### Balancing redox reactions in basic solutionProblems 26 - 45

 Ten Examples Problems 11-25 Balancing in acidic solution Problems 1-10 Only the examples and problems Return to Redox menu

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: HPO32¯ + N2H4 ---> H2PO2¯ + N2

Solution:

1) Half-reactions:

HPO32¯ ---> H2PO2¯
N2H4 ---> N2

2) Balance in acidic solution:

2e¯ + 3H+ + HPO32¯ ---> H2PO2¯ + H2O
N2H4 ---> N2 + 4H+ + 4e¯

3) Equalize electrons:

4e¯ + 6H+ + 2HPO32¯ ---> 2H2PO2¯ + 2H2O
N2H4 ---> N2 + 4H+ + 4e¯

2H+ + 2HPO32¯ + N2H4 ---> 2H2PO2¯ + N2 + 2H2O

5) Convert to basic:

2HPO32¯ + N2H4 ---> 2H2PO2¯ + N2 + 2OH¯

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¯

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

5) Change to base:

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

Problem #41: HXeO4¯ + O3 ---> XeO64¯

Comments: the Xe is oxidized from +6 to +8, so the ozone (the oxidizing agent) gets reduced. It turns out that ozone (in acidic solution) can reduce to water or to a combination of oxygen and water. I propose (1) that the same products occur under basic conditions and (2) to balance the above equation under each set of ozone products.

Solution using oxygen/water:

1) Half-reactions:

XeO42¯ ---> XeO64¯
O3 ---> O2 + H2O

I took out the hydrogen from HXeO4¯. I'll put it back in a moment.

2) Balance in acidic solution:

2H2O + XeO42¯ ---> XeO64¯ + 4H+ + 2e¯
2e¯ + 2H+ + O3 ---> O2 + H2O

3) Since the electrons are equal, add the half-reactions:

H2O + HXeO4¯ + O3 ---> XeO64¯ + O2 + 3H+

I remade the HXeO4¯. I also eliminated one water and one hydrogen ion from each side.

4) Add three hydroxides to each side:

3OH¯ + HXeO4¯ + O3 ---> XeO64¯ + O2 + 2H2O

This equation appears in a Wikipedia discussion of xenon tetroxide. It is in the section titled Synthesis.

Solution using water:

1) Half-reactions:

XeO42¯ ---> XeO64¯
O3 ---> H2O

2) Balance in acidic solution:

2H2O + XeO42¯ ---> XeO64¯ + 4H+ + 2e¯
6e¯ + 6H+ + O3 ---> 3H2O

3) Equalize electrons:

6H2O + 3XeO42¯ ---> 3XeO64¯ + 12H+ + 6e¯
6e¯ + 6H+ + O3 ---> 3H2O

3H2O + 3HXeO4¯ + O3 ---> 3XeO64¯ + 9H+

I reconstituted the three 3HXeO4¯ as well as eliminating three hydrogen ion and three water.

5) Convert to basic by adding nine hydroxide ions:

9OH¯ + 3HXeO4¯ + O3 ---> 3XeO64¯ + 6H2O

Three waters were eliminated from each side.

This equation is the most commonly seen answer.

Problem #42: HCN ---> CO32¯ + NH3

Is this an oxidation or reduction?

Solution:

1) Balance in acidic:

HCN ---> CO32¯ + NH3

3H2O + HCN ---> CO32¯ + NH3

3H2O + HCN ---> CO32¯ + NH3 + 4H+

3H2O + HCN ---> CO32¯ + NH3 + 4H+ + 2e¯

2) Change to basic:

4OH¯ + 3H2O + HCN ---> CO32¯ + NH3 + 4H2O + 2e¯

4OH¯ + HCN ---> CO32¯ + NH3 + H2O + 2e¯

It's an oxidation. The C goes from +2 in the HCN to +4 in the carbonate. The N in HCN and in NH3 stays unchanged at -3.

I changed over to basic at the end because of the presence of the ammonia.

Problem #43: MnO4¯ + SO32¯ ---> S2O82¯ + Mn2O3

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 after adding the hydroxides):

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

Problem #44: Al + KOH + H2SO4 + H2O ---> KAl(SO4)2 12H2O + H2

1) Balance the redox part of the reaction first:

Al ---> Al3+ + 3e¯
2H2O + 2e¯ ---> H2 + 2OH¯

2) Balancing electrons and adding gives:

2Al + 6H2O ---> 2Al3+ + 3H2 + 6 OH¯

3) The formula of the product is KAl(SO4)2 (ignoring the waters of hydration for a moment). Since you are forming two of those (because you have 2Al3+ ions), you will need on the left:

2Al + 2KOH + 4H2SO4 + 6H2O <--- the KOH provides the need K and the H2SO4 gives us the sulfate we need

4) On the right, at this point, you have 2KAl(SO4)2 + 8H+ + 8OH¯ (the last two from the 4H2SO4 and the 2KOH). The 8H+ + 8OH¯ on the right react to form 8H2O. So, at this point you have:

2Al + 2KOH + 4H2SO4 + 6H2O ---> 3H2 + 2 KAl(SO4)2 + 8H2O

5) To form 2 molecules of the hydrated compound, you need a total of 24H2O on the right, so we need to add 16 more waters to both sides. That gives:

2Al + 2KOH + 4H2SO4 + 22H2O ---> 3H2 + 2KAl(SO4)2 12H2O

Problem #45: Al(s) + H2O(ℓ) ---> Al(OH)4¯(aq) + H2(g)

Solution #1:

1) Half-reactions:

Al ---> Al(OH)4¯
H2O ---> H2

2) Balance:

4OH¯ + Al ---> Al(OH)4¯ + 3e¯
2e¯ + 2H2O ---> H2 + 2OH¯

Notice that I balanced the second half-reaction using hydroxide rather than using acidic first. Balancing it in acidic still gets the correct answer. See solution #2.

3) Equalize electrons:

8OH¯ + 2Al ---> 2Al(OH)4¯ + 6e¯
6e¯ + 6H2O ---> 3H2 + 6OH¯

2OH¯ + 2Al + 6H2O ---> 2Al(OH)4¯ + 3H2

Solution #2:

1) I'll balance the second half-reaction in acidic solution:

H2O ---> H2

H2O ---> H2 + H2O <--- balance the oxygen

2H+ + H2O ---> H2 + H2O <--- balance the hydrogen

2e¯ + 2H+ + H2O ---> H2 + H2O <--- balance the charge

2e¯ + 2H+ ---> H2 <--- eliminate the water

2) Here are the two balanced half-reactions:

4OH¯ + Al ---> Al(OH)4¯ + 3e¯
2e¯ + 2H+ ---> H2

3) Equalize electrons:

8OH¯ + 2Al ---> 2Al(OH)4¯ + 6e¯
6e¯ + 6H+ ---> 3H2

6H+ + 8OH¯ + 2Al ---> 2Al(OH)4¯ + 3H2

5) Allow six hydrogen ion and six hydroxide ion to react to form water:

6H2O + 2OH¯ + 2Al ---> 2Al(OH)4¯ + 3H2

Problem #46: Ag2O(s) + HPO32¯(aq) ---> AgO(s) + H2PO2¯(aq)

Solution:

1) Half-reactions:

HPO32¯ ---> H2PO2¯
Ag2O(s) ---> AgO(s)

2) Balance in acidic solution:

2e¯ + 3H+ + HPO32¯ ---> H2PO2¯ + H2O
H2O + Ag2O(s) ---> 2AgO(s) + 2H+ + 2e¯