Chapter 21: Chemistry of the Nonmetals
18. Refer to Sections 21.1 and 21.4 and Chapter 4.
First write the half-reactions, then combine them and cancel common elements.
a. In this reaction, one HClO molecule oxidizes a second HClO molecule.
Cl+ → Cl3+ + 2eHClO(aq) → HClO2(aq) + 2eHClO(aq) → HClO2(aq) + 2e- + 2H+(aq)
H2O(l) + HClO(aq) → HClO2(aq) + 2e- + 2H+(aq)
oxidation half-reaction
Cl+ + e- → Cl0
2HClO(aq) + 2e- → Cl2(g)
2H+(aq) + 2HClO(aq) + 2e- → Cl2(g)
2H+(aq) + 2HClO(aq) + 2e- → Cl2(g) + 2H2O(l)
reduction half-reaction
2H+(aq) + 2HClO(aq) + 2e- + H2O(l) + HClO(aq)
→ HClO2(aq) + 2e- + 2H+(aq) + Cl2(g) + 2H2O(l)
3HClO(aq) → HClO2(aq) + Cl2(g) + H2O(l)
(Note: this equation could also be written with the anions ClO- and ClO2- instead of the acids.
However, since these are weak acids, they exist in solution mainly as the acid.)
b. Cl5+ → Cl7+ + 2eClO3-(aq) → ClO4-(aq) + 2eClO3-(aq) → ClO4-(aq) + 2e- + 2H+(aq)
H2O(l) + ClO3-(aq) → ClO4-(aq) + 2e- + 2H+(aq)
oxidation half-reaction
Cl5+ + 2e- → Cl3+
ClO3-(aq) + 2e- → ClO2-(aq)
2H+(aq) + ClO3-(aq) + 2e- → ClO2-(aq)
2H+(aq) + ClO3-(aq) + 2e- → ClO2-(aq) + H2O(l)
reduction half-reaction
2H+(aq) + ClO3-(aq) + 2e- + H2O(l) + ClO3-(aq)
→ ClO4-(aq) + 2e- + 2H+(aq) + ClO2-(aq) + H2O(l)
2ClO3-(aq) → ClO4-(aq) + ClO2-(aq)
28. Refer to Section 21.3 and Chapter 7.
a.
Cl
O
Cl
P
c.
P
P
b.
O
N
d.
N
N
N
P
46. Refer to Chapter 13.
The equilibrium equation for the final reaction is:
-
K overall =
[H + ][HF2 ]
[HF]2
eq. 1
Write the equilibrium equation for the second reaction, and solve for HF2-.
-
K1 = 2.7 =
[HF2 ]
⇒ [HF2 ] = (2.7)[HF][F - ]
[HF][F ]
Substituting this equation into equation 1 gives:
K overall =
[H + ](2.7)[HF][F - ] [H + ](2.7)[F- ]
=
[HF]2
[HF]
eq. 2
Write the equilibrium equation for the first reaction, and solve for HF.
K a = 6.9 x 10 -4 =
[H + ][F- ]
[H + ][F - ]
⇒ [HF] =
[HF]
6.9 x 10 -4
Substituting this equation into equation 2 gives:
K overall =
[H + ](2.7)[F- ]
= (2.7)(6.9 x 10 -4 ) = 1.9 x 10 -3
[H + ][F- ]
6.9 x 10 -4
Thus we see that the equilibrium constant for the overall process is the product of the
individual constants.
56. Refer to Chapter 18.
Calculate the moles of electrons produced by the current. Then calculate the mass of F2
produced with the given efficiency.
7.00 x 10 3 C/s x
3600 s 24 hr
1 mol.
x
x2dx
= 1.25 x 10 4 mol. e 4
1 hr
1d
9.648 x 10 C
2F- → F2 + 2e1.25 x 10 4 mol. e - x 0.95 x
1 mol. F2 37.996 g F2
x
= 2.26 x 105 g F2
1 mol. F2
2 mol. e -