### Several interesting redox reactions to balance

If you came to this page before studying redox much, I'd ask you to consider coming here after some more study. Please trust me on this point.

Problems 2, 3 and 4 are variations on a theme, involving reactants of somewhat similar character in each.

Problem #1: P2H4 ---> PH3 + P4H2

Solution #1:

1) Break into half-reactions:

P2H4 ---> PH3

P2H4 ---> P4H2

2) Balance in acidic solution:

2e¯ + 2H+ + P2H4 ---> 2PH3

2P2H4 ---> P4H2 + 6H+ + 6e¯

3) Make electrons equal:

6e¯ + 6H+ + 3P2H4 ---> 6PH3

2P2H4 ---> P4H2 + 6H+ + 6e¯

4) Add the two half-reactions together:

5P2H4 ---> 6PH3 + P4H2

Solution #2:

1) remove the hydrogen from each compound and make phosphorous "ions:"

P24¯ ---> P3¯ + P42¯

2) Break into half-reactions:

P24¯ ---> P3¯

P24¯ ---> P42¯

3) Balance each half-reaction:

2e¯ + P24¯ ---> 2 P3¯

2P24¯ ---> P42¯ + 6e¯

4) Make electrons equal:

6e¯ + 3P24¯ ---> 6P3¯

2 P24¯ ---> P42¯ + 6 e¯

5) Add the two half-reactions together:

5P24¯ ---> 6P3¯ + P42¯

6) Add hydrogens back in:

5P2H4 ---> 6PH3 + P4H2

The first method is easy, the second is rather cool (at least I think so).

Problem #2: ClO3¯(aq) + Cl¯(aq) ---> Cl2(g) + ClO2(aq) [acidic sol.]

Before balancing, allow me to point out the oxidation numbers for the Cl in each compound:

ClO3¯ (+5)
Cl¯ (-1)
Cl2 (0)
ClO2 (+4)

Solution #1:

1) Half-reactions:

Cl¯ ---> Cl2
ClO3¯ ---> ClO2

2) Balance them:

2Cl¯ ---> Cl2 + 2e¯
e¯ + 2H+ + ClO3¯ ---> ClO2 + H2O

3) Equalize electrons and combine the half-reactions:

4H+ + 2ClO3¯ + 2Cl¯ ---> Cl2 + 2ClO2 + 2H2O

4) If, perchance, you wanted a full molecular equation:

2HClO3 + 2HCl ---> Cl2 + 2ClO2 + 2H2O

Solution #2:

1) Half-reactions:

ClO3¯ ---> Cl2
Cl¯ ---> ClO2

2) Balance them:

10e¯ + 12H+ + 2ClO3¯ ---> Cl2 + 6H2O
2H2O + Cl¯ ---> ClO2 + 4H+ + 5e¯

3) Equalize electrons and combine the half-reactions:

I will leave you to satisfy yourself that the same net ionic equation results as that in Solution #1.

Problem #3: ClO4¯(aq) + Cl¯(aq) ---> Cl2(g) + ClO¯(aq) [acidic sol.]

Solution #1:

1) Half-reactions:

Cl¯ ---> Cl2
ClO4¯ ---> ClO¯

2) Balance them:

2Cl¯ ---> Cl2 + 2e¯
6e¯ + 6H+ + ClO4¯ ---> ClO¯ + 3H2O

3) Equalize electrons:

6Cl¯ ---> 3Cl2 + 6e¯
6e¯ + 6H+ + ClO4¯ ---> ClO¯ + 3H2O

4) Add:

6HCl + ClO4¯ ---> 3Cl2 + ClO¯ + 3H2O

Solution #2:

1) Half-reactions:

Cl¯ ---> ClO¯
ClO4¯ ---> Cl2

2) Balance them:

H2O + Cl¯ ---> ClO¯ + 2H+ + 2e¯
14e¯ + 16H+ + 2ClO4¯ ---> Cl2 + 8H2O

3) Equalize electrons:

7H2O + 7Cl¯ ---> 7ClO¯ + 14H+ + 14e¯
14e¯ + 16H+ + 2ClO4¯ ---> Cl2 + 8H2O

4) Add:

7Cl¯ + 2ClO4¯ + 2H+ ---> 7ClO¯ + Cl2 + H2O

So, which one is the "correct" answer, the one that actually takes place in the real world? I believe the answer can be demonstrated to be Solution #1 based on Ecell values. However, the ChemTeam has never checked on this point (and probably never will).

Problem #4: ClO4-(aq) + Cl-(aq) ---> ClO3-(aq) + Cl2(g)

Solution #1:

1) Oxidation numbers:

ClO4- = +7
Cl-(aq) = −1

ClO3-(aq) = +5
Cl2(g) = 0

2) Half reactions:

ClO4-(aq) ---> ClO3-(aq)
Cl-(aq) ---> Cl2(g)

3) Balance in acidic solution:

2e- + 2H+ + ClO4-(aq) ---> ClO3-(aq) + H2O
2Cl-(aq) ---> Cl2(g) + 2e-

4) Add:

2H+ + ClO4-(aq) + 2Cl-(aq) ---> ClO3-(aq) + Cl2(g) + H2O

Solution #2:

1) Half reactions:

ClO4-(aq) ---> Cl2(g)
Cl-(aq) ---> ClO3-(aq)

2) Balance:

14e- + 16H+ + 2ClO4-(aq) ---> Cl2(g) + 8H2O
3H2O + Cl-(aq) ---> ClO3-(aq) + 6H+ + 6e-

3) Equalize electrons:

42e- + 48H+ + 6ClO4-(aq) ---> 3Cl2(g) + 24H2O
21H2O + 7Cl-(aq) ---> 7ClO3-(aq) + 42H+ + 42e-

4) Add:

6H+ + 6ClO4-(aq) + 7Cl-(aq) ---> 7ClO3-(aq) + 3Cl2(g) + 3H2O

Why are they different? Frankly, the ChemTeam does not know. Perhaps there is something in taking the two reactions to the molecular state:

NaClO4(aq) + 2HCl(aq) ---> NaClO3(aq) + Cl2(g) + H2O

and

6HClO4(aq) + 7NaCl(aq) ---> 7NaClO3(aq) + 3Cl2(g) + 3H2O

In the first case, hydrochloric acid is attacking sodium perchlorate and in the second case perchloric acid is attacking sodium chloride. Is this difference a realistic thing, enough so that the products are affected?

If someone knows enough chemistry to give a definite answer, I'd sure like to know.

Problem #5: We could balance this chemical reaction:

KMnO4 + H2SO4 + H2S ---> K2SO4 + MnSO4 + H2O + S

in these two ways

2KMnO4 + 2H2SO4 + 2H2S ---> K2SO4 + 2MnSO4 + 4H2O + S

2KMnO4 + 3H2SO4 + 5H2S ---> K2SO4 + 2MnSO4 + 8H2O + 5S

In both cases, they are balanced equations. How is this accomplished?

Solution:

1) Write the net ionic equation:

MnO4¯ + H2S ---> Mn2+ + S

2) Balanced half-reactions:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2S ---> S + 2H+ + 2e¯

3) Equalize elections:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2S ---> 5S + 10H+ + 10e¯

4) Add:

6H+ + 2MnO4¯ + 5H2S ---> 2Mn2+ + 8H2O + 5S

from which results the second balanced equation:

3H2SO4 + 2KMnO4 + 5H2S ---> 2MnSO4 + K2SO4 + 8H2O + 5S

The first balanced equation is arrived at by assuming some sulfide (in the H2S) is oxidized to elemental sulfur AND some sulfide is oxidized to the sulfur in sulfate. The second balanced equation (done just above) was arrived at by assuming all the sulfide went to elemental sulfur and none to sulfate.

1) Three half-reactions (already balanced) are required:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2S ---> S + 2H+ + 2e¯
4H2O + H2S ---> SO42¯ + 10H+ + 8e¯

2) Multiply first half-reaction by two (to balance electrons):

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
H2S ---> S + 2H+ + 2e¯
4H2O + H2S ---> SO42¯ + 10H+ + 8e¯

3) Add:

4H+ + 2MnO4¯ + 2H2S ---> 2Mn2+ + SO42¯ + 4H2O + S

4) Which leads to the first balanced equation:

2KMnO4 + 2H2SO4 + 2H2S ---> K2SO4 + 2MnSO4 + 4H2O + S