Balancing redox reactions in acidic aolution
Problems #11-30

Examples 1-5      Only the examples and problems
Problems 1-10      Balancing in basic solution      Return to Redox menu

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¯

4) Add:

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

4) Add:

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¯

4) Add:

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

4) Add:

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]

5) The almost final answer:

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

3) Add in the chloride:

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-

4) Add:

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¯

4) Add:

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¯

4) Add:

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

4) Add:

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

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: Cr(s) + O2(g) --> Cr3+(aq)

Solution:

1) Balance the oxidation half-reaction, while ignoring any possible reduction half-reactions:

Cr ---> Cr3+

Cr ---> Cr3+ + 3e¯

2) We need a reduction half-reaction:

O2 ---> ????

O2 needs to reduce from zero to something. That something could be this:

O2 ---> H2O

3) H2O2 might also a reduction from O2, but there are reasons for it to not be the product. You'll get to those reasons later. (I discuss going to the oxide below at step 7.) Balance the proposed reduction half-reaction:

2e¯ + 2H+ + 12O2 ---> H2O

4) Write the two half-reactions:

2e¯ + 2H+ + 12O2 ---> H2O
Cr ---> Cr3+ + 3e¯

5) Equalize electrons:

6e¯ + 6H+ + 32O2 ---> 3H2O
2Cr ---> 2Cr3+ + 6e¯

6) Add and clear fraction:

6H+ + 2Cr + 32O2 ---> 2Cr3+ + 3H2O

12H+ + 4Cr + 3O2 ---> 4Cr3+ + 6H2O

7) You could also go to oxide for the reduction half-reaction:

O2 ---> O2¯

8) The two half-reactions:

Cr ---> Cr3+
O2 ---> O2¯

9) Balanced:

Cr ---> Cr3+ + 3e¯
4e¯ + O2 ---> 2O2¯

10) Equalize elections and add:

4Cr ---> 4Cr3+ + 12e¯
12e¯ + 3O2 ---> 6O2¯

4Cr + 3O2 ---> 4Cr3+ + 6O2¯

More commonly written as:

4Cr(s) + 3O2(g) ---> 2Cr2O3(s)

However, this last does not give the Cr3+ in aqueous solution, so the version with water as the product is the best choice.


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