Several interesting redox reactions to balance

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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:

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

2) Equalize electrons:

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

3) Add:

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

Why are the 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.


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