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

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A list of all the three-equation problems minus the solutions.

Redox equations where four half-reactions are required

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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¯]

5) Add the half-reactions:

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¯

b) Add in the carbon:

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:

Cr₂O₇^2- + Cl^- ---> Cr⁺ + Cr³⁺ + Cl₂

2) Three half-reactions:

Cr₂O₇^2- ---> Cr⁺
Cr₂O₇^2- ---> Cr³⁺
Cl^- ---> Cl₂

3) Balance them:

10e^- + 14H⁺ + Cr₂O₇^2- ---> 2Cr⁺ + 7H₂O
6e^- + 14H⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O
2Cl^- ---> Cl₂ + 2e^-

4) Equalize the electrons:

10e^- + 14H⁺ + Cr₂O₇^2- ---> 2Cr⁺ + 7H₂O
6e^- + 14H⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O
16Cl^- ---> 8Cl₂ + 16e^-

5) Add:

28H⁺ + 2Cr₂O₇^2- + 16Cl^- ---> 2Cr⁺ + 2Cr³⁺ + 14H₂O + 8Cl₂

6) Reduce:

14H⁺ + Cr₂O₇^2- + 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⁺ + Cr₂O₇^2- + 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

5) Add thre three half-reactions:

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

6) Add:

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¯

5) Add, then eliminate:

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¯

5) Add:

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.

5) Add:

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¯

6) Add:

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

4) Add:

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¯

5) Add:

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

5) Add:

3Al + 3NH₄ClO₄ ---> 3NO + Al₂O₃ + AlCl₃ + 6H₂O

Ten Examples

A list of all the three-equation problems minus the solutions.

Redox equations where four half-reactions are required

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