding the carbon did not change the number of electrons because carbon was not oxidized or reduced. Also, adding water and hydrogen ion did not affect the electrons because neither H nor O was reduced or oxidized.

Comment 2: another was to recreate the thiocyanate would be to construct a fourth half-reaction:

2H2O + C4+ ---> CO2 + 4H+

It is neither a reduction nor an oxidation, but you can use it in conjunction with the half-reactions involving N and S to recover the thiocyanate ion. I do that in Problem #21 below.

5) Rewrite our now two half-reactions:

5H2O + CNS¯ ---> CO2 + NO + SO2 + 10H+ + 11e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

6) Equalize electrons:

25H2O + 5CNS¯ ---> 5CO2 + 5NO + 5SO2 + 50H+ + 55e¯
55e¯ + 88H+ + 11MnO4¯ ---> 11Mn2+ + 44H2O
38H+ + 5CNS¯ + 11MnO4¯ ---> 5CO2 + 5NO + 5SO2 + 11Mn2+ + 19H2O

Problem #3: K2Cr2O7 + HCl -----> KCl + CrCl + CrCl3 + H2O + Cl2

Solution

1) Net ionic:

Cr2O72- + Cl- ---> Cr+ + Cr3+ + Cl2

2) Three half-reactions:

Cr2O72- ---> Cr+
Cr2O72- ---> Cr3+
Cl- ---> Cl2

3) Balance them:

10e- + 14H+ + Cr2O72- ---> 2Cr+ + 7H2O
6e- + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O
2Cl- ---> Cl2 + 2e-

4) Equalize the electrons:

10e- + 14H+ + Cr2O72- ---> 2Cr+ + 7H2O
6e- + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O
16Cl- ---> 8Cl2 + 16e-

28H+ + 2Cr2O72- + 16Cl- ---> 2Cr+ + 2Cr3+ + 14H2O + 8Cl2

6) Reduce:

14H+ + Cr2O72- + 8Cl- ---> Cr+ + Cr3+ + 7H2O + 4Cl2

7) To recover molecular equation, I need to beef up the chloride because I eventually want to make HCl. I'll add 4 chlorides to each side:

14H+ + Cr2O72- + 12Cl- ---> CrCl + CrCl3 + 7H2O + 4Cl2

8) Now, I'll add two potassium:

14H+ + K2Cr2O7 + 12Cl- ---> 2K^+ + CrCl + CrCl3 + 7H2O + 4Cl2

9) Two more chloride:

14H+ + K2Cr2O7 + 14Cl- ---> 2KCl + CrCl + CrCl3 + 7H2O + 4Cl2

10) Last step:

K2Cr2O7 + 14HCl ---> 2KCl + CrCl + CrCl3 + 7H2O + 4Cl2

Problem #4: Zn + H3AsO4 + HNO3 ---> AsH3 + Zn(NO3)2 + H2O

Solution

1) The change in zinc's oxidation state should be rather obvious. It makes the first half-reaction:

Zn ---> Zn2+ + 2e¯

Notice that I already balanced it. In addition, note that the nitrate has been removed. The nitrogen in the nitrate is neither oxidized nor reduced.

2) The arsenic should be a fairly obvious candidate for something and, in fact, it gets reduced. In the arsenate, the arsenic is a +5 and in arsine it is a +3. Here's the unbalanced half-reaction:

AsO43¯ ---> As3+

Notice that I used the full polyatomic ion for arsenate. This is common practice in going from the molecular equation to the net ionic. Use the full polyatomic ion.

3) The third half-reaction (already balanced) is this one:

2e¯ + H+ ---> H¯

How do I know this is the third half-reaction? Things that I know: (1) AsH3 is a hydride, therefore the hydrogen is a −1 oxidation state, (2) it is a product and (3) hydrogen is in a +1 oxidation state as a (4) reactant.

How did I know to split up the AsH3 into two pieces? I know: (1) you can only have one reduction or one oxidation in a half-reaction and (2) As gets reduced AND H also gets reduced. The solution? Split AsH3 into two pieces, each with its own half-reaction.

Here is a warning: I will have to recombine the two pieces of AsH3 in the final answer and there MUST be three H for every one As. Look for how I do that.

4) Here are the three balanced half-reactions:

Zn ---> Zn2+ + 2e¯
2e¯ + 8H+ + AsO43¯ ---> As3+ + 4H2O
2e¯ + H+ ---> H¯

5) Here's how to equalize the electrons:

Note: at the end of the solution, I have added comments about a better way to recreate the AsH3.

4 [Zn ---> Zn2+ + 2e¯]
2e¯ + 8H+ + AsO43¯ ---> As3+ + 4H2O
3 [2e¯ + H+ ---> H¯]

The key? There MUST be a three in front of the third half-reaction. This gives me three H¯ to pair up with the one As3+. Since that means 8e¯ on the left, I have to put a 4 in front of the first half-reaction, thereby equalizing the electrons. Also, please note that I have to keep the second half-reaction as is because I need to have one As3+ for my three H¯.

6) Add up the three half-reactions to give:

4Zn + 8H+ + AsO43¯ + 3H+ ---> 4Zn2+ + AsH3 + 4H2O

7) Notice that I combined the As3¯ and the 3H¯. Also notice that I did not combine the H+ on the left-hand side. I shall now combine the 3H+ with the arsenate ion:

4Zn + 8H+ + H3AsO4 ---> 4Zn2+ + AsH3 + 4H2O

8) The only remaining task is to re-introduce the nitrate ion; we will need 8 of them:

4Zn + 8HNO3 + H3AsO4 ---> 4Zn(NO3)2 + AsH3 + 4H2O

The better way to recreate the arsine is to combine the second and third half-reactions first:

2e¯ + 8H+ + AsO43¯ ---> As3+ + 4H2O
3 [2e¯ + H+ ---> H¯]

which gives:

8e¯ + 11H+ + AsO43¯ ---> AsH3 + 4H2O

You then use the above with the zinc half-reaction. You equalize the electrons by multiplying the zinc half-reaction by 4 and then add the two half-reactions together

Problem #5: Al + NH4ClO4 ---> Al2O3 + HCl + N2 + H2O

Solution:

1) Three half-reactions:

Al ---> Al2O3
NH4+ ---> N2
ClO4¯ ---> Cl¯

2) Balance in acidic solution (the hint is in the ammonium, which is acidic in solution, the chlorate is neutral):

3H2O + Al ---> Al2O3 + 6H+ + 6e¯
2NH4+ ---> N2 + 8H+ + 6e¯
8e¯ + 8H+ + ClO4¯ ---> Cl¯ + 4H2O

3) Multiply third half-reaction by two:

3H2O + Al ---> Al2O3 + 6H+ + 6e¯
2NH4+ ---> N2 + 8H+ + 6e¯
16e¯ + 16H+ + 2ClO4¯ ---> 2Cl¯ + 8H2O

This is done to make the ammonium and perchlorate have the same coefficient.

4) A bit of explanation: you have to equalize the electrons, but you MUST use the same factor for the second and third half-reactions. This is because you will reunite the ammonium and the perchlorate and they have to be the same amounts. The least common multiple of 6 and 16 is 48, so do this:

15H2O + 10Al ---> 5Al2O3 + 30H+ + 30e¯
6NH4+ ---> 3N2 + 24H+ + 18e¯
48e¯ + 48H+ + 6ClO4¯ ---> 6Cl¯ + 24H2O

6NH4+ + 48H+ + 6ClO4¯ + 15H2O + 10Al ---> 5Al2O3 + 3N2 + 54H+ + 6Cl¯ + 24H2O

6) Eliminate like items, then recombine to obtain the final answer:

6NH4+ + 6ClO4¯ + 10Al ---> 5Al2O3 + 3N2 + 6H+ + 6Cl¯ + 9H2O

6NH4ClO4 + 10Al ---> 5Al2O3 + 3N2 + 6HCl + 9H2O

A point about step 4: you can take a different route by combining the second and third half-reactions:

10e¯ + 8H+ + 2NH4ClO4 ---> N2 + 2Cl¯ + 8H2O

This gives you two half-reactions:

15H2O + 10Al ---> 5Al2O3 + 30H+ + 30e¯
10e¯ + 8H+ + 2NH4ClO4 ---> N2 + 2Cl¯ + 8H2O

You multiply the second half-reaction by three and then on to the final answer.

Problem #6: Cu + HNO3 ---> Cu(NO3)2 + HNO2 + NO + H2O

Solution:

1) Write the net ionic equation:

Cu + NO3¯ ---> Cu2+ + NO2¯ + NO

Even though HNO2 is a weak acid and normally written in molecular form, I decided to include just the part that is involved in the redox. The hdrogen will be added back in later. As will the water.

2) Three half-reactions:

Cu ---> Cu2+

NO3¯ ---> NO2¯

NO3¯ ---> NO

3) Balance in acidic solution:

Cu ---> Cu2+ + 2e¯

2e¯ + 2H+ + NO3¯ ---> NO2¯ + H2O

3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

4) Combine the two reductions:

Cu ---> Cu2+ + 2e¯

5e¯ + 6H+ + 2NO3¯ ---> NO2¯ + NO + 3H2O

5) Equalize the electrons:

5Cu ---> 5Cu2+ + 10e¯

10e¯ + 12H+ + 4NO3¯ ---> 2NO2¯ + 2NO + 6H2O

12H+ + 4NO3¯ + 5Cu ---> 5Cu2+ + 2NO2¯ + 2NO + 6H2O

7) Add 10 nitrates and two hydrogen ions to both sides. First, to the right-hand side:

12H+ + 4NO3¯ + 5Cu ---> 5Cu(NO3)2 + 2HNO2 + 2NO + 6H2O

and then to the left:

14H+ + 14NO3¯ + 5Cu ---> 5Cu(NO3)2 + 2HNO2 + 2NO + 6H2O

8) Combine the nitric acid for the final answer:

5Cu + 14HNO3 ---> 5Cu(NO3)2 + 2HNO2 + 2NO + 6H2O

Problem #7: H3AsO4 + H2SO4 + NO ---> As2S3 + HNO3 + H2O

Solution:

1) Here are the three unbalanced half-reactions:

AsO43¯ ---> As26+
SO42¯ ---> S36¯
NO ---> NO3¯

2) Reunite As2S3 before balancing:

AsO43¯ + SO42¯ ---> As2S3
NO ---> NO3¯

I could have balanced the two half-reactions before reuniting to form As2S3. I do that in problem #19.

3) Balance in acidic solution:

28e¯ + 40H+ + 2AsO43¯ + 3SO42¯ ---> As2S3 + 20H2O
2H2O + NO ---> NO3¯ + 4H+ + 3e¯

4) Equalize the electrons:

84e¯ + 120H+ + 6AsO43¯ + 9SO42¯ ---> 3As2S3 + 60H2O
56H2O + 28NO ---> 28NO3¯ + 112H+ + 84e¯

8H+ + 6AsO43¯ + 9SO42¯ + 28NO ---> 3As2S3 + 28NO3¯ + 4H2O

6) I will add 28 hydrogen ions to each side and make nitric acid on the right-hand side:

36H+ + 6AsO43¯ + 9SO42¯ + 28NO ---> 3As2S3 + 28HNO3 + 4H2O

7) Distribute the 36H+ to make H3AsO4 and H2SO4:

6H3AsO4 + 9H2SO4 + 28NO ---> 3As2S3 + 28HNO3 + 4H2O

Problem #8: NaHSO3 + HCl + H2O2 ---> NaHSO4 + NaCl + SO2 + H2O

Solution:

1) Three half-reactions, done as net-ionic:

HSO3¯ ---> HSO4¯
HSO3¯ ---> SO2
H2O2 ---> H2O

2) Balance the half-reactions in acidic solution:

H2O + HSO3¯ ---> HSO4¯ + 2H+ + 2e¯
H+ + HSO3¯ ---> SO2 + H2O <--- not redox, but needed to get the SO2 into play
2e¯ + 2H+ + H2O2 ---> 2H2O

3) Add 'em up for the net-ionic:

H+ + 2HSO3¯ + H2O2 ---> HSO4¯ + SO2 + 2H2O

4) Add sodium ion and chloride ion to bring it back to molecular

HCl + 2NaHSO3 + H2O2 ---> NaHSO4 + NaCl + SO2 + 2H2O

Problem #9: ClO3¯ + As2S3 ---> Cl¯ + H2AsO4¯ + HSO4¯

Solution:

1) Three half-reactions:

ClO3¯ ---> Cl¯
As26+ ---> H2AsO4¯
S36¯ ---> HSO4¯

2) Balance:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
8H2O + As26+ ---> 2H2AsO4¯ + 12H+ + 4e¯
12H2O + S36¯ ---> 3HSO4¯ + 21H+ + 24e¯

3) Add the second and third half-reactions:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
20H2O + As2S3 ---> 2H2AsO4¯ + 3HSO4¯ + 33H+ + 28e¯

4) Multiply first half-reaction by 14 and second half-reaction by 3:

84e¯ + 84H+ + 14ClO3¯ ---> 14Cl¯ + 42H2O
60H2O + 3As2S3 ---> 6H2AsO4¯ + 9HSO4¯ + 99H+ + 84e¯

18H2O + 14ClO3¯ + 3As2S3 ---> 14Cl¯ + 6H2AsO4¯ + 9HSO4¯ + 15H+

Problem #10: FeC2O4 + KMnO4 + H2SO4 ---> Fe2(SO4)3 + CO2 + MnSO4 + K2SO4 + H2O

Solution:

1) Write the net-ionic equation:

Fe2+ + C2O42¯ + MnO4¯ ---> Fe3+ + CO2 + Mn2+

I split the iron(II) oxalate apart because I knew it would go into two different half-reactions.

2) Write the half-reactions:

MnO4¯ ---> Mn2+
Fe2+ ---> Fe3+
C2O42¯ ---> CO2

3) Using electrons only, balance the change in oxidation number:

5e¯ + MnO4¯ ---> Mn2+
Fe2+ ---> Fe3+ + e¯
C2O42¯ ---> 2CO2 + 2e¯

1) Mn went from +7 to +2, so five electrons gained to reduce the +7 to a +2.
2) Fe went from +2 to +3, so one electron lost to oxidize the +2 to a +3.
3) Two carbons oxidized from +3 to +4, so two electrons removed, one for each atom of C.

4) Equalize the electrons:

5e¯ + MnO4¯ ---> Mn2+
FeC2O4 ---> Fe3+ + 2CO2 + 3e¯

15e¯ + 3MnO4¯ ---> 3Mn2+
5FeC2O4 ---> 5Fe3+ + 10CO2 + 15e¯

Notice that I unified the two oxidations first.

5FeC2O4 + 3MnO4¯ ---> 5Fe3+ + 10CO2 + 3Mn2+

6) Balance oxygens with water, then balance hydrogens with H+:

Add water: 5FeC2O4 + 3MnO4¯ ---> 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O

Add H+: 5FeC2O4 + 3MnO4¯ + 24H+ ---> 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O

7) Check to make sure charges are balanced:

+21 on each side. Balanced. Yay!

8) To reconstitute the molecular equation:

You should first multiply through by two. This is because you will have an odd number of potassiums on the left (as in 3KMnO4), but the potassium on the right will only come out to be an even number (because of the K2SO4).

You can finish it from there.

Problem #11: K3Fe(SCN)6 + Na2Cr2O7 + H2SO4 ---> Fe(NO3)3 + Cr2(SO4)3 + Na2SO4 + KNO3 + CO2 + H2O

Solution:

Some initial comments: this looks a lot like a four half-reaction problem, but upon closer inspection, it a three half-reaction problem. First off, the oxidation number of the Fe remains unchanged. It's +3 in both of its compounds. The same for carbon, it's +4 in the thiocyanate and in the carbon dioxide.

That leaves us with these changes:

Cr ---> +6 in dichromate to +3 in the Cr(III) ion
N ---> −3 in thiocyanate to +5 in nitrate
S ---> −2 in thiocyanate to +6 in sulfate

1) The first task will be to create a net-ionic equation of just the redox portion:

SCN¯ + Cr2O72¯ ---> NO3¯ + Cr3+ + SO42¯ + CO2

I kept the carbon dioxide in because of the thiocyanate ion. I will include it in its own "half-reaction" so as to be able to reunite the S, the C, and the N of the thiocyanate ion. Yes, in a moment, I'm going to break the thiocyanate up.

2) The half-reactions:

Cr2O72¯ ---> Cr3+
S2¯ ---> SO42¯
C4+ ---> CO2
N3¯ ---> NO3¯

3) Balance the half-reactions:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯
2H2O + C4+ ---> CO2 + 4H+
3H2O + N3¯ ---> NO3¯ + 6H+ + 8e¯

Notice that there are no electrons in the third half-reaction because it is not a reduction or an oxidation.

4) I'm going to re-form the thiocyanate ion:

9H2O + SCN¯ ---> SO42¯ + CO2 + NO3¯ + 18H+ + 16e¯

54H2O + (SCN)66¯ ---> 6SO42¯ + 6CO2 + 6NO3¯ + 108H+ + 96e¯

I decided to add in the subscript of 6. That will make it easier to re-form the ferrithiocyanate ion a bit further down in the solution.

5) Equalize the electrons:

96e¯ + 224H+ + 16Cr2O72¯ ---> 32Cr3+ + 112H2O
54H2O + (SCN)66¯ ---> 6SO42¯ + 6CO2 + 6NO3¯ + 108H+ + 96e¯

116H+ + (SCN)66¯ + 16Cr2O72¯ ---> 32Cr3+ + 6SO42¯ + 6CO2 + 6NO3¯ + 58H2O

7) Now, we have to rebuild the molecular equation:

58H2SO4 + K3Fe(SCN)6 + 16Na2Cr2O7 ---> 32Cr3+ + 6SO42¯ + 6CO2 + 6NO3¯ + 58H2O

Added 58 sulfates, three potassium, one iron, and 32 sodium to the left-hand side

8) I'll not add everything to the right-hand side in one step:

58H2SO4 + K3Fe(SCN)6 + 16Na2Cr2O7 ---> 32Cr3+ + 6SO42¯ + Fe(NO3)3 + 6CO2 + 3KNO3 + 58H2O

Added three potassium and one iron to the right-hand side. Still have sulfate and sodium left to go.

9) I will add 48 sulfates to the RHS and make chromium(III) sulfate:

58H2SO4 + K3Fe(SCN)6 + 16Na2Cr2O7 ---> 16Cr2(SO4)3 + 6SO42¯ + Fe(NO3)3 + 6CO2 + 3KNO3 + 58H2O

I still have 10 sulfate and 32 sodium left to be added in.

Notice that I left the six sulfates uncombined. I'm going to use them to make the sodium sulfate in the last step.

10) Make 16 sodium sulfate:

58H2SO4 + K3Fe(SCN)6 + 16Na2Cr2O7 ---> 16Cr2(SO4)3 + 16Na2SO4 + Fe(NO3)3 + 6CO2 + 3KNO3 + 58H2O

Problem #12: Fe + H2SO4 ---> FeSO4 + Fe2(SO4)3 + H2O + SO2

Solution:

1) The three half-reactions are these:

Fe ---> Fe2+
Fe ---> Fe26+
SO42¯ ---> SO2

Notice the subscript of 2 on the Fe(III) ion. I did that because I knew I would make Fe2(SO4)3 at the end of the problem.

2) Balance:

Fe ---> Fe2+ + 2e¯
2Fe ---> Fe26+ + 6e¯
2e¯ + 4H+ + SO42¯ ---> SO2 + 2H2O

3) Equalize electrons:

Fe ---> Fe2+ + 2e¯
2Fe ---> Fe26+ + 6e¯
8e¯ + 16H+ + 4SO42¯ ---> 4SO2 + 8H2O

3Fe + 16H+ + 4SO42¯ ---> Fe2+ + Fe26+ + 4SO2 + 8H2O

5) Sometimes it seems best to address the reactant side first, sometimes the product side. It seems best to address the product side first in this problem.

3Fe + 16H+ + 4SO42¯ ---> FeSO4 + Fe2(SO4)3 + 4SO2 + 8H2O

6) I added four sulfates to the right side, so I will add four to the left side:

3Fe + 16H+ + 8SO42¯ ---> FeSO4 + Fe2(SO4)3 + 4SO2 + 8H2O

7) Form the sulfuric acid:

3Fe + 8H2SO4 ---> FeSO4 + Fe2(SO4)3 + 4SO2 + 8H2O

Problem #13: As2S3 + K2Cr2O7 + H2SO4 ---> H3AsO4 + K2SO4 + Cr2(SO4)3 + H2O

Solution:

1) Eliminate spectator ions:

As2S3 + Cr2O72¯ ---> H3AsO4 + Cr3+ + SO42¯

2) Write half-reactions:

As26+ ---> H3AsO4
S36¯ ---> SO42¯
Cr2O72¯ ---> Cr3+

I could have written the Cr3+ as Cr26+ in order to set up re-forming the Cr2(SO4)3. I decided to not do it and re-create the Cr2(SO4)3 a bit further down in the solution. There's no particular reason other than I just decided to do it that way.

3) Balance:

8H2O + As26+ ---> 2H3AsO4 + 10H+ + 4e¯
12H2O + S36¯ ---> 3SO42¯ + 24H+ + 24e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

4) Add the first two half-reactions together:

20H2O + As2S3 ---> 2H3AsO4 + 3SO42¯ + 34H+ + 28e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

I re-created the As2S3

5) Equalize electrons:

After some study, it is found that 3 x 28 = 6 x 14 = 84.

60H2O + 3As2S3 ---> 6H3AsO4 + 9SO42¯ + 102H+ + 84e¯
84e¯ + 196H+ + 14Cr2O72¯ ---> 28Cr3+ + 98H2O

6) Add the two half-reactions and eliminate duplicate items:

94H+ + 3As2S3 + 14Cr2O72¯ ---> 6H3AsO4 + 14Cr26+ + 9SO42¯ + 38H2O

I re-created Cr26+ but I have not yet created Cr2(SO4)3.

7) Add 47 sulfates to each side:

47H2SO4 + 3As2S3 + 14Cr2O72¯ ---> 6H3AsO4 + 14Cr26+ + 56SO42¯ + 38H2O

8) Add 28 potassium ions to each side:

47H2SO4 + 3As2S3 + 14K2Cr2O7 ---> 6H3AsO4 + 14Cr26+ + 42SO42¯ + 14K2SO4 + 38H2O

9) Make some chromium(III) sulfate:

47H2SO4 + 3As2S3 + 14K2Cr2O7 ---> 6H3AsO4 + 14Cr2(SO4)3 + 14K2SO4 + 38H2O

14Cr2(SO4)3 contains 42 sulfates! Everything works!

Problem #14: KNO3 + S8 ---> K2S + N2 + O2

Solution:

1) Separate into half-reactions:

N5+ ---> N2
O36¯ ---> O2
S8 ---> S2¯

2) Balance:

10e¯ + 2N5+ ---> N2
2O36¯ ---> 3O2 + 12e¯
16e¯ + S8 ---> 8S2¯

3) Form nitrate from the first two half-reactions:

2NO3¯ ---> N2 + 3O2 + 2e¯
16e¯ + S8 ---> 8S2¯

4) Equalize electrons:

16NO3¯ ---> 8N2 + 24O2 + 16e¯
16e¯ + S8 ---> 8S2¯

16NO3¯ + S8 ---> 8S2¯ + 8N2 + 24O2

6) Add sixteen potassium ions to each side:

16KNO3 + S8 ---> 8K2S + 8N2 + 24O2

7) You might see the equation this way:

2KNO3 + S ---> K2S + N2 + 3O2

Problem #15: Al + NH4ClO4 ---> Al2O3 + AlCl3 + H2O + NO

Solution:

1) Half-reactions:

Al ---> Al2O3 + Al3+
NH4+ ---> NO
ClO4¯ ---> Cl¯

2) Balance:

3H2O + 3Al ---> Al2O3 + Al3+ + 6H+ + 9e¯
H2O + NH4+ ---> NO + 6H+ + 5e¯
8e¯ + 8H+ + ClO4¯ ---> Cl¯ + 4H2O

3) Add second and third equations:

3H2O + 3Al ---> Al2O3 + Al3+ + 6H+ + 9e¯
3e¯ + 2H+ + NH4ClO4 ---> NO + Cl¯ + 3H2O

4) Equalize electrons:

3H2O + 3Al ---> Al2O3 + Al3+ + 6H+ + 9e¯
9e¯ + 6H+ + 3NH4ClO4 ---> 3NO + 3Cl¯ + 9H2O