Balancing redox reactions in basic solution
Problems 26 - 50

Problem #26a: 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¯

4) Add:

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

5) Convert to basic:

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

One water removed from each side.

Problem #26b: MnO₄¯ + SO₃²¯ ---> MnO₄²¯ + SO₄²¯

Solution:

1) Half-reactions:

MnO₄¯ ---> MnO₄²¯
SO₃²¯ ---> SO₄²¯

2) Balance:

MnO₄¯ + e¯ ---> MnO₄²¯
H₂O + SO₃²¯ ---> SO₄²¯ + 2H⁺ + 2e¯

3) Convert second half-reaction to basic by adding two hydroxides to each side:

2OH¯ + H₂O + SO₃²¯ ---> SO₄²¯ + 2H₂O + 2e¯

then:

2OH¯ + SO₃²¯ ---> SO₄²¯ + H₂O + 2e¯

4) Equalize electrons:

2MnO₄¯ + 2e¯ ---> 2MnO₄²¯
2OH¯ + SO₃²¯ ---> SO₄²¯ + H₂O + 2e¯

5) Add:

2MnO₄¯ + SO₃²¯ + 2OH¯ ---> 2MnO₄²¯ + SO₄²¯ + H₂O

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¯

4) Add:

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¯

3) Electrons already equal, so add:

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¯

4) Add:

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

4) Add:

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

4) Add:

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.

3) Add (since the electrons are already equal):

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¯

4) Add:

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¯

4) Add:

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

4) Add:

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¯

4) Add:

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¯

4) Add:

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¯

4) Add:

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¯

4) Add:

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

4) Add:

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¯

4) Add:

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¯

4) Add:

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

4) Add:

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¯

3) Add (since electrons are already equal):

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

4) Convert to basic:

H₂O + Ag₂O(s) + HPO₃²¯ ---> 2AgO(s) + H₂PO₂¯ + OH¯

Problem #47: V + ClO₃¯ ---> HV₂O₇³¯ + Cl¯

Solution:

1) Half-reactions:

V ---> HV₂O₇³¯
ClO₃¯ ---> Cl¯

2) Balance in acidic:

7H₂O + 2V ---> HV₂O₇³¯ + 13H⁺ + 10e¯
6e¯ + 6H⁺ + ClO₃¯ ---> Cl¯ + 3H₂O

3) Equalize electrons:

21H₂O + 6V ---> 3HV₂O₇³¯ + 39H⁺ + 30e¯
30e¯ + 30H⁺ + 5ClO₃¯ ---> 5Cl¯ + 15H₂O

4) Add:

6H₂O + 6V + 5ClO₃¯ ---> 3HV₂O₇³¯ + 5Cl¯ + 9H⁺

5) Convert to basic by adding nine hydroxides to each side:

9OH¯ + 6V + 5ClO₃¯ ---> 3HV₂O₇³¯ + 5Cl¯ + 3H₂O

Six duplicate waters were eliminated from each side.

Problem #48: AsO₄³¯ + NO₂¯ ---> AsO₂¯ + NO₃¯

Solution:

1) Half-reactions:

AsO₄³¯ ---> AsO₂¯
NO₂¯ ---> NO₃¯

2) Balance in acidic:

2e¯ + 4H⁺ + AsO₄³¯ ---> AsO₂¯ + 2H₂O
H₂O + NO₂¯ ---> NO₃¯ + 2H⁺ + 2e¯

3) Electrons already equalized, so add them together:

2H⁺ + AsO₄³¯ + NO₂¯ ---> AsO₂¯ + NO₃¯ + H₂O

4) Convert to basic by adding two hydroxides to each side:

H₂O + AsO₄³¯ + NO₂¯ ---> AsO₂¯ + NO₃¯ + 2OH¯

One duplicate water was removed.

Problem #49: Ag₂S(s) + Cr(OH)₃(s) ---> Ag(s) + HS¯(aq) + CrO₄²¯(aq)

Solution:

1) Half-reactions:

Ag₂S(s) ---> Ag(s) + HS¯(aq)
Cr(OH)₃(s) ---> CrO₄²¯(aq)

The sulfide is not reduced or oxidized.

2) Balance in acid:

2e¯ + H⁺ + Ag₂S(s) ---> 2Ag(s) + HS¯(aq)
H₂O + Cr(OH)₃(s) ---> CrO₄²¯(aq) + 5H⁺ + 3e¯

3) Equalize electrons:

6e¯ + 3H⁺ + 3Ag₂S(s) ---> 6Ag(s) + 3HS¯(aq)
2H₂O + 2Cr(OH)₃(s) ---> 2CrO₄²¯(aq) + 10H⁺ + 6e¯

4) Add:

2H₂O + 2Cr(OH)₃(s) + 3Ag₂S(s) ---> 6Ag(s) + 3HS¯(aq) + 2CrO₄²¯(aq) + 7H⁺

5) Make basic with the addition of seven hydroxides on each side:

7OH¯ + 2Cr(OH)₃(s) + 3Ag₂S(s) ---> 6Ag(s) + 3HS¯(aq) + 2CrO₄²¯(aq) + 5H₂O

Two duplicate waters were removed.

Problem #50: Co(OH)₃(aq) + Sn(s) ---> Co(OH)₂(aq) + HSnO₂¯(aq)

Solution:

1) Half-reactions:

Co(OH)₃ ---> Co(OH)₂
Sn ---> HSnO₂¯

2) Balance:

e¯ + Co(OH)₃ ---> Co(OH)₂ + OH¯
2H₂O + Sn ---> HSnO₂¯ + 3H⁺ + 2e¯

Notice how one is balanced in basic and one in acidic. This does not affect the final answer.

3) Equalize electrons:

2e¯ + 2Co(OH)₃ ---> 2Co(OH)₂ + 2OH¯
2H₂O + Sn ---> HSnO₂¯ + 3H⁺ + 2e¯

4) Add:

2H₂O + 2Co(OH)₃ + Sn ---> 2Co(OH)₂ + HSnO₂¯ + 3H⁺ + 2OH¯

5) 3H⁺ + 2OH¯ creates two waters and a hydrogen ion:

2Co(OH)₃ + Sn ---> 2Co(OH)₂ + HSnO₂¯ + H⁺

I also eliminated two duplicate waters.

6) Change to basic:

OH¯ + 2Co(OH)₃ + Sn ---> 2Co(OH)₂ + HSnO₂¯ + H₂O