### Balancing redox equations when three half-reactions are requiredTen Examples

Example #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.

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

Solution:

Go to this video for the solution

Example #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!

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.

Example #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]

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

Example #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.

Example #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.

Example #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

Example #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.

Example #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

Example #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:

18S8 ---> 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:

18S8 ---> 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¯:

18S8 + 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.

Bonus Example: Cr3+ + I¯ + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ (two answers given, both correct)

Solution #1:

1) Half-reactions:

Cr3+ ---> CrO42¯
I¯ ---> IO4¯
Cl2 ---> Cl¯

2) Balance:

4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯
4H2O + I¯ ---> IO4¯ + 8H+ + 8e¯
2e¯ + Cl2 ---> 2Cl¯

8H2O + Cr3+ + I¯ ---> CrO42¯ + IO4¯ + 16H+ + 11e¯
2e¯ + Cl2 ---> 2Cl¯

4) Equalize electrons:

16H2O + 2Cr3+ + 2I¯ ---> 2CrO42¯ + 2IO4¯ + 32H+ + 22e¯
22e¯ + 11Cl2 ---> 22Cl¯

16H2O + 2Cr3+ + 2I¯ + 11Cl2 ---> 2CrO42¯ + 2IO4¯ + 22Cl¯ + 32H+

Solution #2:

1) Reproduce the three balanced half-reactions:

4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯
4H2O + I¯ ---> IO4¯ + 8H+ + 8e¯
2e¯ + Cl2 ---> 2Cl¯

2) Multiply first by 2 and third by 7:

8H2O + 2Cr3+ ---> 2CrO42¯ + 16H+ + 6e¯
4H2O + I¯ ---> IO4¯ + 8H+ + 8e¯
14e¯ + 7Cl2 ---> 14Cl¯