### Balancing redox reactions in basic solutionProblems 11 - 30

Problem #11: ClO3¯ + N2H4 ---> NO + Cl¯

Solution:

1) Half-reactions:

ClO3¯ ---> Cl¯
N2H4 ---> NO

2) Balance in acidic solution first:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
2H2O + N2H4 ---> 2NO + 8H+ + 8e¯

4) Equalize electrons:

24e¯ + 24H+ + 4ClO3¯ ---> 4Cl¯ + 12H2O
6H2O + 3N2H4 ---> 6NO + 24H+ + 24e¯

4ClO3¯ + 3N2H4 ---> 4Cl¯ + 6NO + 6H2O

Note that there is no hydroxide in the final answer. Does that mean this reaction does not require a basic solution? The answer is no. The hydroxide is functioning as a catalyst to the reaction. Some hydroxide is used up to start the rection, but it is regenerated by the end of the reaction. So, while hydroxide is necessary for the reaction, it does not appear in the overall equation.

Problem #12: H2O2 + Cl2O7 ---> ClO2¯ + O2

Problem #13: HXeO4¯ + Pb ---> Xe + HPbO2¯

Solution:

1) Half-reactions:

HXeO4¯ ---> Xe
Pb ---> HPbO2¯

2) Balance in acid:

6e¯ + 7H+ + HXeO4¯ ---> Xe + 4H2O
2H2O + Pb ---> HPbO2¯ + 3H+ + 2e¯

3) Equalize electrons:

6e¯ + 7H+ + HXeO4¯ ---> Xe + 4H2O
6H2O + 3Pb ---> 3HPbO2¯ + 9H+ + 6e¯

2H2O + HXeO4¯ + 3Pb ---> Xe + 3HPbO2¯ + 2H+

5) Convert to basic with two hydroxides on each side:

2OH¯ + HXeO4¯ + 3Pb ---> Xe + 3HPbO2¯

Problem #14: HPbO2¯ + Cr(OH)3 ---> Pb + CrO42¯

Solution:

1) Half-reactions:

HPbO2¯ ---> Pb
Cr(OH)3 ---> CrO42¯

2) Balance in acidic solution:

2e¯ + 3H+ + HPbO2¯ ---> Pb + 2H2O
H2O + Cr(OH)3 ---> CrO42¯ + 5H+ + 3e¯

3) Equalize electrons:

6e¯ + 9H+ + 3HPbO2¯ ---> 3Pb + 6H2O
2H2O + 2Cr(OH)3 ---> 2CrO42¯ + 10H+ + 6e¯

3HPbO2¯ + 2Cr(OH)3 ---> 3Pb + 2CrO42¯ + H+ + 4H2O

5) Add one hydroxide to each side:

3HPbO2¯ + 2Cr(OH)3 + OH¯ ---> 3Pb + 2CrO42¯ + 5H2O

Problem #15: Basic solutions of thallium(III) oxide and hydroxylamine are mixed together to produce thallium(I) hydroxide and nitrogen gas

Solution:

1) The full equation followed by the half-reactions:

Tl2O3 + NH2OH ---> TlOH + N2

Tl2O3 ---> TlOH
NH2OH ---> N2

2) Balance in acidic solution:

4e¯ + 4H+ + Tl2O3 ---> 2TlOH + H2O
2NH2OH ---> N2 + 2H2O + 2H+ + 2e¯

3) Equalize electrons:

4e¯ + 4H+ + Tl2O3 ---> 2TlOH + H2O
4NH2OH ---> 2N2 + 4H2O + 4H+ + 4e¯

Tl2O3 + 4NH2OH ---> 2TlOH + 2N2 + 5H2O

Problem #16: Mg + NiO2 ---> Mg(OH)2 + Ni(OH)2

Solution:

1) Half-reactions:

Mg ---> Mg(OH)2
NiO2 ---> Ni(OH)2

2) Balance first in basic and second in acidic:

2OH¯ + Mg ---> Mg(OH)2 + 2e-
2e¯ + 2H+ + NiO2 ---> Ni(OH)2

3) Note that the electrons are already equalized. Note also that hydrogen ion and hydroxide ion (both on the left-hand side of the two half-reactions) react immediately to produce water. That leads to the final answer:

Mg + NiO2 + 2H2O ---> Mg(OH)2 + Ni(OH)2

Problem #17: Cr(OH)4¯ + H2O2 ---> CrO42¯ + H2O

Solution:

1) Half-reactions:

Cr(OH)4¯ ---> CrO42¯
H2O2 ---> H2O

2) Balance as if in acidic solution:

Cr(OH)4¯ ---> CrO42¯ + 4H+ + 3e¯
2e¯ + 2H+ + H2O2 ---> 2H2O

3) Equalize electrons:

2Cr(OH)4¯ ---> 2CrO42¯ + 8H+ + 6e¯
6e¯ + 6H+ + 3H2O2 ---> 6H2O

2Cr(OH)4¯ + 3H2O2 ---> 2CrO42¯ + 2H+ + 6H2O

5) Change to basic by adding two hydroxides to each side:

2OH¯ + 2Cr(OH)4¯ + 3H2O2 ---> 2CrO42¯ + 8H2O

The two H+ and the two OH¯ on the right combine to form two waters to make a total of eight waters.

Problem #18: CrO42¯ + HSnO2¯ + H2O ---> CrO2(s) + H2SnO3(aq) + OH¯(aq)

1) Balanced half-reactions:

4H+ + CrO42¯ + 2e¯ ---> CrO2 + 2H2O
HSnO2¯ + H2O ---> H2SnO3 + H+ + 2e¯

4H+ + CrO42¯ + HSnO2¯ + H2O ---> CrO2 + H2SnO3 + H+ + 2H2O

3H+ + CrO42¯ + HSnO2¯ ---> CrO2 + H2SnO3 + H2O

3) Add 3OH¯ to each side to neutralize H+ since the reaction is in basic solution, then simplify:

3H2O + CrO42¯ + HSnO2¯ ---> CrO2 + H2SnO3 + H2O + 3OH¯

2H2O + CrO42¯ + HSnO2¯ ---> CrO2 + H2SnO3 + 3OH¯ <--- that's the balanced equation

Problem #19: HXeO4¯(aq) ---> XeO64¯(aq) + Xe(g)

Solution:

1) Half-reactions:

HXeO4¯(aq) ---> XeO64¯(aq)
HXeO4¯(aq) ---> Xe(g)

2) Balance in acidic:

2H2O + HXeO4¯(aq) ---> XeO64¯(aq) + 5H+ + 2e¯
6e¯ + 7H+ + HXeO4¯(aq) ---> Xe(g) + 4H2O

3) Equalize electrons:

6H2O + 3HXeO4¯(aq) ---> 3XeO64¯(aq) + 15H+ + 6e¯
6e¯ + 7H+ + HXeO4¯(aq) ---> Xe(g) + 4H2O

2H2O + 4HXeO4¯(aq) ---> 3XeO64¯(aq) + Xe(g) + 8H+

5) Add eight hydroxides to each side:

8OH¯ + 2H2O + 4HXeO4¯(aq) ---> 3XeO64¯(aq) + Xe(g) + 8H2O

6) Remove water for the final answer (with state symbols for everything):

8OH¯(aq) + 4HXeO4¯(aq) ---> 3XeO64¯(aq) + Xe(g) + 6H2O(ℓ)

Problem #20: S2O42¯ + O2(g) ---> SO42¯

Solution:

1) Half-reactions:

S2O42¯ ---> SO42¯
O2(g) ---> OH¯

No hydroxide is indicated in the problem but, since we know the reaction to be in basic solution, we can add it in.

2) Balance:

4H2O + S2O42¯ ---> 2SO42¯ + 8H+ + 6e¯
4e¯ + 2H+ + O2(g) ---> 2OH¯ <--- note the hydrogen ion on one side and the hydroxide on the other. Unusual.

3) Equalize electrons:

8H2O + 2S2O42¯ ---> 4SO42¯ + 16H+ + 12e¯
12e¯ + 6H+ + 3O2(g) ---> 6OH¯

8H2O + 2S2O42¯ + 3O2(g) ---> 4SO42¯ + 10H+ + 6OH¯

4OH¯ + 8H2O + 2S2O42¯ + 3O2(g) ---> 4SO42¯ + 10H2O

Note that I combined the hydrogen ion and hydroxide ion on the right into water.

6) Remove water for the final answer:

4OH¯ + 2S2O42¯ + 3O2(g) ---> 4SO42¯ + 2H2O

Here's an incorrect answer to this problem. The answerer failed to reduce the O2 properly and wound up with an answer that is not balanced for charge or for oxygen

Problem #21: SO32- + Cr2O72- ---> Cr3+ + SO42-

Solution:

1) Half-reactions:

SO32- ---> SO42-
Cr2O72- ---> Cr3+

2) Balance in acidic solution first:

H2O + SO32- ---> SO42- + 2H+ + 2e¯
6e¯ + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O

3) Equalize electrons:

3H2O + 3SO32- ---> 3SO42- + 6H+ + 6e¯
6e¯ + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O

8H+ + 3SO32- + Cr2O72- ---> 2Cr3+ + 3SO42- + 4H2O

5) Convert to basic:

4H2O + 3SO32- + Cr2O72- ---> 2Cr3+ + 3SO42- + 8OH¯

6) Since chromium(III) hydroxide precipitates, this may the desired answer:

4H2O + 3SO32- + Cr2O72- ---> 2Cr(OH)3 + 3SO42- + 2OH¯

Problem #22: Sb2O3(s) + NO3¯(aq) ---> H3SbO4(aq) + NO(g)

Solution:

1) Half-reactions:

Sb2O3(s) ---> H3SbO4(aq)
NO3¯(aq) ---> NO(g)

2) Balance in acidic:

5H2O + Sb2O3(s) ---> 2H3SbO4(aq) + 4H+ + 4e¯
3e¯ + 4H+ + NO3¯(aq) ---> NO(g) + 2H2O

3) Equalize electrons:

15H2O + 3Sb2O3(s) ---> 6H3SbO4(aq) + 12H+ + 12e¯
12e¯ + 16H+ + 4NO3¯(aq) ---> 4NO(g) + 8H2O

4H+ + 7H2O + 3Sb2O3(s) + 4NO3¯(aq) ---> 6H3SbO4(aq) + 4NO(g)

5) Convert to basic:

11H2O + 3Sb2O3(s) + 4NO3¯(aq) ---> 6H3SbO4(aq) + 4NO(g) + 4OH¯

Problem #23: Al(s) + NO3¯ ---> Al(OH)4¯ + NH3(g)

Solution:

I propose to balance one half-reaction in basic, the other in acidic. I'll then add and change over to basic at the end.

1) Half-reactions:

Al(s) ---> Al(OH)4¯
NO3¯ ---> NH3(g)

2) Balance:

Al(s) + 4OH¯ ---> Al(OH)4¯ + 3e¯
8e¯ + 9H+ + NO3¯ ---> NH3(g) + 3H2O

3) Equalize electrons:

8Al(s) + 32OH¯ ---> 8Al(OH)4¯ + 24e¯
24e¯ + 27H+ + 3NO3¯ ---> 3NH3(g) + 9H2O

8Al(s) + 32OH¯ + 27H+ + 3NO3¯ ---> 8Al(OH)4¯ + 3NH3(g) + 9H2O

5) Make twenty-seven waters from 27 hydrogen ions and 27 hydroxide ions:

8Al(s) + 5OH¯ + 27H2O + 3NO3¯ ---> 8Al(OH)4¯ + 3NH3(g) + 9H2O

6) Eliminate nine waters for the final answer:

8Al(s) + 3NO3¯ + 5OH¯ + 18H2O ---> 8Al(OH)4¯ + 3NH3(g)

I rearranged the left-hand side a bit, simply because it looks better to me. Also, note how I never had to change over to basic, it just happened naturally.

Problem #24: MnO4¯ + I¯ ----> MnO42¯ + IO3¯

Solution:

1) Half-reactions:

MnO4¯ ---> MnO42¯
I¯ ---> IO3¯

2) Balance in acidic solution:

e¯ + MnO4¯ ---> MnO42¯
3H2O + I¯ ---> IO3¯ + 6H+ + 6e¯

3) Equalize electrons:

6e¯ + 6MnO4¯ ---> 6MnO42¯
3H2O + I¯ ---> IO3¯ + 6H+ + 6e¯

3H2O + I¯ + 6MnO4¯ ---> 6MnO42¯ + IO3¯ + 6H+

5) Change to basic by adding six hydroxides to each side:

6OH¯ + 3H2O + I¯ + 6MnO4¯ ---> 6MnO42¯ + IO3¯ + 6H2O

6) Remove three H2O:

6OH¯ + I¯ + 6MnO4¯ ---> 6MnO42¯ + IO3¯ + 3H2O

Problem #25: Cu2+ + SO32¯ ---> SO42¯ + Cu

Solution:

1) Half-reactions:

SO32¯ ---> SO42¯
Cu2+ ---> Cu

2) Balance them:

H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯
2e¯ + Cu2+ ---> Cu

3) Note that I balanced the first half-reaction in acidic solution. I'll change it to basic in a moment. Since there are already an equal number of electrons on each side, we may add the two half-reactions:

H2O + SO32¯ + Cu2+ ---> SO42¯ + Cu + 2H+

4) Add two hydroxide ions to each side to make the conversion from acidic to basic:

2OH¯ + H2O + SO32¯ + Cu2+ ---> SO42¯ + Cu + 2H2O

Note that the hydrogen ions and hydroxide ions form water.

5) Cancel out one water:

2OH¯ + SO32¯ + Cu2+ ---> SO42¯ + Cu + H2O

Problem #26: MnO4¯ + SO32¯ ---> MnO2 + SO42¯

Solution:

1) Half-reactions:

MnO4¯ ---> MnO2
SO32¯ ---> SO42¯

2) Balance:

3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O
3H2O + 3SO32¯ ---> 3SO42¯ + 6H+ + 6e¯

2H+ + 2MnO4¯ + 3SO32¯ ---> 2MnO2 + 3SO42¯ + H2O

5) Convert to basic:

H2O + 2MnO4¯ + 3SO32¯ ---> 2MnO2 + 3SO42¯ + 2OH¯

One water removed from each side.

Problem #27: MnO4¯ + SO32¯ ---> Mn2O3 + S2O82¯

Mn2O3 is an unlikely product, but it is an actual compound and, using it, we can make an equation to balance. The non-real-worldedness of Mn2O3 being a product is beside the point.

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2O3
SO32¯ ---> S2O82¯

2) Balance in acidic:

8e¯ + 10H+ + 2MnO4¯ ---> Mn2O3 + 5H2O
2H2O + 2SO32¯ ---> S2O82¯ + 4H+ + 6e¯

3) Equalize electrons:

24e¯ + 30H+ + 6MnO4¯ ---> 3Mn2O3 + 15H2O
8H2O + 8SO32¯ ---> 4S2O82¯ + 16H+ + 24e¯

14H+ + 6MnO4¯ + 8SO32¯ ---> 4S2O82¯ + 3Mn2O3 + 7H2O

5) Add 14 hydroxides to each side (also removed 7 waters from each side):

7H2O + 6MnO4¯ + 8SO32¯ ---> 4S2O82¯ + 3Mn2O3 + 14OH¯

Problem #28: MnO4¯ + SbH3 ---> MnO2 + Sb

Solution:

1) Half-reactions:

MnO4¯ ---> MnO2
SbH3 ---> Sb

2) Balance:

3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
SbH3 ---> Sb + 3H+ + 3e¯

H+ + MnO4¯ + SbH3 ---> MnO2 + Sb + 2H2O

4) Convert to basic:

MnO4¯ + SbH3 ---> MnO2 + Sb + H2O + OH¯

Note: one excess water was removed.

Problem #29: N2H4 + Cu(OH)2 ---> Cu + N2

Solution:

1) Half-reactions:

N2H4 ---> N2
Cu(OH)2 ---> Cu

2) Balance:

N2H4 ---> N2 + 4H+ + 4e¯
2e¯ + Cu(OH)2 ---> Cu + 2OH¯

Note how one equation is in acidic and one in basic. That will not create a problem.

3) Equalize electrons:

N2H4 ---> N2 + 4H+ + 4e¯
4e¯ + 2Cu(OH)2 ---> 2Cu + 4OH¯

N2H4 + 2Cu(OH)2 ---> 2Cu + N2 + 4H2O

The four hydrogen ion and the four hydroxide combined to make four waters.

Problem #30: Fe(OH)2 + CrO42¯ ---> Fe2O3 + Cr(OH)4¯

Solution:

1) Half-reactions:

Fe(OH)2 ---> Fe2O3
CrO42¯ ---> Cr(OH)4¯

2) Balance in acidic:

2Fe(OH)2 ---> Fe2O3 + H2O + 2H+ + 2e¯
3e¯ + 4H+ + CrO42¯ ---> Cr(OH)4¯

3) Equalize electrons:

6Fe(OH)2 ---> 3Fe2O3 + 3H2O + 6H+ + 6e¯
6e¯ + 8H+ + 2CrO42¯ ---> 2Cr(OH)4¯

4) Add and eliminate like items:

6Fe(OH)2 + 2CrO42¯ + 2H+ ---> 3Fe2O3 + 2Cr(OH)4¯ + 3H2O

5) Change to basic and eliminate like items:

6Fe(OH)2 + 2CrO42¯ ---> 3Fe2O3 + 2Cr(OH)4¯ + H2O + 2OH¯

Problem #31: H2O(ℓ) + S2¯(aq) + MnO4¯(aq) ---> S8(s) + MnS(s) + OH¯(aq)

Solution:

1) Separate into half-reactions:

S2¯(aq) ---> S8(s)
MnO4¯(aq) ---> Mn2+(s) + S2¯

Notice that I split the MnS. When I finish balancing the half-reactions, I'll recombine it. I did this so as to make the fact that the Mn is being reduced a bit more apparent.

2) Balance the half-reactions:

8S2¯(aq) ---> S8(s) + 16e¯
5e¯ + 8H+ + MnO4¯(aq) + S2¯ ---> MnS + 4H2O

3) Equalize the electrons:

40S2¯(aq) ---> 5S8(s) + 80e¯
80e¯ + 128H+ + 16MnO4¯(aq) + 16S2¯ ---> 16MnS + 64H2O

128H+ + 56S2¯(aq) + 16MnO4¯(aq) ---> 5S8(s) + 16MnS + 64H2O

5) Convert to basic:

128H2O + 56S2¯(aq) + 16MnO4¯(aq) ---> 5S8(s) + 16MnS + 64H2O + 128OH¯

6) Eliminate water (and add in state symbols):

64H2O(ℓ) + 56S2¯(aq) + 16MnO4¯(aq) ---> 5S8(s) + 16MnS(s) + 128OH¯(aq)

Problem #32: H2O(ℓ) + CN¯(aq) + MnO4¯(aq) ---> CNO¯(aq) + MnO2(s) + OH¯(aq)

Solution:

1) Separate into half-reactions:

CN¯(aq) ---> CNO¯(aq)
MnO4¯(aq) ---> MnO2(s)

2) Balance the half-reactions:

H2O + CN¯(aq) ---> CNO¯(aq) + 2H+ + 2e¯
3e¯ + 4H+ + MnO4¯(aq) ---> MnO2(s) + 2H2O

3) Equalize electrons:

3H2O + 3CN¯(aq) ---> 3CNO¯(aq) + 6H+ + 6e¯
6e¯ + 8H+ + 2MnO4¯(aq) ---> 2MnO2(s) + 4H2O

2H+ + 2MnO4¯(aq) + 3CN¯(aq) ---> 2MnO2(s) + 3CNO¯(aq) + H2O

5) Change to basic:

H2O(ℓ) + 2MnO4¯(aq) + 3CN¯(aq) ---> 2MnO2(s) + 3CNO¯(aq) + 2OH¯(aq)

One water was removed from both sides.

Problem #33: Cr(s) + CrO42¯(aq) ---> Cr(OH)3(s) + OH¯(aq)

Solution:

1) Separate into half-reactions:

Cr(s) ---> Cr(OH)3(s)
CrO42¯(aq) ---> Cr(OH)3(s)

2) Balance the half-reactions:

3OH¯ + Cr(s) ---> Cr(OH)3(s) + 3e¯
3e¯ + 5H+ + CrO42¯(aq) ---> Cr(OH)3(s) + H2O

Notice how one is balanced in basic and one in acid. You can change the acid one over to basic right now or wait until the end. This will not make any difference in the final answer.

2H+ + 3H2O + Cr(s) + CrO42¯(aq) ---> 2Cr(OH)3(s) + H2O

I combined three hydrogen ions and three hydroxide to make 3 waters on the left-hand side. I'll now cancel one water from both sides:

2H+ + 2H2O + Cr(s) + CrO42¯(aq) ---> 2Cr(OH)3(s)

4) Add two hydroxides to each side to convert from acidic to basic:

4H2O(ℓ) + Cr(s) + CrO42¯(aq) ---> 2Cr(OH)3(s) + 2OH¯(aq)

Problem #34: CrO42¯(aq) + S2O42¯(aq) ---> Cr(OH)3(s) + SO42¯(aq) (basic)

Solution:

1) Half-reactions:

CrO42¯ ---> Cr(OH)3
S2O42¯ ---> SO42¯

2) Balance both half-reactions in acid, add them and then change over to basic at the end:

3e¯ + 5H+ + CrO42¯ ---> Cr(OH)3 + H2O
4H2O + S2O42¯ ---> 2SO42¯ + 8H+ + 6e¯

3) Equalize electrons:

6e¯ + 10H+ + 2CrO42¯ ---> 2Cr(OH)3 + 2H2O
4H2O + S2O42¯ ---> 2SO42¯ + 8H+ + 6e¯

2H2O + 2H+ + 2CrO42¯ + S2O42¯ ---> 2Cr(OH)3 + 2SO42¯

5) Add two hydroxide to each side:

4H2O + 2CrO42¯ + S2O42¯ ---> 2Cr(OH)3 + 2SO42¯ + 2OH¯

Problem #35: Sn(OH)3¯(aq) + Bi(OH)3(s) + OH¯(aq) ---> Sn(OH)62¯(aq) + Bi(s)

Solution:

1) Half-reactions:

Sn(OH)3¯(aq) ---> Sn(OH)62¯(aq)
Bi(OH)3(s) ---> Bi(s)

2) Balance:

3OH¯ + Sn(OH)3¯(aq) ---> Sn(OH)62¯(aq) + 2e¯
3e¯ + Bi(OH)3(s) ---> Bi(s) + 3OH¯

Notice that wasn't really a need to balance in acid first.

3) Equalize electrons:

9OH¯ + 3Sn(OH)3¯(aq) ---> 3Sn(OH)62¯(aq) + 6e¯
6e¯ + 2Bi(OH)3(s) ---> 2Bi(s) + 6OH¯

3OH¯(aq) + 3Sn(OH)3¯(aq) + 2Bi(OH)3(s) ---> 3Sn(OH)62¯(aq) + 2Bi(s)

Problem #36: MnO2 + O2 ---> MnO4¯ + H2O

1) Half-reactions:

MnO2 ---> MnO4¯
O2 ---> H2O

2) Balance in acidic:

MnO2 + 2H2O ---> MnO4¯ + 4H+ + 3e¯
4e¯ + 4H+ + O2 ---> 2H2O

3) Equalize electrons:

4MnO2 + 8H2O ---> 4MnO4¯ + 16H+ + 12e¯
12e¯ + 12H+ + 3O2 ---> 6H2O

4MnO2 + 3O2 + 2H2O ---> 4MnO4¯ + 4H+

5) Change to basic:

4OH- + 4MnO2 + 3O2 ---> 4MnO4¯ + 2H2O

Problem #37: CrO42¯ + S2¯ ---> Cr2O3 + S8

1) Half-reactions:

CrO42¯ ---> Cr2O3
S2¯ ---> S8

2) Balance in acidic:

6e¯ + 10H+ + 2CrO42¯ ---> Cr2O3 + 5H2O
8S2¯ ---> S8 + 16e¯

3) Equalize electrons:

48e¯ + 80H+ + 16CrO42¯ ---> 8Cr2O3 + 40H2O
24S2¯ ---> 3S8 + 48e¯

80H+ + 16CrO42¯ + 24S2¯ ---> 8Cr2O3 + 3S8 + 40H2O

5) Change to basic and eliminate excess water:

40H2O + 16CrO42¯ + 24S2¯ ---> 8Cr2O3 + 3S8 + 80OH¯

6) This problem can also be seen in textbooks as:

CrO42¯ + S2¯ ---> Cr2O3 + S

Solving it yields this as an answer:

5H2O + 2CrO42¯ + 3S2¯ ---> Cr2O3 + 3S + 10OH¯

When you compare to my answer in step 5 (with one little change):

40H2O + 16CrO42¯ + 24S2¯ ---> 8Cr2O3 + 24S + 80OH¯

you can see that the two approaches (S or S8) yield the same answer.

Problem #38: MnO4¯(aq) + I¯(aq) ---> MnO2(s) + I2(s)

Solution:

1) Half-reactions:

MnO4¯(aq) ---> MnO2(s)
I¯(aq) ---> I2(s)

2) Balance in acidic:

3e¯ + 4H+ + MnO4¯(aq) ---> MnO2(s) + 2H2O
2I¯(aq) ---> I2(s) + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2MnO4¯(aq) ---> 2MnO2(s) + 4H2O
6I¯(aq) ---> 3I2(s) + 6e¯

8H+ + 2MnO4¯(aq) = 6I¯(aq) ---> 2MnO2(s) + 3I2(s) + 4H2O

5) Add eight OH¯ to each side, then eliminate four excess waters:

4H2O + 2MnO4¯(aq) + 6I¯(aq) ---> 2MnO2(s) + 3I2(s) + 8OH¯

Problem #39: ClO2 + SbO2¯ ---> ClO2¯ + Sb(OH)6¯ [base]

Solution:

1) Half-reactions:

ClO2 ---> ClO2¯
SbO2¯ ---> Sb(OH)6¯

2) Balance in acidic solution:

e¯ + ClO2 ---> ClO2¯
4H2O + SbO2¯ ---> Sb(OH)6¯ + 2H+ + 2e¯

3) Equalize electrons:

2e¯ + 2ClO2 ---> 2ClO2¯
4H2O + SbO2¯ ---> Sb(OH)6¯ + 2H+ + 2e¯

2ClO2 + 4H2O + SbO2¯ ---> Sb(OH)6¯ + 2H+ + 2ClO2¯

5) Change to base:

2ClO2 + 4H2O + SbO2¯ + 2OH¯ ---> Sb(OH)6¯ + 2H2O + 2ClO2¯

6) Remove two H2O from each side:

2ClO2 + 2H2O + SbO2¯ + 2OH¯ ---> Sb(OH)6¯ + 2ClO2¯

Problem #40: MnO4¯ + C2O42¯ ---> CO2 + MnO2

Solution:

1) Half-reactions:

MnO4¯ ---> MnO2
C2O42¯ ---> CO2

2) Balance in acidic solution:

3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
C2O42¯ ---> 2CO2 + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O
3C2O42¯ ---> 6CO2 + 6e¯