AP Chemistry Chapter 4 Answers – Zumdahl
4.67
Apply the rules in Table 4.2.
a.
KMnO4 is composed of K+ and MnO4- ions. Assign oxygen a value of -2, which gives
manganese a +7 oxidation state since the sum of oxidation states for all atoms in MnO4must equal the -1 charge on MnO4-. K, +1; O, -2; Mn, +7.
b.
Assign O a -2 oxidation state, which gives nickel a +4 oxidation state. Ni, +4; O, -2.
c.
Na4Fe(OH)6 is composed of Na+ cations and Fe(OH)64+ anions. Fe(OH)64- is composed
of an iron cation and 6 OH- anions. For an overall anion charge of -4, iron must have a
+2 oxidation state. As is usually the case in compounds, assign O a -2 oxidation state and
H a +1 oxidation state. Na, +1; Fe, +2; O, -2; H, +1.
d.
(NH4)2HPO4 is made of NH4+ cations and HPO42- anions. Assign +1 as the oxidation
state of H and -2 as the oxidation state of O. In NH4+, x + 4(+1) = +1, x = -3 = oxidation
state of N. In HPO42-, +1 + y + 4(-2) = -2, y = +5 = oxidation state of P.
e.
O, -2; P, +3
f.
O, -2; Fe, +8/3
g.
O, -2; F, -1; Xe, +6
h.
F, -1; S, +4
i.
O, -2; C, +2
j.
H, +1; O, -2; C, 0
4.71
To determine if the reaction is an oxidation-reduction reaction, assign oxidation numbers. If the
oxidation numbers change for some elements, then the reaction is a redox reaction. If the
oxidation numbers do not change, then the reaction is not a redox reaction. In redox reactions,
the species oxidized (called the reducing agent) shows an increase in oxidation numbers and the
species reduced (called the oxidizing agent) shows a decrease in oxidation numbers.
a.
b.
c.
d.
e.
Redox?
Oxidizing
Agent___
Reducing
Agent___
Substance
Oxidized__
Substance
Reduced__
Yes
No
No
Yes
No
Ag+
Cu
Cu
Ag+
SiCl4
Mg
Mg
SiCl4 (Si)
4.73
Use the method of half-reactions described in Section 4.10 of the text to balance these redox
reactions. The first step always is to separate the reaction into the two half-reactions, then
balance each half-reaction separately.
a.
Zn → Zn2+ + 2e-
2e- + 2HCl → H2 + 2Cl-
Adding the two balanced half-reactions:
Zn(s) + 2HCl(aq) → H2(g) + Zn2+(aq) + 2Cl-(aq)
b.
3I- → I3- + 2e-
ClO- → Cl2e- + 2H+ + ClO- → Cl- + H2O
Adding the two balanced half-reactions so electrons cancel:
3I-(aq) + 2H+(aq) + ClO-(aq) → I3-(aq) + Cl-(aq) + H2O(l)
c.
As2O3 → H3AsO4
NO3- → NO + 2H2O
As2O3 → 2H3AsO4
4H+ + NO3- → NO + 2H2O
Left 3 – O; Right 8 – O
(3e- + 4H+ + NO3- → NO + 2H2O) x 4
Right hand side has 5 extra O.
Balance the oxygen atoms first using H2O, then balance H using H+, and finally balance
charge using electrons. This gives:
(5H2O + As2O3 → 2H3AsO4 + 4H+ + 4e-) x 3
Common factor is a transfer of 12e-. Add half-reactions so electrons cancel.
12e- + 16H+ + 4NO3- → 4NO + 8H2O
15H2O + 3As2O3 → 6H3AsO4 + 12H+ + 12e_____________________________________________________________________
7H2O(l) + 4H+(aq) + 3As2O3(s) + 4NO3-(aq) → 4NO(g) + 6H3AsO4(aq)
d.
MnO4- → Mn2+ + 4H2O
(5e- + 8H+ + MnO4- → Mn2+ + 4H2O) x 2
(2Br → Br2 + 2e-) x 5
Common factor is a transfer of 10e-.
10Br- → 5Br2 + 10e10e + 16H + 2MnO4- → 2Mn2+ + 8H2O
______________________________________________________________
16H+ (aq) + 2MnO4-(aq) + 10Br-(aq) → 5Br2(l) + 2Mn2+(aq) + 8H2O(l)
-
e.
+
CH3OH → CH2O
(CH3OH → CH2O + 2H+ + 2e-) x 3
Cr2O72- → 2Cr3+
14H+ + Cr2O72- → 2Cr3+ + 7H2O
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
Common factor is a transfer of 6e-.
3CH3OH → 3CH2O + 6H+ + 6e6e + 14H + Cr2O72- → 2Cr3+ + 7H2O
___________________________________________________________
8H+(aq) + 3CH3OH(aq) + Cr2O72-(aq) → 2Cr3+ (aq) + 3CH2O(aq) + 7H2O(l)
-
4.75
+
Use the same method as with acidic solutions. After the final balanced equation, convert H+ to
OH- as described in section 4.10 of the text. The extra step involves converting H+ into H2O by
adding equal moles of OH- to each side of the equation. This converts the reaction to a basic
solution while still keeping it balanced.
a.
Al → Al(OH)44H2O + Al → Al(OH)4- + 4H+
4H2O + Al → Al(OH)4- + 4H+ + 3e-
MnO4- → MnO2
3e- + 4H+ + MnO4- → MnO2 + 2H2O
4H2O + Al → Al(OH)4- + 4H+ + 3e3e + 4H+ + MnO4- → MnO2 + 2H2O
__________________________________________________________
2H2O(l) + Al(s) + MnO4-(aq) → Al(OH)4- (aq) + MnO2(s)
-
H+ doesn’t appear in the final balanced reaction, so we are done.
b.
Cl2 → Cl2e- + Cl2 → 2Cl-
Cl2 → OCl2H2O + Cl2 → 2OCl- + 4H+ + 2e-
2e- + Cl2 → 2Cl2H2O + Cl2 → 2OCl- + 4H+ + 2e____________________________________
2H2O + 2Cl2 → 2Cl- + 2OCl- + 4H+
Now convert to a basic solution. Add 4 OH- to both sides of the equation. The 4 OHwill react with the 4H+ on the product side to give 4 H2O. After this step, cancel identical
species on both sides (2 H2O). Applying these steps gives: 4OH- + 2Cl2 →2Cl- + 2OCl+ 2H2O, which can be further simplified to:
2OH-(aq) + Cl2(g) → Cl-(aq) + OCl-(aq) + H2O(l)
c.
NO2- → NH3
6e- + 7H+ + NO2- → NH3 + 2H2O
Al → AlO2(2H2O + Al → AlO2- + 4H+ + 3e-) x 2
Common factor is a transfer of 6e-.
6e- + 7H+ + NO2- → NH3 + 2H2O
4H2O + 2Al → 2AlO2- + 8H+ + 6e___________________________________________________________
OH- + 2H2O + NO2- + 2Al → NH3 + 2AlO2- + H+ + OHReducing gives: OH-(aq) + H2O(l) + NO2-(aq) + 2Al(s) → NH3(g) + 2AlO2-(aq)