### Balancing redox reactions in acidic aolutionProblems #11-30

Problem #11: Cr2O72¯ + CH3CHO ---> CH3COOH + Cr3+

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr3+
CH3CHO ---> CH3COOH

2) Balance:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
H2O + CH3CHO ---> CH3COOH + 2H+ + 2e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
3H2O + 3CH3CHO ---> 3CH3COOH + 6H+ + 6e¯

4) Add and eliminate like items:

8H+ + Cr2O72¯ + 3CH3CHO ---> 3CH3COOH + 2Cr3+ + 4H2O

Problem #12: MnO4¯ + C2H4O ---> CH3COOH + MnO2

Solution:

1) Half-reactions:

MnO4¯ ---> MnO2
C2H4O ---> CH3COOH

2) Balance:

3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
H2O + C2H4O ---> CH3COOH + 2H+ + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O
3H2O + 3C2H4O ---> 3CH3COOH + 6H+ + 6e¯

2H+ + 2MnO4¯ + 3C2H4O ---> 3CH3COOH + 2MnO2 + H2O

5) You could make some permanganic acid on the left:

2HMnO4 + 3C2H4O ---> 3CH3COOH + 2MnO2 + H2O

Problem #13: Zn + NO3¯ ---> NH4+ + Zn2+

Solution:

1) Half-reactions:

Zn ---> Zn2+
NO3¯ ---> NH4+

2) Balance:

Zn ---> Zn2+ + 2e¯
8e¯ + 10H+ + NO3¯ ---> NH4+ + 3H2O

3) Equalize electrons:

4Zn ---> 4Zn2+ + 8e¯
8e¯ + 10H+ + NO3¯ ---> NH4+ + 3H2O

4Zn + 10H+ + NO3¯ ---> NH4+ + 3H2O + 4Zn2+

Problem #14: NO2 + H2O ---> HNO3 + NO

Solution:

1) Half-reactions:

NO2 ---> NO
NO2 ---> HNO3

2) Balance:

2e¯ + 2H+ + NO2 ---> NO + H2O
H2O + NO2 ---> HNO3 + H+ + e¯

3) Equalize electrons:

2e¯ + 2H+ + NO2 ---> NO + H2O
2H2O + 2NO2 ---> 2HNO3 + 2H+ + 2e¯

3NO2 + H2O ---> 2HNO3 + NO

Problem #15: S2¯ + NO3¯ ----> NO + S8

Solution:

1) Half-reactions:

S2¯ ----> S8
NO3¯ ----> NO

2) Balance:

8S2¯ ----> S8 + 16e¯
3e¯ + 4H+ + NO3¯ ----> NO + 2H2O

3) Equalize electrons:

24S2¯ ----> 3S8 + 48e¯ <--- multiplied by a factor of 3
48e¯ + 64H+ + 16NO3¯ ----> 16NO + 32H2O <--- multiplied by a factor of 16

64H+ + 24S2¯ + 16NO3¯ ----> 16NO + 3S8 + 32H2O

Problem #16: CuS + NO3¯ ---> NO + Cu2+ + HSO4¯

Solution:

In this reaction, CuS is a solid substance with a nitric acid solution being poured on it. Chemically, both the Cu and the S must be accounted for, but ONLY at the end, in the final answer.

In this problem, we can eliminate the Cu from both sides of the equation and make things a bit simpler. The reason we can do this has to do with what happens to the Cu during the reaction. Note that it is a +2 ion in solution as a product. However, what is its charge in the CuS? The answer is +2, so the Cu was neither reduced nor oxidized in the reaction. That means we can eliminate it during the balancing and act like only S2¯ is in solution. However, we must add the Cu back in at the end.

1) Drop the Cu to get:

S2¯ + NO3¯ ---> NO + HSO4¯

2) The two half-reactions are as follows:

 S2¯ ---> HSO4¯ NO3¯ ---> NO

3) Balancing them gives:

 4H2O + S2¯ ---> HSO4¯ + 7H+ + 8e¯ 3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

4) Make the number of electrons equal:

 3 [4H2O + S2¯ ---> HSO4¯ + 7H+ + 8e¯] 8 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

3S2¯ + 11H+ + 8NO3¯ ---> 3HSO4¯ + 8NO + 4H2O

6) Put the Cu2+ back in.

3CuS + 11H+ + 8NO3¯ ---> 3 Cu2+ + 3HSO4¯ + 8NO + 4H2O

Problem #17: Ru(s) + Cl¯(aq) + NO3¯(aq) ---> RuCl63¯(aq) + NO2(g)

Solution

1) Eliminate the Cl¯ for the moment:

Ru ---> Ru3+ + 3e¯
e¯ + 2H+ + NO3¯ ---> NO2 + H2O

2) Multiply second half-reaction by 3 and add:

6H+ + 3NO3¯ + Ru ---> Ru3+ + 3NO2 + 3H2O

6HCl + 3NO3¯ + Ru ---> RuCl63¯ + 3NO2 + 3H2O

4) Add in three more H+, if you wish:

6HCl + 3HNO3 + Ru ---> H3RuCl6 + 3NO2 + 3H2O

Comment: good ole aqua regia to the rescue!

Problem #18: IO3¯(aq) + I2(s) ---> I2(s) + HIO(aq)

Solution

1) Half-reactions:

10e¯ + 12H+ + 2IO3¯(aq) ---> I2(s) + 6H2O
2H2O + I2(s) ---> 2HIO(aq) + 2H+ + 2e¯

2) Multiply bottom by 5, then add:

10H2O + 5I2(s) + 12H+ + 2IO3¯(aq) ---> I2(s) + 6H2O + 10HIO(aq) + 10H+

3) Eliminate like items:

4H2O + 4I2(s) + 2H+ + 2IO3¯(aq) ---> 10HIO(aq)

4) Combine on the left-hand side to make some iodic acid:

4H2O + 4I2(s) + 2HIO3(aq) ---> 10HIO(aq)

Problem #19: PH3 + I2 ---> H3PO2 + I¯

Solution:

PH3 ---> H3PO2
I2 ---> I¯

2H2O + PH3 ---> H3PO2 + 4H+ + 4e¯
2e¯ + I2 ---> 2I¯

2H2O + PH3 ---> H3PO2 + 4H+ + 4e¯
4e¯ + 2I2 ---> 4I¯

2H2O + PH3 + 2I2 ---> H3PO2 + 4I¯ + 4H+

Problem #20: P4 + HNO3 ---> H3PO4 + NO

Solution:

P4 ---> H3PO4
HNO3 ---> NO

16H2O + P4 ---> 4H3PO4 + 20H+ + 20e¯
3e¯ + 3H+ + HNO3 ---> NO + 2H2O

need to equalize electrons:

48H2O + 3P4 ---> 12H3PO4 + 60H+ + 60e¯
60e¯ + 60H+ + 20HNO3 ---> 20NO + 40H2O

8H2O + 3P4 + 20HNO3 ---> 12H3PO4 + 20NO

Problem #21: BrO3-(aq) + N2H4(aq) ---> Br2(ℓ) + N2(g)

Solution:

1) Half-reactions:

BrO3-(aq) ---> Br2(ℓ)
N2H4(aq) ---> N2(g)

2) Balance:

10e- + 12H+ + 2BrO3-(aq) ---> Br2(ℓ) + 6H2O
N2H4(aq) ---> N2(g) + 4H+ + 4e-

3) Equalize electrons:

20e- + 24H+ + 4BrO3-(aq) ---> 2Br2(ℓ) + 12H2O
5N2H4(aq) ---> 5N2(g) + 20H+ + 4e-

4H+ + 4BrO3-(aq) + 5N2H4(aq) ---> 2Br2(ℓ) + 5N2(g) + 12H2O

5) You could put the ions together to make some bromic acid:

4HBrO3(aq) + 5N2H4(aq) ---> 2Br2(ℓ) + 5N2(g) + 12H2O

Problem #22: FeSO4 + H2O2 ---> Fe2O3 + SO42¯ + H2O

Solution:

1) The balancing for only one of the half-reactions (notice how I kept the sulfate rather than eliminating it and then adding it back later):

FeSO4 ---> Fe2O3 + SO42¯

balance the iron and the sulfate:

2FeSO4 ---> Fe2O3 + 2SO42¯

balance the O:

3H2O + 2FeSO4 ---> Fe2O3 + 2SO42¯

balance the H:

3H2O + 2FeSO4 ---> Fe2O3 + 2SO42¯ + 6H+

balance the electrons:

3H2O + 2FeSO4 ---> Fe2O3 + 2SO42¯ + 6H+ + 2e¯

2) Second half-reaction:

H2O2 ---> H2O

2e¯ + 2H+ + H2O2 ---> 2H2O

3) The two balanced half-reactions already have the electrons equalized, so add them:

H2O + 2FeSO4 + H2O2 ---> Fe2O3 + 2SO42¯ + 4H+

4) Put the ions together to make sulfuric acid:

H2O + 2FeSO4 + H2O2 ---> Fe2O3 + 2H2SO4

Note: the water was present on the right-hand side to signal to you what the H2O2 should decompose to.

Problem #23: Cr2O72¯ + H+ + I¯ ---> Cr3+ + I2 + H2O

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr3+
I¯ ---> I2

2) Balance in acidic solution:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
2I¯ ---> I2 + 2e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6I¯ ---> 3I2 + 6e¯

14H+ + Cr2O72¯ + 6I¯ ---> 2Cr3+ + 3I2 + 7H2O

Problem #24: S2O82¯ + Mn2+ ---> SO42¯ + MnO4¯

Solution:

1) Half reactions:

S2O82¯ ---> SO42¯
Mn2+ ---> MnO4¯

2) Balanced:

2e¯ + S2O82¯ ---> 2SO42¯
4H2O + Mn2+ ---> MnO4¯ + 8H+ + 5e¯

3) Equalize electrons:

10e¯ + 5S2O82¯ ---> 10SO42¯
8H2O + 2Mn2+ ---> 2MnO4¯ + 16H+ + 10e¯

8H2O + 5S2O82¯ + 2Mn2+ ---> 2MnO4¯ + 10SO42¯ + 16H+

Problem #25: HNO3 + Cu2O --> Cu(NO3)2 + NO + H2O

Solution:

1) Half-reactions:

Cu2O ---> Cu2+
NO3¯ ---> NO

2) Balance:

2H+ + Cu2O ---> 2Cu2+ + H2O + 2e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Equalize electrons:

6H+ + 3Cu2O ---> 6Cu2+ + 3H2O + 6e¯
6e¯ + 8H+ + 2NO3¯ ---> 2NO + 4H2O

14H+ + 2NO3¯ + 3Cu2O ---> 6Cu2+ + 2NO + 7H2O

5) Add twelve nitrates to each side to restore the molecular equation:

14HNO3 + 3Cu2O ---> 6Cu(NO3)2 + 2NO + 7H2O

Problem #26: As2O3 + NO3¯ ---> H3AsO4 + N2O3

Solution:

1) Half-reactions:

As2O3 ---> H3AsO4
NO3¯ ---> N2O3

2) Balance:

5H2O + As2O3 ---> 2H3AsO4 + 4H+ + 4e¯
4e¯ + 6H+ + 2NO3¯ ---> N2O3+ 3H2O

3) The electrons are already equal, so add the two half reactions:

2H+ + 2H2O + As2O3 + 2NO3¯ ---> 2H3AsO4 + N2O3

4) If you want to, you can form nitric acid:

2H2O + As2O3 + 2HNO3 ---> 2H3AsO4 + N2O3

Problem #27: MnO4¯ + S2O32¯ ---> Mn2+ + SO42¯

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
S2O32¯ ---> SO42¯

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
5H2O + S2O32¯ ---> 2SO42¯ + 10H+ + 8e¯

3) Equalize electrons:

40e¯ + 64H+ + 8MnO4¯ ---> 8Mn2+ + 32H2O
25H2O + 5S2O32¯ ---> 10SO42¯ + 50H+ + 40e¯

4) Add and eliminate like items:

14H+ + 8MnO4¯ + 5S2O32¯ ---> 8Mn2+ + 10SO42¯ + 7H2O

Problem #28: MnO4¯ + H2SO3 ---> Mn2+ + HSO4¯

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
H2SO3 ---> HSO4¯

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + H2SO3 ---> HSO4¯ + 3H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2O + 5H2SO3 ---> 5HSO4¯ + 15H+ + 10e¯

4) Add and eliminate like items:

H+ + 2MnO4¯ + 5H2SO3 ---> 2Mn2+ + 5HSO4¯ + 3H2O

Problem #29: MnO4¯ + HSO3¯ ---> Mn2+ + SO42¯

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
HSO3¯ ---> SO42¯

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
HSO3¯ + H2O ---> SO42¯ + 3H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5HSO3¯ + 5H2O ---> 5SO42¯ + 15H+ + 10e¯

4) Add and eliminate like items:

H+ + 2MnO4¯ + 5HSO3¯ ---> 2Mn2+ + 5SO42¯ + 3H2O

Problem #30: H2O2 + Cr2O72¯ ---> Cr3+ + O2 + H2O

Solution:

1) Half-reactions:

H2O2 ---> O2
Cr2O72¯ ---> Cr3+

2) Balance:

H2O2 ---> O2 + 2H+ + 2e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

3) Equalize electrons:

3H2O2 ---> 3O2 + 6H+ + 6e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

8H+ + 3H2O2 + Cr2O72¯ ---> 2Cr3+ + 3O2 + 7H2O

Problem #31: Cl2 + CrO42¯ + H+ ---> Cr3+ + ClO3¯ + H2O

Solution:

1) Half-reactions:

Cl2 ---> ClO3¯
CrO42¯ ---> Cr3+

2) Balance in acidic solution:

6H2O + Cl2 ---> 2ClO3¯ + 12H+ + 10e¯
3e¯ + 8H+ + CrO42¯ ---> Cr3+ + 4H2O

3) Equalize electrons:

18H2O + 3Cl2 ---> 6ClO3¯ + 36H+ + 30e¯
30e¯ + 80H+ + 10CrO42¯ ---> 10Cr3+ + 40H2O

4) Add & eliminate only electrons:

80H+ + 18H2O + 3Cl2 + 10CrO42¯ ---> 10Cr3+ + 6ClO3¯ + 36H+ + 40H2O

5) Eliminate excess water and hydrogen ion:

44H+ + 3Cl2 + 10CrO42¯ ---> 10Cr3+ + 6ClO3¯ + 22H2O

Problem #32: IO3¯(aq) + SO2(g) ---> I2(s) + SO42¯(aq)

Solution:

1) Half-reactions:

IO3¯(aq) ---> I2(s)
SO2(g) ---> SO42¯(aq)

2) Balance the half-reactions:

10e¯ + 12H+ + 2IO3¯(aq) ---> I2(s) + 6H2O
2H2O + SO2(g) ---> SO42¯(aq) + 4H+ + 2e¯

3) Equalize electrons:

10e¯ + 12H+ + 2IO3¯(aq) ---> I2(s) + 6H2O
10H2O + 5SO2(g) ---> 5SO42¯(aq) + 20H+ + 10e¯

4H2O(ℓ) + 2IO3¯(aq) + 5SO2(g) ---> I2(s) + 5SO42¯(aq) + 8H+(aq)

Problem #33: Cr2O72¯ + Mn2+ --> Cr2+ + MnO4¯

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr2+
Mn2+ ---> MnO4¯

2) Balance 'em in acidic solution:

8e¯ + 14H+ + Cr2O72¯ ---> 2Cr2+ + 7H2O
4H2O + Mn2+ ---> MnO4¯ + 8H+ + 5e¯

3) The least-common multiple between 8 and 5 is 40:

40e¯ + 70H+ + 5Cr2O72¯ ---> 10Cr2+ + 35H2O
32H2O + 8Mn2+ ---> 8MnO4¯ + 64H+ + 40e¯

6H+ + 5Cr2O72¯ + 8Mn2+ ---> 10Cr2+) + 8MnO4¯ + 3H2O

Problem #34: Mn2+(aq) + BiO3¯(aq) ---> MnO4¯(aq) + Bi3+(aq)

Solution:

1) Half-reactions:

Mn2+(aq) ---> MnO4¯(aq)
BiO3¯(aq) ---> Bi3+(aq)

2) Balance:

4H2O + Mn2+(aq) ---> MnO4¯(aq) + 8H+ + 5e¯
2e¯ + 6H+ + BiO3¯(aq) ---> Bi3+(aq) + 3H2O

3) Equalize electrons:

8H2O + 2Mn2+(aq) ---> 2MnO4¯(aq) + 16H+ + 10e¯
10e¯ + 30H+ + 5BiO3¯(aq) ---> 5Bi3+(aq) + 15H2O

4) Add and eliminate duplicate items:

14H+ + 2Mn2+(aq) + 5BiO3¯(aq) ---> 5Bi3+(aq) + 2MnO4¯(aq) + 7H2O

Problem #35: Cr2O72¯ + CH3CH2OH ---> Cr3+ + CH3CHO

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr3+
CH3CH2OH ---> CH3CHO

2) Balance:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
CH3CH2OH ---> CH3CHO + 2H+ + 2e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
3CH3CH2OH ---> 3CH3CHO + 6H+ + 6e¯

4) Add and eliminate duplicate items (six electrons and six hydrogen ion):

8H+ + Cr2O72¯ + 3CH3CH2OH ---> 2Cr3+ + 3CH3CHO + 7H2O

Problem #36: S2O32¯ + OCl¯ ---> Cl¯ + S4O62¯

Solution:

1) Half-reactions:

S2O32¯ ---> S4O62¯
OCl¯ ---> Cl¯

2) Balance:

2S2O32¯ ---> S4O62¯ + 2e¯
2e¯ + 2H+ + OCl¯ ---> Cl¯ + H2O

2H+ + 2S2O32¯ + OCl¯ ---> Cl¯ + S4O62¯ + H2O

Problem #37: Cr3+ + BiO3¯ ---> Cr2O72¯ + Bi3+

Solution:

1) Half-reactions:

Cr3+ ---> Cr2O72¯
BiO3¯ ---> Bi3+

2) Balance:

7H2O + 2Cr3+ ---> Cr2O72¯ + 14H+ + 6e¯
2e¯ + 6H+ + BiO3¯ ---> Bi3+ + 3H2O

3) Equalize electrons:

7H2O + 2Cr3+ ---> Cr2O72¯ + 14H+ + 6e¯
6e¯ + 18H+ + 3BiO3¯ ---> 3Bi3+ + 9H2O

4H+ + 2Cr3+ + 3BiO3¯ ---> Cr2O72¯ + 3Bi3+ + 2H2O

Problem #38: HNO2 + MnO4¯ ---> Mn2+ + NO3¯

Solution:

1) Half-reactions:

HNO2 ---> NO3¯
MnO4¯ ---> Mn2+

2) Balance:

H2O + HNO2 ---> NO3¯ + 3H+ + 2e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Equalize electrons:

5H2O + 5HNO2 ---> 5NO3¯ + 15H+ + 10e¯
10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O

H+ + 5HNO2 + 2MnO4¯ ---> 2Mn2+ + 5NO3¯ + 3H2O

Problem #39: V ---> H2V2O42¯ + VH3

Solution:

1) Half-reactions:

V ---> H2V2O42¯
V ---> VH3

2) Balance:

4H2O + 2V ---> H2V2O42¯ + 6H+ + 4e¯
3e¯ + 3H+ + V ---> VH3

3) Equalize electrons:

12H2O + 6V ---> 3H2V2O42¯ + 18H+ + 12e¯
12e¯ + 12H+ + 4V ---> 4VH3

12H2O + 10V ---> 3H2V2O42¯ + 4VH3 + 6H+

Problem #40: Bi2O4 + MoO2+ ---> BiO+ + H2MoO4

Solution:

1) Half-reactions:

Bi2O4 ---> BiO+
MoO2+ ---> H2MoO4

2) Balance:

2e¯ + 4H+ + Bi2O4 ---> 2BiO+ + 2H2O
2H2O + MoO2+ ---> H2MoO4 + 2H+ + e¯

3) Equalize electrons:

2e¯ + 4H+ + Bi2O4 ---> 2BiO+ + 2H2O
4H2O + 2MoO2+ ---> 2H2MoO4 + 4H+ + 2e¯

4) Add and eliminate like items (2 electrons, 4 hydrogen ion, 2 water):

2H2O + Bi2O4 + 2MoO2+ ---> 2BiO+ + 2H2MoO4