Balancing redox reactions in acidic aolution
Problems #26 - 50

Ten Examples      Problems 11-25      Balancing in basic solution
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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

4) Add:

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¯

4) Add:

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¯

4) Add and delete duplicates:

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

3) Add (since electrons are already equal):

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

4) Add:

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

4) Add:

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

4) Add:

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

Problem #41: As2O3 + NO3¯ ---> H3AsO4 + NO

Solution:

1) Separate into half-reactions:

As2O3 ---> AsO43¯
NO3¯ ---> NO

2) Balance in acidic solution:

5H2O + As2O3 ---> 2AsO43¯ + 10H+ + 4e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Equalize electrons:

15H2O + 3As2O3 ---> 6AsO43¯ + 30H+ + 12e¯
12e¯ + 16H+ + 4NO3¯ ---> 4NO + 8H2O

4) Add:

7H2O + 3As2O3 + 4NO3¯ ---> 6AsO43¯ + 14H+ + 4NO

5) I would like to reconstitute the arsenic acid as a molecular species, but I need four more hydrogen ions to do so. I can do this by adding four hydrogen ions to both sides:

7H2O + 3As2O3 + 4HNO3 ---> 6H3AsO4 + 4NO

Notice that I also formed nitric acid in a molecular way, as opposed to showing it ionized.


Problem #42: Cr2O72¯ + Cu+ ---> Cr3+ + Cu2+

Solution:

1) Separate into half reactions:

Cr2O72¯ ---> Cr3+
Cu+ ---> Cu2+

2) You balance in acidic solution if the problem is silent on acidic/basic:

6e¯ + 14H+ + Cr2O72¯- ---> 2Cr3+ + 7H2O
Cu+ ---> Cu2+ + e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6Cu+ ---> 6Cu2+ + 6e¯

4) Add:

14H+ + Cr2O72¯ + 6Cu+ ---> 2Cr3+ + 6Cu2+ + 7H2O

Problem #43: IO3¯ + H2SO3 ---> I2 + SO42¯

Solution:

1) Half reactions:

IO3¯ ---> I2
H2SO3 ---> SO42¯

2) Balance:

10e¯ + 12H+ + 2IO3¯ ---> I2 + 6H2O
H2O + H2SO3 ---> SO42¯ + 4H+ + 2e¯

3) Equalize electrons:

10e¯ + 12H+ + 2IO3¯ ---> I2 + 6H2O
5H2O + 5H2SO3 ---> 5SO42¯ + 20H+ + 10e¯

4) Add:

2IO3¯ + 5H2SO3 ---> I2 + 5SO42¯ + 8H+ + H2O

Problem #44: NO3¯ + Sn2+ ---> NO + Sn4+

Solution:

1) Half reactions:

NO3¯ ---> NO
Sn2+ ---> Sn4+

2) You balance in acidic solution if the problem is silent on acidic/basic:

3e¯ + 4H+ + NO3¯ ---> NO + 2H2O
Sn2+ ---> Sn4+ + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2NO3¯ ---> 2NO + 4H2O
3Sn2+ ---> 3Sn4+ + 6e¯

4) Add:

8H+ + 2NO3¯ + 3Sn2+ ---> 2NO + 3Sn4+ + 4H2O

Problem #45: NO3¯ + S2¯ ---> S8 + NO2

Solution:

1) Half reactions:

NO3¯ ---> NO2
S2¯ ---> S8

2) Balance:

e¯ + 2H+ + NO3¯ ---> NO2 + H2O
8S2¯ ---> S8 + 16e¯

3) Equalize electrons:

16e¯ + 32H+ + 16NO3¯ ---> 16NO2 + 16H2O
8S2¯ ---> S8 + 16e¯

4) Add:

32H+ + 16NO3¯ + 8S2¯ ---> 16NO2 + S8 + 16H2O

Problem #46: HNO2 + In+ ---> NO + In3+

Solution:

1) Half reactions:

HNO2 ---> NO
In+ ---> In3+

2) Balance:

e¯ + H+ + HNO2 ---> NO + H2O
In+ ---> In3+ + 2e¯

3) Equalize electrons:

2e¯ + 2H+ + 2HNO2 ---> 2NO + 2H2O
In+ ---> In3+ + 2e¯

4) Add:

2H+ + 2HNO2 + In+ ---> 2NO + In3+ + 2H2O

Problem #47: HNO3 + I2 ---> HIO3 + NO2 + H2O

Solution:

1) Write the net-ionic equation:

NO3¯ + I2 ---> IO3¯ + NO2

2) Half-reactions:

NO3¯ ---> NO2
I2 ---> IO3¯

3) Balance:

e¯ + 2H+ + NO3¯ ---> NO2 + H2O
6H2O + I2 ---> 2IO3¯ + 12H+ + 10e¯

4) Equalize electrons:

10e¯ + 20H+ + 10NO3¯ ---> 10NO2 + 10H2O
6H2O + I2 ---> 2IO3¯ + 12H+ + 10e¯

5) Add:

8H+ + 10NO3¯ + I2 ---> 2IO3¯ + 10NO2 + 4H2O

6) Add two hydrogen ion and make nitric acid as well as iodic acid:

10HNO3 + I2 ---> 2HIO3 + 10NO2 + 4H2O

Problem #48: VO2+ + Sn2+ ---> VO2+ + Sn4+

Solution:

1) Half-reactions:

VO2+ ---> VO2+
Sn2+ ---> Sn4+

2) Balance:

e¯ + 2H+ + VO2+ ---> VO2+ + H2O
Sn2+ ---> Sn4+ + 2e¯

3) Equalize electrons:

2e¯ + 4H+ + 2VO2+ ---> 2VO2+ + 2H2O
Sn2+ ---> Sn4+ + 2e¯

4) Add:

4H+ + 2VO2+ + Sn2+ ---> 2VO2+ + Sn4+ + 2H2O

Problem #49: As2O3 + NO3¯ ---> H3AsO4 + NO

This equation is balanced as part of a presentation here.

Solution:

1) Half-reactions:

As2O3 ---> H3AsO4
NO3¯ ---> NO

2) Balance:

5H2O + As2O3 ---> 2H3AsO4 + 4H+ + 4e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Equalize electrons:

15H2O + 3As2O3 ---> 6H3AsO4 + 12H+ + 12e¯
12e¯ + 16H+ + 4NO3¯ ---> 4NO + 8H2O

4) Add:

4H+ + 7H2O + 3As2O3 + 4NO3¯ ---> 6H3AsO4 + 4NO

5) You could make nitric acid in order to present a balanced, molecular equation:

7H2O + 3As2O3 + 4HNO3 ---> 6H3AsO4 + 4NO

Problem #50: H3PO2 + Cr2O72¯ ---> H3PO4 + Cr3+

Solution:

1) Half-reactions:

H3PO2 ---> H3PO4
Cr2O72¯ ---> Cr3+

2) Balance:

2H2O + H3PO2 ---> H3PO4 + 4H+ + 4e¯
6e¯+ 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

3) Equalize electrons:

6H2O + 3H3PO2 ---> 3H3PO4 + 12H+ + 12e¯
12e¯+ 28H+ + 2Cr2O72¯ ---> 4Cr3+ + 14H2O

4) Add:

16H+ + 3H3PO2 + 2Cr2O72¯ ---> 3H3PO4 + 4Cr3+ + 8H2O

5) This can be put into molecular form by adding sixteen chlorides and four potassium ions to the left-hand side . . .

16HCl + 3H3PO2 + 2K2Cr2O7 ---> 3H3PO4 + 4Cr3+ + 8H2O

6) . . . and to the right-hand side:

16HCl + 3H3PO2 + 2K2Cr2O7 ---> 3H3PO4 + 4CrCl3 + 8H2O + 4KCl

Sodium ions would have also worked.


Bonus Problem: As(s) + ClO3¯(aq) ---> H3AsO3(aq) + HClO(aq) (acidic solution)

Solution:

1) Half-reactions:

As ---> H3AsO3
ClO3¯ ---> HClO

2) Balance:

As + 3H2O ---> H3AsO3 + 3H+ + 3e¯
4e¯ + 5H+ + ClO3¯ ---> HClO + 2H2O

3) Equalize electrons:

4As + 12H2O ---> 4H3AsO3 + 12H+ + 12e¯
12e¯ + 15H+ + 3ClO3¯ ---> 3HClO + 6H2O

4) Add:

3H+ + 4As + 3ClO3¯ + 6H2O ---> 4H3AsO3 + 3HClO

5) Form chloric acid on the left to make it a molecular equation:

4As + 3HClO3 + 6H2O ---> 4H3AsO3 + 3HClO

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