Balancing redox equations when four half-reactions are required

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Redox equations where three half-reactions are required


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 four-equation problems minus the solutions.


Problem #1: K4Fe(CN)6 + Ce(NO3)4 + KOH ---> Ce(OH)3 + Fe(OH)3 + H2O + K2CO3 + KNO3

Comment: I balance the above equation via net ionic half reactions and wind up at a full molecular equation for the answer. Sometimes, a net ionic equation is what is given to you:

Fe(CN)64¯ + Ce4+ ---> Ce(OH)3 + Fe(OH)3 + CO32¯ + NO3¯

Take a look at step 7, which is almost a balanced net ionic equation. I'll put the correct net ionic at the end of the solution.

1) This reaction has four half-reactions:

1) Fe2+ ---> Fe3+

2) Ce4+ ---> Ce3+

3) CN¯ ---> NO3¯ (the N goes from -3 to +5)

4) The C in CN¯ is +2, in CO32¯ it is +4

2) Balance the half-reactions as if in acidic solution:

Fe2+ ---> Fe3+ + e¯

e¯ + Ce4+ ---> Ce3+

N3¯ + 3H2O ---> NO3¯ + 6H+ + 8e¯

C2+ + 3H2O ---> CO32¯ + 6H+ + 2e¯

3) Combine the last two half-reactions and convert it to basic:

CN¯ + 6H2O ---> NO3¯ + CO32¯ + 12H+ + 10e¯

CN¯ + 12OH¯ ---> NO3¯ + CO32¯ + 6H2O + 10e¯

6CN¯ + 72OH¯ ---> 6NO3¯ + 6CO32¯ + 36H2O + 60e¯

Why the last step? To set up the eventual return to the full molecular equation and the formula of K4Fe(CN)6.

4) The three half reactions (with electrons equalized):

Fe2+ ---> Fe3+ + e¯

61e¯ + 61Ce4+ ---> 61Ce3+

6CN¯ + 72OH¯ ---> 6NO3¯ + 6CO32¯ + 36H2O + 60e¯

5) Add 'em up:

Fe2+ + 61Ce4+ + 6CN¯ + 72OH¯ ---> Fe3+ + 61Ce3+ + 6NO3¯ + 6CO32¯ + 36H2O

6) Start the process of going back to the molecular equation by putting back 4 potassium ions:

K4Fe(CN)6 + 61Ce4+ + 72OH¯ ---> Fe3+ + 61Ce3+ + 2NO3¯ + 4KNO3 + 6CO32¯ + 36H2O

This makes the K4Fe(CN)6.

7) Add 186 hydroxide ions:

K4Fe(CN)6 + 61Ce4+ + 258OH¯ ---> Fe(OH)3 + 61Ce(OH)3 + 2NO3¯ + 4KNO3 + 6CO32¯ + 36H2O

This makes Fe(OH)3 and Ce(OH)3 on the right-hand side.

8) Add 244 nitrate ions:

K4Fe(CN)6 + 61Ce(NO3)4 + 258OH¯ ---> Fe(OH)3 + 61Ce(OH)3 + 246NO3¯ + 4KNO3 + 6CO32¯ + 36H2O

This makes the cerium(IV) nitrate on the left. The 246 nitrate on the right includes 2 nitrates already present.

9) Add 258 potassium ions:

K4Fe(CN)6 + 61Ce(NO3)4 + 258KOH ---> Fe(OH)3 + 61Ce(OH)3 + 250KNO3 + 6K2CO3 + 36H2O

On the right-hand side, I added 12 potassium to the carbonate and 246 to the nitrate, then combined the KNO3 to get 250.

And it's done. Whew!

Comment: the balanced net ionic equation:

Fe(CN)64¯ + 61Ce4+ + 258OH¯ ---> Fe(OH)3 + 61Ce(OH)3 + 6NO3¯ + 6CO32¯ + 36H2O

Problem #2: K4Fe(CN)6 + H2SO4 + KMnO4 ---> MnSO4 + Fe2(SO4)3 + K2SO4 + HNO3 + CO2 + H2O

1) This reaction has four half-reactions:

1) Fe2+ ---> Fe3+

2) MnO4¯ ---> Mn2+

3) CN¯ ---> NO3¯ (the N goes from -3 to +5)

4) The C in CN¯ is +2, in CO2 it is +4

2) Balance the half-reactions in acidic solution:

Fe2+ ---> Fe3+ + e¯

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

N3¯ + 3H2O ---> NO3¯ + 6H+ + 8e¯

C2+ + 2H2O ---> CO2 + 4H+ + 2e¯

3) Combine the last two half-reactions and multiply by 6:

CN¯ + 5H2O ---> NO3¯ + CO2 + 10H+ + 10e¯

6CN¯ + 30H2O ---> 6NO3¯ + 6CO2 + 60H+ + 60e¯

Why the last step? To set up the eventual return to the full molecular equation and the formula of K4Fe(CN)6.

4) The three half reactions (with electrons equalized):

10Fe2+ ---> 10Fe3+ + 10e¯

610e¯ + 976H+ + 122MnO4¯ ---> 122Mn2+ + 488H2O

60CN¯ + 300H2O ---> 60NO3¯ + 60CO2 + 600H+ + 600e¯

What????? What happened?

5) Add 'em up (eliminate duplicate water and hydrogen ion):

10Fe2+ + 376H+ + 122MnO4¯ + 60CN¯ ---> 10Fe3+ + 122Mn2+ + 188H2O + 60NO3¯ + 60CO2

6) Add 40 potassium ions:

10K4Fe(CN)6 + 376H+ + 122MnO4¯ ---> 10Fe3+ + 122Mn2+ + 188H2O + 40K+ + 60NO3¯ + 60CO2

7) Add 60 hydrogen ions:

10K4Fe(CN)6 + 436H+ + 122MnO4¯ ---> 10Fe3+ + 122Mn2+ + 188H2O + 40K+ + 60HNO3 + 60CO2

8) Add 122 potassium ions:

10K4Fe(CN)6 + 436H+ + 122KMnO4 ---> 10Fe3+ + 122Mn2+ + 188H2O + 162K+ + 60HNO3 + 60CO2

9) Add 218 sulfate ions:

10K4Fe(CN)6 + 218H2SO4 + 122KMnO4 ---> 5Fe2(SO4)3 + 122MnSO4 + 188H2O + 81K2SO4 + 60HNO3 + 60CO2

On the right, the sulfates are distributed thus: 15 to the iron(III) ion, 122 to the managanese(II) ion and 81 to the potassium ion.


Problem #3: K4Fe(CN)6 + H2SO4 + KMnO4 ---> MnSO4 + Fe2(SO4)3 + K2SO4 + KNO3 + CO2 + H2O

Go to the answer.


Problem #4: K4Fe(CN)6 + H2SO4 + KMnO4 ---> MnSO4 + Fe2(SO4)3 + KHSO4 + HNO3 + CO2 + H2O

Review the solution to problem #2 from step one through step eight. Here's some hints before the answer:

1) Change step nine to add something else, but not sulfate.
2) Make a new step ten and add sulfate. The amount will be 218 plus half of whatever amount of the item added in the changed step nine.

Go to the answer.


Problem #5: CuCNS + KIO3 + HCl ---> HCN + CuSO4 + ICl + H2O + KCl

Solution:

1) Get net ionic:

Cu+ + CNS¯ + IO3¯ ---> CN¯ + Cu2+ + SO42¯ + I+

I split the CuCNS and CuSO4 because they will go into separate half-reactions.

2) Let's look at oxidation nmbers:

a) copper: +1 and +2
b) thiocyanate ion: C = +4, N = -3 and S = -2
c) cyanide ion: C = +2, N = -3
d) sulfate ion: S = +6
e) iodate ion: I = +5

3) The half-reactions:

Cu+ ---> Cu2+
C4+ ---> C2+
S2¯ ---> SO42¯
IO3¯ ---> I+

4) Balance them:

Cu+ ---> Cu2+ + e¯
2e¯ + C4+ ---> C2+
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯
4e¯ + 6H+ + IO3¯ ---> I+ + 3H2O

5) Make the thiocyanate and cyanide ions:

4H2O + C4+ + S2¯ ---> C2+ + SO42¯ + 8H+ + 6e¯

Add N3¯ to each side

4H2O + CNS¯ ---> CN¯ + SO42¯ + 8H+ + 6e¯

6) Rewrite the three half-reactions of interest, equalize electrons and add:

4 [Cu+ ---> Cu2+ + e¯]
4 [4H2O + CNS¯ ---> CN¯ + SO42¯ + 8H+ + 6e¯]
7 [4e¯ + 6H+ + IO3¯ ---> I+ + 3H2O]

4Cu+ + 4CNS¯ + 10H+ + 7IO3¯ ---> 4CN¯ + 4Cu2+ + 4SO42¯ + 7I+ + 5H2O

7) Recombine:

4CuCNS + 10H+ + 7IO3¯ ---> 4CN¯ + 4CuSO4 + 7I+ + 5H2O

8) Make KIO3 (add 7 K+) and HCN (add 4 H+):

4CuCNS + 14H+ + 7KIO3 ---> 4HCN + 4CuSO4 + 7I+ + 5H2O + 7K+

9) Add 14 chloride ions:

4CuCNS + 14HCl + 7KIO3 ---> 4HCN + 4CuSO4 + 7ICl + 5H2O + 7KCl

Problem #6: CuCNS + KIO3 + H3PO4 ---> Cu3(PO4)2 + KH2PO4 + HCN + I2 + KHSO4 + H2O

Solution:

1) Half-reactions:

Cu+ ---> Cu2+
IO3¯ ---> I2
C4+ ---> C2+ (the C in thiocyanate to the C in cyanide)
S2¯ ---> SO42¯ (the S in thiocyanate to the S in sulfate)

Warning: a change in the copper half-reaction will suddenly occur in step 4. Watch for it.

2) Balance in acidic solution (the HCN tells us that):

Cu+ ---> Cu2+ + e¯
10e¯ + 12H+ + 2IO3¯ ---> I2 + 6H2O
2e¯ + C4+ ---> C2+
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯

3) Make the thiocyanate and cyanide ions:

4H2O + C4+ + S2¯ ---> C2+ + SO42¯ + 8H+ + 6e¯

Add one N3¯ to each side:

4H2O + CNS¯ ---> CN¯ + SO42¯ + 8H+ + 6e¯

4) Rewrite the three half-reactions of interest:

10 [3Cu+ ---> Cu36+ + 3e¯] <--- here's the changed half-reaction
21 [10e¯ + 12H+ + 2IO3¯ ---> I2 + 6H2O]
30 [4H2O + CNS¯ ---> CN¯ + SO42¯ + 8H+ + 6e¯]

Why did I make the change in the copper half-reaction? If you do not make the change, you will eventally wind up with a coefficient of ten in the unchanged copper half-reaction:

10Cu+ ---> 10Cu2+ + 10e¯

This will cause problems when you come to make Cu3(PO4)2 since 10Cu2+ cannot be evenly distributed taking them three at a time.

There is nothing that tells you this in advance. In fact, when I went to balance this problem (Sept. 2012), I first used the unmodifed half-reaction in step 1 and then realized what would happen as I worked through the problem.

Also, note the coefficients I have to use for the first and third half-reactions. This is because I have to remake CuCNS at some point in order to finish the molecular equation. The choice of 10 and 30 is for that reason and the 21 for the second half-reaction is driven by my need to eventually remake the CuCNS.

5) The balanced net-ionic equation with all duplicates removed:

30CuCNS + 12H+ + 42IO3¯ ---> 10Cu36+ + 21I2 + 6H2O + 30CN¯ + 30SO42¯

6) There are many ways to rebuild the molecular equation. Mine is only one way to do it. Add phosphate to the right, but only to the cuprous ion:

30CuCNS + 12H+ + 42IO3¯ ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30CN¯ + 30SO42¯

That's 20 phosphates. Add them to the left:

30CuCNS + 4H3PO4 + 16PO43¯ + 42IO3¯ ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30CN¯ + 30SO42¯

7) I know I have 30 cyanides, so add 30 hydrogen ion to the right:

30CuCNS + 14H3PO4 + 6PO43¯ + 42IO3¯ ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30HCN + 30SO42¯

See how I added 30 hydrogen to the left? Look at the changed numbers for the phosporic acid and the phosphate.

8) Add 42 potassum ion to each side:

30CuCNS + 14H3PO4 + 6PO43¯ + 42KIO3 ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30HCN + 30KSO4¯ + 12K+

Notice how I distributed the 42 potassium ions on the right. I did not make any K2SO4 since I'm reconstructing the original molecular equation.

9) Add 30 hydrogen ions to the right:

30CuCNS + 14H3PO4 + 30H+ + 6PO43¯ + 42KIO3 ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30HCN + 30KHSO4 + 12K+

Make some more phosphoric acid:

30CuCNS + 20H3PO4 + 12H+ + 42KIO3 ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30HCN + 30KHSO4 + 12K+

10) Add 12 dihydrogen phosphate:

30CuCNS + 20H3PO4 + 12H+ + 12H2PO4¯ + 42KIO3 ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30HCN + 30KHSO4 + 12KH2PO4

Make some more phosphoric acid:

30CuCNS + 32H3PO4 + 42KIO3 ---> 10Cu3(PO4)2 + 21I2 + 6H2O + 30HCN + 30KHSO4 + 12KH2PO4


Problem #7: Cr(NCS)64-(aq) + Ce4+(aq) ---> Cr3+(aq) + Ce3+(aq) + NO3-(aq) + CO2(g) + SO42-(aq) [acidic sol.]

Solution:

Comment: the use of NCS is simply an alternate way of writing the thiocyanate ion, SCN-. A mention of NCS is made in the Wikipedia entry on thiocyanate, where a concept called 'linkage isomerism' is mentioned.

1) Let's look at the oxidation state changes:

Cr goes from +2 to +3 (remember that thiocyanate has a -1 charge)
Ce goes from +4 to +3
N goes from -3 (in the thiocyanate) to +5 in the nitrate
S goes from -2 (in the thiocyanate) to +6 in the sulfate

The oxidation state of the carbon (+4 in NCS- and CO2) remains unchanged.

2) The four half reactions:

Cr2+ ---> Cr3+
Ce4+ ---> Ce3+
N3- ---> NO3-
S2- ---> SO42-

3) Balance them in acidic solution:

Cr2+ ---> Cr3+ + e-
e- + Ce4+ ---> Ce3+
3H2O + N3- ---> NO3- + 6H+ + 8e-
4H2O + S2- ---> SO42- + 8H+ + 8e-

4) Reconstitue the NCS- anion before continuing:

3H2O + N3- ---> NO3- + 6H+ + 8e-
2H2O + C4+ ---> CO2 + 4H+
4H2O + S2- ---> SO42- + 8H+ + 8e-

The second reaction is not a redox half-reaction (no electrons transferred!). It is there to introduce the carbon portion of the thiocyanate anion.

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

5) Reconstitute Cr(NCS)64-:

Cr2+ ---> Cr3+ + e-
54H2O + 6NCS- ---> 6NO3- + 6CO2 + 6SO42- + 108H+ + 96e-

54H2O + Cr(NCS)64- ---> Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e-

6) Write the (now two) half-reactions and balance:

54H2O + Cr(NCS)64- ---> Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e-
e- + Ce4+ ---> Ce3+

equalize electrons:

54H2O + Cr(NCS)64- ---> Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e-
97e- + 97Ce4+ ---> 97Ce3+

and add:

54H2O + Cr(NCS)64- + 97Ce4+ ---> 97Ce3+ + Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+

and we are done!


Problem #8: Ce2(SO4)3 + HgSO4 + H2SO4 + HNO3 + CO2 ---> Ce(SO4)2 + Hg2(CN)2 + H2O

Solution:

1) Write the net ionic equation:

Ce3+ + Hg2+ + NO3- + CO2 ---> Ce4+ + Hg22+ + 2CN-

2) Write the four half-reactions:

Ce3+ ---> Ce4+
Hg2+ ---> Hg22+
NO3- ---> N3- (in the CN- ion)
CO2 ---> C2+ (in the CN- ion)

3) Balance them:

Ce3+ ---> Ce4+ + e-
2e- + 2Hg2+ ---> Hg22+
8e- + 6H+ + NO3- ---> N3- + 3H2O
2e- + 4H+ + CO2 ---> C2+ + 2H2O

4) Reconstitute the cyanide ion:

10e- + 10H+ + NO3- + CO2 ---> CN- + 5H2O

In the next step, I will have multiplied the above half-reaction by 2. I'm doing this to set myself up to reconstitute Hg2(CN)2.

5) Balance for charge:

22Ce3+ ---> 22Ce4+ + 22e-
2e- + 2Hg2+ ---> Hg22+
20e- + 20H+ + 2NO3- + 2CO2 ---> 2CN- + 10H2O

5) Add (and reconstitute mercury(I) cyanide):

22Ce3+ + 2Hg2+ + 2NO3- + 2CO2 + 20H+ ---> 22Ce4+ + Hg2(CN)2 + 10H2O

6) Add 44 sulfates on the right-hand side:

22Ce3+ + 2Hg2+ + 2HNO3 + 2CO2 + 18H+ ---> 22Ce(SO4)2 + Hg2(CN)2 + 10H2O

I also used 2 hydrogen ions to make 2 molecules of nitric acid.

7) Add 44 sulfates to the left-hand side:

11Ce2(SO4)3 + 2HgSO4 + 2HNO3 + 2CO2 + 9H2SO4 ---> 22Ce(SO4)2 + Hg2(CN)2 + 10H2O

Ce2(SO4)3 ---> 33 sulfates
H2SO4 ---> 9 sulfates
HgSO4 ---> 2 sulfates

33 + 9 + 2 = 44

Here is a solution to the above problem that uses a different style than mine. Examine his answer to get a sense of a different explanation that is just as correct as mine, as well as being more compact. I tend to be a bit long-winded in my explanations.


Problem #9: As2S3 + Mn(NO3)2 + K2CO3 ---> K3AsO4 + K2SO4 + K2MnO4 + NO + CO2

Solution:

1) The first step is to identify what is reduced or oxidized:

arsenic ---> goes from +3 to +5
sulfur ---> goes from -2 to +6
manganese ---> goes from +2 to +6
nitrogen ---> goes from +5 to +2

I will leave you to verify the above oxidation numbers. The carbon remains at +4 from carbonate to carbon dioxide. However, the transition from carbonate to carbon dioxide will play a role.

2) Now the unbalanced half-reactions:

2As3+ ---> 2AsO43¯

3S2¯ ---> 3SO42¯

Mn2+ ---> MnO42¯

2NO3¯ ---> 2NO

I wrote the half-reactions as above in order to eventually combine them so as to recover As2S3 and Mn(NO3)2. Please note that I wrote everything in net ionic form. Also note that the potassium carbonate and the carbon dioxide will reappear.

3) Balance the half-reactions (in acidic solution):

8H2O + 2As3+ ---> 2AsO43¯ + 16H+ + 4e¯

12H2O + 3S2¯ ---> 3SO42¯ + 24H+ + 24e¯

4H2O + Mn2+ ---> MnO42¯ + 8H+ + 4e¯

6e¯ + 8H+ + 2NO3¯ ---> 2NO + 4H2O

4) Combine the first two to make As2S3 and the last two to make Mn(NO3)2:

20H2O + As2S3 ---> 2AsO43¯ + 3SO42¯ + 40H+ + 28e¯

2e¯ + Mn(NO3)2 ---> MnO42¯ + 2NO

5) Equalize electrons and add:

20H2O + As2S3 ---> 2AsO43¯ + 3SO42¯ + 40H+ + 28e¯

28e¯ + 14Mn(NO3)2 ---> 14MnO42¯ + 28NO

Added together:

20H2O + As2S3 + 14Mn(NO3)2 ---> 2AsO43¯ + 3SO42¯ + 14MnO42¯ + 28NO + 40H+

6) Now, I'm going to add 40 potassium ions to the right-hand side and, for the moment, I will add nothing to the left-hand side:

20H2O + As2S3 + 14Mn(NO3)2 ---> 2K3AsO4 + 3K2SO4 + 14K2MnO4 + 28NO + 40H+

7) I will balance the 40 potassium ions by using 20 potassium carbonate on the left-hand side. Look for where I put the 20 carbonate anions on the right-hand side:

20H2O + As2S3 + 14Mn(NO3)2 + 20K2CO3 ---> 2K3AsO4 + 3K2SO4 + 14K2MnO4 + 28NO + 20H2CO3

I made 20 molecules of carbonic acid to balance the carbonates.

8) However, carbonic acid is unstable and it is always broken down immediately into carbon dioxide and water (20 molecules of each):

As2S3 + 14Mn(NO3)2 + 20K2CO3 ---> 2K3AsO4 + 3K2SO4 + 14K2MnO4 + 28NO + 20CO2

That's the final answer.

Want to double-check? Here it is using an online chemical equation balancer.


Problem #10: Pb(N3)2 + Cr(MnO4)2 ---> Cr2O3 + MnO2 + Pb3O4 + NO

Solution:

This is an arbitrary equation that has no basis in reality. However, it can be balanced. I balanced it below as if in acidic solution, since nothing was specified. Notice how, at the end, all the hydrogen ions and water cancel out.

1) Half-reactions:

Pb2+ ---> Pb3O4

2N3- ---> NO

Cr2+ ---> Cr2O3

2MnO4- ---> MnO2

Notice how I preserve the ratios between the plumbous ion and the azide ion, as well as between chromous and permanganate. I'm doing this because I know I have to reunite those formulas.

2) Balance half-reactions:

4H2O + 3Pb2+ ---> Pb3O4 + 8H+ + 2e-

18H2O + 6N3- ---> 18NO + 36H+ + 42e-

3H2O + 2Cr2+ ---> Cr2O3 + 6H+ + 2e-

12e- + 16H+ + 4MnO4- ---> 4MnO2 + 8H2O

Comments on the above:

a) When I put in 3Pb2+ to start the balancing (as part of preparing this answer), I immediately put in 6N3- in the second half-reaction and then balanced for 6N3-. That keeps the 1:2 ratio for the Pb(N3)2 formula.
b) Same for the 1:2 ratio between Cr2+ and MnO4- in the Cr(MnO4)2

3) Combine into two half-reactions:

22H2O + 3Pb(N3)2 ---> Pb3O4 + 18NO + 44H+ + 44e-

10e- + 10H+ + 2Cr(MnO4)2 ---> Cr2O3 + 4MnO2 + 5H2O

5) Equalize electrons:

110H2O + 15Pb(N3)2 ---> 5Pb3O4 + 90NO + 220H+ + 220e-

220e- + 220H+ + 44Cr(MnO4)2 ---> 22Cr2O3 + 88MnO2 + 110H2O

6) Add:

15Pb(N3)2 + 44Cr(MnO4)2 ---> 22Cr2O3 + 88MnO2 + 5Pb3O4 + 90NO

Problem #11: Pb(N3)2 + Cr(MnO4)2 ---> Cr2O3 + MnO2 + PbO4 + NO

Solution:

This problem has PbO4, a compound that does not exist. I found the equation in a Yahoo Answers question and I wonder if the PbO4 resulted from a mis-copy of Pb3O4. It turns out that, even though PbO4 is not real, the equation can still be balanced.

1) Half-reactions:

Pb2+ ---> PbO4

2N3- ---> NO

Cr2+ ---> Cr2O3

2MnO4- ---> MnO2

2) Balance half-reactions:

4H2O + Pb2+ ---> PbO4 + 8H+ + 6e-

6H2O + 2N3- ---> 6NO + 12H+ + 14e-

3H2O + 2Cr2+ ---> Cr2O3 + 6H+ + 2e-

12e- + 16H+ + 4MnO4- ---> 4MnO2 + 8H2O

3) Combine into two half-reactions:

10H2O + Pb(N3)2 ---> PbO4 + 6NO + 20H+ + 20e-

10e- + 10H+ + 2Cr(MnO4)2 ---> Cr2O3 + 4MnO2 + 5H2O

4) Equalize the electrons:

10H2O + Pb(N3)2 ---> PbO4 + 6NO + 20H+ + 20e-

20e- + 20H+ + 4Cr(MnO4)2 ---> 2Cr2O3 + 8MnO2 + 10H2O

5) Add:

Pb(N3)2 + 4Cr(MnO4)2 ---> 2Cr2O3 + 8MnO2 + PbO4 + 6NO

Problem #12: Pb(N3)2 + Cr(MnO4)2 ---> Cr2O3 + MnO2 + PbO2 + NO

Solution:

When I first came across the equation with PbO4 in it, I modified the formula to PbO2 (a real compound) and balanced it for inclusion here.

1) Half-reactions:

Pb2+ ---> PbO2

2N3- ---> NO

Cr2+ ---> Cr2O3

2MnO4- ---> MnO2

2) Balance half-reactions:

2H2O + Pb2+ ---> PbO2 + 4H+ + 2e-

6H2O + 2N3- ---> 6NO + 12H+ + 14e-

3H2O + 2Cr2+ ---> Cr2O3 + 6H+ + 2e-

12e- + 16H+ + 4MnO4- ---> 4MnO2 + 8H2O

3) Combine into two half-reactions:

8H2O + Pb(N3)2 ---> PbO2 + 6NO + 16H+ + 16e-

10e- + 10H+ + 2Cr(MnO4)2 ---> Cr2O3 + 4MnO2 + 5H2O

4) Equalize electrons:

40H2O + 5Pb(N3)2 ---> 5PbO2 + 30NO + 80H+ + 80e-

80e- + 80H+ + 16Cr(MnO4)2 ---> 8Cr2O3 + 32MnO2 + 40H2O

5) Add:

5Pb(N3)2 + 16Cr(MnO4)2 ---> 8Cr2O3 + 32MnO2 + 5PbO2 + 30NO

Comment: you could modify the equation by replacing NO with NO2 or N2O. I will leave the exercise to you, the student.


Problem #13:

K4Fe(SCN)6 + K2Cr2O7 + H2SO4 ---> Fe2(SO4)3 + Cr2(SO4)3 + CO2 + H2O + K2SO4 + KNO3

Solution:

1) Identify the items reduced/oxidized:

Fe ---> from +2 to +3
Cr (in dichromate) to chromium(III) ---> +6 to +3
S (in thiocyanate) to sulfate ---> -2 to +6
C (in thiocyanate) to carbon dioxide ---> +4 to +4
N (in thiocyanate) to nitrate ---> -3 to +5

I included the carbon because I know, from experience, that I will need it to re-form the thiocyanate ion.

2) Write the half-reactions:

Fe2+ ---> Fe3+
Cr2O72¯ ---> Cr3+
S2¯ ---> SO42¯
C4+ ---> CO2
N3¯ ---> NO3¯

3) Balance the half-reactions:

2Fe2+ ---> 2Fe3+ + 2e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯
2H2O + C4+ ---> CO2 + 4H+
3H2O + N3¯ ---> NO3¯ + 6H+ + 8e¯

I know I wil need to make Fe2(SO4)3, so I doubled the Fe half-reaction.

4) Re-form the entire ferrocyanate ion into one half-reaction:

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

We need 12 SCN¯:

108H2O + 12SCN¯ ---> 12SO42¯ + 12CO2 + 12NO3¯ + 216H+ + 192e¯

Add in two ferrous ions:

108H2O + 2Fe(SCN)64¯ ---> 2Fe3+ + 12SO42¯ + 12CO2 + 12NO3¯ + 216H+ + 194e¯

5) Equalize the electrons:

582e¯ + 1358H+ + 97Cr2O72¯ ---> 194Cr3+ + 679H2O
324H2O + 6Fe(SCN)64¯ ---> 6Fe3+ + 36SO42¯ + 36CO2 + 36NO3¯ + 648H+ + 582e¯

First half-reaction by 97 and second half-reaction by 3.

6) Add the two half-reactions:

710H+ + 6Fe(SCN)64¯ + 97Cr2O72¯ ---> 6Fe3+ + 36SO42¯ + 36CO2 + 36NO3¯ + 194Cr3+ + 355H2O

7) I'm going to skip any explanations about re-forming the molecular equation. Here's the final answer:

6K4Fe(SCN)6 + 97K2Cr2O7 + 355H2SO4 ---> 3Fe2(SO4)3 + 97Cr2(SO4)3 + 36CO2 + 355H2O + 91K2SO4 + 36KNO3

Problem #14: K4Fe(SCN)6 + K2Cr2O7 + H2SO4 ---> Fe2(SO4)3 + Cr2(SO4)3 + CO + H2O + K2SO4 + KNO3

This problem uses five half-reactions. I changed the CO2 from problem #13 to CO in order to create the fifth half-reaction.

Solution:

1) Identify the items reduced/oxidized:

Fe ---> from +2 to +3
Cr (in dichromate) to chromium(III) ---> +6 to +3
S (in thiocyanate) to sulfate ---> -2 to +6
C (in thiocyanate) to carbon monoxide ---> +4 to +2
N (in thiocyanate) to nitrate ---> -3 to +5

2) Write the half-reactions:

Fe2+ ---> Fe3+
Cr2O72¯ ---> Cr3+
S2¯ ---> SO42¯
C4+ ---> CO
N3¯ ---> NO3¯

3) Balance the half-reactions:

2Fe2+ ---> 2Fe3+ + 2e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯
H2O + C4+ + 2e¯ ---> CO + 2H+
3H2O + N3¯ ---> NO3¯ + 6H+ + 8e¯

I know I wil need to make Fe2(SO4)3, so I doubled the Fe half-reaction.

4) Re-form the entire ferrocyanate ion into one half-reaction:

8H2O + SCN¯ ---> SO42¯ + CO + NO3¯ + 16H+ + 14e¯

We need 12 SCN¯:

96H2O + 12SCN¯ ---> 12SO42¯ + 12CO + 12NO3¯ + 192H+ + 168e¯

Add in two ferrous ions:

96H2O + 2Fe(SCN)64¯ ---> 2Fe3+ + 12SO42¯ + 12CO + 12NO3¯ + 192H+ + 170e¯

When I first wrote this part of the answer (late Sept. 2016), I forgot to add in the two electrons when I added in the Fe2+ and Fe3+. Oopsie!!!

5) Equalize the electrons:

510e¯ + 1190H+ + 85Cr2O72¯ ---> 170Cr3+ + 595H2O
288H2O + 6Fe(SCN)64¯ ---> 12Fe3+ + 36SO42¯ + 36CO + 36NO3¯ + 576H+ + 510e¯

First half-reaction by 85 and second half-reaction by 3.

6) Add the two half-reactions:

614H+ + 6Fe(SCN)64¯ + 85Cr2O72¯ ---> 12Fe3+ + 170Cr3+ + 36SO42¯ + 36CO + 36NO3¯ + 307H2O

7) I'm going to skip any explanations about re-forming the molecular equation. Here's the final answer:

6K4Fe(SCN)6 + 85K2Cr2O7 + 307H2SO4 ---> 3Fe2(SO4)3 + 85Cr2(SO4)3 + 36CO + 307H2O + 79K2SO4 + 36KNO3

Some of the other four half-reaction problems can be modified to produce five half-reactions. I will leave you to follow up on that, if so desired.


Redox equations where three half-reactions are required

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