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.
Here are the problems:
| Problem #1: K4Fe(CN)6 + Ce(NO3)4 + KOH ---> Ce(OH)3 + Fe(OH)3 + H2O + K2CO3 + KNO3 |
| Problem #2: K4Fe(CN)6 + H2SO4 + KMnO4 ---> MnSO4 + Fe2(SO4)3 + K2SO4 + HNO3 + CO2 + H2O |
| Problem #3: go to problem #2 and replace HNO3 with KNO3. |
| Problem #4: go to problem #2 and replace K2SO4 with KHSO4 |
| Problem #5: CuCNS + KIO3 + HCl ---> HCN + CuSO4 + ICl + H2O + KCl |
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¯
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: Go to problem #2 and replace HNO3 with KNO3.
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.
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