Balancing redox equations when three half-reactions are required

Redox equations where four half-reactions are required

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Before starting the examples, be assured that at least one half-reaction will be a reduction and at least one half-reaction will be an oxidation.

This link will take you to a list of all the three-equation problems minus the solutions.


Problem #1: FeS + NO3¯ ---> NO + SO42¯ + Fe3+ [acidic sol.]

Solution:

1) Separate out the half-reactions. The only issue is that there are three of them.

Fe2+ ---> Fe3+
S2¯ ---> SO42¯
NO3¯ ---> NO

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going -2 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe2+ with the S2¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe2+ to S2¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

Fe2+ ---> Fe3+ + e¯
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.


Problem #2: CrI3 + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ [basic sol.]

Solution:

Go to this video for the solution

Problem #3: Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Solution:

1) Remove all the spectator ions:

Sb26+ + CO32- + C ---> Sb + CO

Notice that I did not write Sb3+. I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

Sb26+ ---> Sb
CO32- ---> CO
C ---> CO

3) Balance as if in acidic solution:

6e¯ + Sb26+ ---> 2Sb
2e¯ + 4H+ + CO32- ---> CO + 2H2O
H2O + C ---> CO + 2H+ + 2e¯

Could you balance in basic? I suppose, but why?

4) Use a factor of three on the second half-reaction and a factor of six on the third.

6e¯ + Sb26+ ---> 2Sb
3 [2e¯ + 4H+ + CO32- ---> CO + 2H2O]
6 [H2O + C ---> CO + 2H+ + 2e¯]

The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H+ on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.

Everything that needs to cancel gets canceled!

5) The answer (with spectator ions added back in):

Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO

6) Here's a slightly different take on the solution just presented.

a) Write the net ionic equation:
Sb26+ + CO32- + C ---> Sb + CO

b) Notice that charges must be balanced and that we have zero charge on the right. So, do this:

Sb26+ + 3CO32- + C ---> Sb + CO

c) Now, balance for atoms:

Sb26+ + 3CO32- + 6C ---> 2Sb + 9CO

d) Add back the sodium ions and sulfide ions to recover the molecular equation.

Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO

7) Here's a discussion of a wrong answer to the above problem.

However, after reading the above wrong answer example, look at problem #10 below for an instance of having to add in a substance not included in the original reaction.


Problem #4: CrI3 + H2O2 ---> CrO42¯ + IO4¯ [basic sol.]

Solution:

1) write the half-reactions:

Cr3+ ---> CrO42¯
I33¯ ---> IO4¯
H2O2 ---> H2O

I wrote the iodide as I33¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯
12H2O + I33¯ ---> 3IO4¯ + 24H+ + 24e¯
2e¯ + 2H+ + H2O2 ---> 2H2O

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯

4) Equalize the electrons:

2 [16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯]
27 [2e¯ + 2H+ + H2O2 ---> 2H2O]

leads to:

32H2O + 2CrI3 ---> 6IO4¯ + 2CrO42¯ + 64H+ + 54e¯
54e¯ + 54H+ + 27H2O2 ---> 54H2O

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

2CrI3 + 27H2O2 ---> 2CrO42¯ + 6IO4¯ + 10H+ + 22H2O

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

2CrI3 + 27H2O2 + 10OH¯ ---> 2CrO42¯ + 6IO4¯ + 32H2O

Problem #5: XO2+ + YO+ ---> X2O43¯ + Y¯ + Y3O72¯ [basic sol.]

Solution:

1) Write the three half-reactions:

XO2+ ---> X2O43¯

YO+ ---> Y¯

YO+ ---> Y3O72¯

2) Balance each half-reaction as if in acidic solution:

5e¯ + 2XO2+ ---> X2O43¯

4e¯ + 2H+ + YO+ ---> Y¯ + H2O

4H2O + 3YO+ ---> Y3O72¯ + 8H+ + 3e¯

3) Multiply the third half-reaction by three:

12H2O + 9YO+ ---> 3Y3O72¯ + 24H+ + 9e¯

This gives nine electrons on each side when the three half-reactions are added together.

4) Add the half-reactions together and eliminate like items (two H+ and one H2O):

2XO2+ + 10YO+ + 11H2O ---> X2O43¯ + Y¯ + 3Y3O72¯ + 22H+

5) Add 22 hydroxides to each side and eliminate like items:

22OH¯ + 2XO2+ + 10YO+ ---> X2O43¯ + Y¯ + 3Y3O72¯ + 11H2O

Eleven waters were eliminated after the addition of the 22 hydroxides.

Comment: I got this equation from a question on Yahoo Answers. Click here to see the original question.

The mistake the person made was not seeing that he had an oxidation and a reduction in his second "half-reaction." If you add the second and third half-reactions from my answer above, you will get what the asker wrote for his second "half-reaction." I have reproduced it below in case this question ever gets deleted on Yahoo.

e¯ + 3H2O + 4YO+ ---> Y¯ + Y3O72¯ + 6H+

Problem #6: Bi(NO3)3 + Al + NaOH ---> Bi + NH3 + NaAlO2

Solution:

1) Write the three half-reactions while also stripping out spectator ions (only the sodium ion!):

Bi3+ ---> Bi
NO3¯ ---> NH3
Al ---> AlO2¯

2) Balance the half-reations as if in acidic solution (we'll change to basic in a moment):

Bi3+ + 3e¯ ---> Bi
8e¯ + 9H+ + NO3¯ ---> NH3 + 3H2O
2H2O + Al ---> AlO2¯ + 4H+ + 3e¯

3) The first step in equalizing the electrons is to see that the Bi and the nitrate MUST be in a 1:3 ratio:

Bi3+ + 3e¯ ---> Bi
3 [8e¯ + 9H+ + NO3¯ ---> NH3 + 3H2O]
2H2O + Al ---> AlO2¯ + 4H+ + 3e¯

4) Now we balance the 27 electrons on the left with 27 on the right:

Bi3+ + 3e¯ ---> Bi
3 [8e¯ + 9H+ + NO3¯ ---> NH3 + 3H2O]
9 [2H2O + Al ---> AlO2¯ + 4H+ + 3e¯]

5) Multiply through and add the three half-reactions:

Bi3+ + 27H+ + 3NO3¯ + 18H2O + 9Al ---> Bi + 3NH3 + 9H2O + 9AlO2¯ + 36H+

6) Eliminate duplicate water and hydrogen ion:

Bi(NO3)3 + 9H2O + 9Al ---> Bi + 3NH3 + 9AlO2¯ + 9H+

7) Change to basic solution and elminate the nine waters that result:

Bi(NO3)3 + 9H2O + 9Al + 9OH¯ ---> Bi + 3NH3 + 9AlO2¯ + 9H2O

Bi(NO3)3 + 9Al + 9OH¯ ---> Bi + 3NH3 + 9AlO2¯

8) Add the sodium ion back in:

Bi(NO3)3 + 9Al + 9NaOH ---> Bi + 3NH3 + 9NaAlO2

Comment: I could have written the nitrate in the same fashion as I wrote the iodide in a problem above, where I wrote I33¯, but I did not. If I had, then there would have been 24 electrons on the left in the final half-reaction.

The above problem was formatted the morning of February 20, 2011 while aboard the Queen Mary, in a stateroom on the "B" level.


Problem #7: Cu3P + Cr2O72¯ ---> Cu2+ + H3PO4 + Cr3+

Solution

1) Split into half-reactions:

Cu33+ ---> Cu2+
P3¯ ---> PO43¯
Cr2O72¯ ---> Cr3+

2) Balance in acidic solution:

Cu33+ ---> 3Cu2+ + 3e¯
4H2O + P3¯ ---> PO43¯ + 8H+ + 8e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

3) Combine first two half-reactions to recover Cu3P:

4H2O + Cu3P ---> 3Cu2+ + PO43¯ + 8H+ + 11e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

4) Equalize electrons:

24H2O + 6Cu3P ---> 18Cu2+ + 6PO43¯ + 48H+ + 66e¯
66e¯ + 154H+ + 11Cr2O72¯ ---> 22Cr3+ + 77H2O

5) Add, then eliminate water and hydrogen ion to get:

106H+ + 11Cr2O72¯ + 6Cu3P ---> 18Cu2+ + 6PO43¯ + 22Cr3+ + 53H2O

6) Recover phosphoric acid:

124H+ + 11Cr2O72¯ + 6Cu3P ---> 18Cu2+ + 6H3PO4 + 22Cr3+ + 53H2O

7) Although not needed, here's a full molecular equation:

124HCl + 11K2Cr2O7 + 6Cu3P ---> 18CuCl2 + 6H3PO4 + 22CrCl3 + 53H2O + 22KCl

Problem #8: As2S3 + NO3¯ ---> H3AsO4 + S + NO

Solution:

1) Write the half-reactions:

As26+ ---> H3AsO4
S36¯ ---> S
NO3¯ ---> NO

Note how I kept the As and the S together in the first two half-reactions. This is because I know I will recombine them in the final answer, so I wanted to easily preserve the 2:3 ratio of the As and the S.

If I had not done this, then I would have had to make sure the As half-rection was multiplied through by 2 and the S half-reaction multiplied through by three. As it is, all I need to do is make sure the two half-reactions are multiplied through by the same factor, if a factor is needed.

Also, I could have eliminated the hydrogen from H3AsO4. There's no need to do so, so I didn't.

2) Balance in acidic solution (because of the H3AsO4):

8H2O + As26+ ---> 2H3AsO4 + 10H+ + 4e¯
S36¯ ---> 3S + 6e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Multiply through by factors selected to balance the electrons:

3 [8H2O + As26+ ---> 2H3AsO4 + 10H+ + 4e¯]
3 [S36¯ ---> 3S + 6e¯]
10 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

This gives 30e¯ on each side, when the half-reactions are combined.

4) Show the three half-reactions with the factors applied:

24H2O + 3As26+ ---> 6H3AsO4 + 30H+ + 12e¯
3S36¯ ---> 9S + 18e¯
30e¯ + 40H+ + 10NO3¯ ---> 10NO + 20H2O

5) Add everything, eliminating only electrons:

24H2O + 3As26+ + 3S36¯ + 40H+ + 10NO3¯ ---> 6H3AsO4 + 9S + 30H+ + 10NO + 20H2O

6) Eliminate hydrogen ion and water:

4H2O + 3As26+ + 3S36¯ + 10H+ + 10NO3¯ ---> 6H3AsO4 + 9S + 10NO

7) Recombine (and rearrange):

3As2S3 + 10HNO3 + 4H2O---> 6H3AsO4 + 9S + 10NO

I put the water at the end to make the final answer correspond a bit closer to the order the substances were in the original problem statement.

The 10 hydrogens on the left-hand side did not magically appear out of nowhere. Keep in mind that this reaction is occuring in acidic solution. If I had removed the hydrogen from the arsenic acid at the beginning, I would have had 8 hydrogen ions on the right-hand side when all was said and done. I would have simply added 10 more H+ on each side, recovering the HNO3 and the H3AsO4.


Problem #9: As2S3 + NO3¯ ---> H3AsO4 + S8 + NO

1) Write the half-reactions:

As26+ ---> H3AsO4
S36¯ ---> S8
NO3¯ ---> NO

2) Balance in acidic solution (because of the H3AsO4):

64H2O + 8As26+ ---> 16H3AsO4 + 80H+ + 32e¯
8S36¯ ---> 3S8 + 48e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Combine the first two half-reactions:

64H2O + 8As2S3 ---> 3S8 + 16H3AsO4 + 80H+ + 80e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

4) Multiply through by factors selected to balance the electrons:

3 [64H2O + 8As2S3 ---> 3S8 + 16H3AsO4 + 80H+ + 80e¯]
80 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

4) Show the two half-reactions with the factors applied:

192H2O + 24As2S3 ---> 9S8 + 48H3AsO4 + 240H+ + 240e¯
240e¯ + 320H+ + 80NO3¯ ---> 80NO + 160H2O

5) Add everything, eliminating only electrons:

320H+ + 80NO3¯ + 192H2O + 24As2S3 ---> 9S8 + 48H3AsO4 + 240H+ + 80NO + 160H2O

6) Eliminate hydrogen ion and water:

80H+ + 80NO3¯ + 32H2O + 24As2S3 ---> 9S8 + 48H3AsO4 + 80NO

7) Recombine (and rearrange):

24As2S3 + 80HNO3 + 32H2O ---> 9S8 + 48H3AsO4 + 80NO

Problem #10: S8 + O2 ---> SO42¯

I'm going to approach the solution in the normal way and let the difficulties with this equation appear during the solution.

Solution:

1) Separate into half-reactions:

Here's the first problem: the S is oxidized and the O is reduced, but they are both in the sulfate. This difficulty is solved by splitting the sulfate between the two half-reactions.

S8 ---> S6+

O2 ---> O48¯

Notice how the four oxides each have a -2 charge, giving the -8 total.

2) Balance the half-reactions:

(1/8) S8 ---> S6+ + 6e¯

8e¯ + 2O2 ---> O48¯

Here arises the second problem and it lies in the number of electrons. If you equalize the number of electrons (multiply first half-reaction by eight, multiply second half-reaction by six), you create problems with the S6+ and the O48¯.

What do I mean by this? Keep in mind that splitting the sulfate was done to separate the reduction from the oxidation, this type of split does not occur in nature. Therefore, we will be forced to re-unite the S6+ and the O48¯ at the end of the balancing process.

That means the the S6+ and the O48¯ MUST remain equal. We cannot do that using the above two half-reactions. This is because we are working under two constraints (the second one being unique to this problem):

1) keep the electron amounts equal
2) keep the S6+ and the O48¯ amounts equal

What to do?

3) The answer is to introduce a third equation, one that does not appear in the original problem. The third equation is this:

H2 ---> 2H+¯ + 2e¯

The reason for this lies in our need to equalize the electrons. This half-reaction gives us what we need.

Also, go back to the original equation. Notice that there is zero charge on the left-hand side and -2 on the right-hand side. Where did the two extra electrons come from? The answer: from the H2 that was not written in the problem. You may suspect that the original statement of the problem was deliberately incomplete. I think that would be a correct feeling. By the way, I did not write this problem.

4) All three half-reactions:

(1/8) S8 ---> S6+ + 6e¯

8e¯ + 2O2 ---> O48¯

H2 ---> 2H+¯ + 2e¯

5) Since there are now eight electrons on each side; we can add the three half-reactions together and recombine the S6+ and the O48¯:

(1/8) S8 + 2O2 + H2 ---> H2SO4

Notice how I combined the hydrogen ion and the sulfate. Do you recognize the equation? It is the formation equation for sulfuric acid, you can probably look up its enthalpy in an appendix in the back of your book. The value given here is -814 kJ/mol.


Problem #11: 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 #12: 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
7) Add for final answer:
38H+ + 5CNS¯ + 11MnO4¯ ---> 5CO2 + 5NO + 5SO2 + 11Mn2+ + 19H2O

Problem #13: 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-

5) Add:

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 #14: 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 #15: 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 #16: 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 #17: 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 did 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 #18: 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 #19: 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 #20: 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 #21: 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 #22: 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

Redox equations where four half-reactions are required

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