Homework 6
14. a. H(+1), O(-2), N(+5)
b. Cl(-1), Cu(+2)
c. O(0)
d. H(+1), O(-1)
e. Mg(+2), O(-2), S(+6)
f. Ag(0)
g. Pb(+2), O(-2), S(+6)
h. O(-2), Pb(+4)
i. Na(+1), O(-2), C(+3)
j. O(-2), C(+4)
k. (NH4)2Ce(SO4)3 contains NH4+ ions and SO42- ions. Thus, cerium exists as the
Ce4+ ion. H(+1), N(-3), Ce(+4), S(+6), O(-2)
l. O(-2), Cr(+3)
16. a. Cr → Cr3+ + 3 e-
NO3- → NO
4H + NO3- → NO + 2 H2O
3 e- + 4H+ + NO3- → NO + 2 H2O
+
Cr → Cr3+ + 3e3 e- + 4H+ + NO3- → NO + 2 H2O
+
4 H (aq) + NO3-(aq) + Cr(s) → Cr3+(aq) + NO(g) + 2H2O(l)
b. (Al → Al3+ + 3 e-)x5
MnO4- → Mn2+
8H+ + MnO4- → Mn2+ + 4H2O
(5 e- + 8H+ + MnO4- → Mn2+ + 4H2O)x3
5Al → 5Al3+ + 15e15e- + 24H+ + 3MnO4- → 3Mn2+ + 12H2O
24H+ + 3MnO4- + 5Al(s) → 5Al3+ + 3Mn2+ + 12H2O
c. (Ce4+ + e → Ce3+ -)x6
CH3OH → CO2
H2O + CH3OH → CO2 + 6H+
H2O + CH3OH → CO2 + 6H+ + 6e-
6Ce4+ + 6e → 6Ce3+
H2O + CH3OH → CO2 + 6H+ + 6eH2O + 6Ce4+ + CH3OH → CO2 + 6H+ + 6Ce3+
d.
PO33- → PO43(H2O + PO33- → PO43- + 2H+ + 2e-)x3
MnO4- → MnO2
(3 e- + 4H+ + MnO4- → MnO2 + 2H2O)x2
3H2O + 3PO33- → 3PO43- + 6H+ + 6e6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
3PO33- + 2H+ + 2MnO4- → 3PO43- + 2MnO2 + H2O
Now convert to a basic solution by adding 2 OH- to both sides. 2 H+ + 2OH- → 2H2O
on the reactant side. After concerting H+ to OH-, simplify the overall equation by
crossing off 1 H2O on each side of the reaction. The overall balanced equation is:
H2O(l) + 3PO33-(aq) + 2MnO4-(aq) → 3PO43-(aq) + 2MnO2(aq) + 2OH-(aq)
OCl- → Cl2e- + 2H+ + OCl- → Cl- + H2O
e. Mg → Mg(OH)2
2H2O + Mg → Mg(OH)2 + 2 H+ + 2e-
2H2O + Mg → Mg(OH)2 + 2 H+ + 2e2e- + 2H+ + OCl- → Cl- + H2O
H2O(l) + OCl-(aq) + Mg(s) → Mg(OH)2(aq) + Cl-(aq)
The final overall reaction does not contain H+, so we are done.
f.
H2CO → HCO32H2O + H2CO → HCO3- + 5H+ + 4e-
Ag(NH3)2+ → Ag + 2NH3
(e- + Ag(NH3)2+ → Ag + 2NH3)x4
2H2O + H2CO → HCO3- + 5H+ + 4e4e- + 4Ag(NH3)2+ → 4Ag + 8NH3
2H2O + H2CO + 4Ag(NH3)2+ → 4Ag + 8NH3 + HCO3- + 5H+
Convert to a basic solution by adding 5 OH- to both sides (5H+ + 5OH- → 5H2O).
Then, cross off 2 H2O on both sides, which gives the overall balanced equation:
5OH-(aq) + H2CO(aq) + Ag(NH3)2+(aq) → Ag(s) + 2NH3(aq) + HCO3-(aq) + 3H2O(l)
30. a. H2O2 + 2H+ + 2e- → 2H2O
H2O2 → O2 + 2H+ +2e2 H2O2(aq) → 2H2O(l) + O2(g)
E° = 1.78 V
-E° = -0.68 V
E°cell = 1.10 V
Cathode: Pt electrode; H2O2 and H+ in solution
Anode: Pt electrode; O2(g) bubbled in, H2O2 and H+ in solution
(Fe3+ + 3e- → Fe) x 2
(Mn → Mn2+ + 2e-) x 3
2 Fe3+(aq) + 3 Mn(s) → 2Fe(s) + 3Mn2+(aq)
b.
E° = -0.036 V
-E° = 1.18 V
E°cell = 1.14 V
E° = 0.96 V
(4H+ + NO3- + 3e- → NO + 2H2O)x2
2+
(Mn → Mn + 2e )x3
-E° = 1.18 V
+
2+
3Mn(s) + 8H (aq) + 2NO3 (aq) → 2NO(g) + 4H2O(l) + 3Mn (aq)
E°cell = 2.14 V
+
E° = 1.60 V
(2e + 2H + IO4 → IO3 + H2O)x5
(Mn2+ + 4H2O → MnO4- + 8H+ + 5e-)x2
-E° = -1.51 V
5IO4-(aq) + 2Mn2+(aq) + 3H2O(l) → 5IO3-(aq) + 2MnO4-(aq) + 6H+(aq) E°cell = 0.09 V
38. a.
54. As is the case for all concentration cells, E°cell = o, and the smaller ion concentration
is always in the anode compartment. The general Nernst equation for the
Ni | Ni2+(xM) | | Ni2+(y M) | Ni concentration cell is:
Ecell = E°cell – 0.0591/n log Q = -0.0591/2 log [Ni2+]anode/[Ni2+]cathode
a. Since both compartments are at standard conditions ([Ni2+] = 1.0 M), then Ecell =
E°cell = 0 V. No electron flow occurs.
b. Cathode = 2.0 M Ni2+; Anode = 1.0 M Ni2+; Electron flow is always from the
anode to the cathode, so electrons flow to the right in the diagram.
Ecell = = -0.0591/2 log [Ni2+]anode/[Ni2+]cathode = -0.0591/2 log 1.0/2.0 = 8.9•10-3 V
c. Cathode = 1.0 M Ni2+; Anode = 0.10 M Ni2+; Electrons flow to the left in the
diagram.
Ecell = = -0.0591/2 log [Ni2+]anode/[Ni2+]cathode = -0.0591/2 log 0.10/1.0 = 0.030 V
d. Cathode = 1.0 M Ni2+; Anode = 4.0•10-5 M Ni2+; Electrons flow to the left in the
diagram.
Ecell = = -0.0591/2 log [Ni2+]anode/[Ni2+]cathode = -0.0591/2 log 4.0•10-5/1.0 = 0.13 V
e. Since both concentrations are equal, log(2.5/2.5) = log 1.0 = 0 and Ecell = 0. No
electron flow occurs.
60. a. Cu+ + e- → Cu
Cu+ → Cu2+ + e2Cu+(aq) → Cu2+(aq) + Cu(s)
E° = 0.52 V
-E° = -0.16 V
E°cell = 0.36 V; Spontaneous
∆G° = -nFE°cell = -(1mol e-)(96,485 C/mol e-)(0.36 J/C) = -34,700 J = -35 kJ
E°cell = 0.0591/n log K, log K = nE°/0.0591 = 1(0.36)/0.0591 = 6.09, K = 106.09 =
1.2•106
b. Fe2+ + 2e- → Fe
(Fe2+ → Fe3+ + e-)x2
3Fe2+(aq) → 2Fe3+(aq) + Fe(s)
E° = -0.44 V
-E° = -0.77 V
E°cell = -1.21 V; Not spontaneous
c. HClO2 + 2H+ + 2e- → HClO + H2O
E° = 1.65 V
+
HClO2 + H2O → ClO3 + 3H + 2e
-E° = -1.21 V
+
2HClO2(aq) → ClO3(aq) + H (aq) + HClO(aq) E°cell = 0.44 V; Spontaneous
∆G° = -nFE°cell = -(2mol e-)(96,485 C/mol e-)(0.44 J/C) = -84,900 J = -85 kJ
E°cell = 0.0591/n log K, log K = nE°/0.0591 = 2(0.44)/0.0591 = 14.89, K = 1014.89 =
7.8•1014
74. The oxidation state of bismuth in BiO+ is +3 because oxygen has a -2 oxidation state
in this ion. Therefore, 3 moles of electrons are required to reduce the bismuth in
BiO+ to Bi(s).
10.0 g Bi x 1 mol Bi/209.0 g Bi x 3 mol e-/mol Bi x 94,485 C/mol e- x 1 s/25.0 C =
554 s = 9.23 min
80. The half-reactions for the electrolysis of water are:
(2 e- + 2 H2O → H2 + 2 OH-) x 2
2 H2O → 4 H+ + O2 + 4 e2 H2O(l) → 2 H2(g) + O2(g)
Note: 4H+ + 4 OH- → 4 H2O and n
= 4 for this reaction s it is written.
15.0 min x 60 s/min x 2.50 C/s x 1 mol e-/94,485 C x 2 mol H2/4 mol e- = 1.17•10-2
mol H2.
At STP, 1 mol of an ideal gas occupies a volume of 22.42 L (see Chapter 5 of the
text).
1.17•10-2 mol H2 x 22.42 L/mol H2 = 0.262 L = 262 mL H2.
1.17•10-2 mol H2 x 1 mol O2/2 mol H2 x 22.42 L/mol O2 = 0.131 L = 131 mL O2.
106. a. We can calculate ∆G° from ∆G° = ∆H° - T ∆S° and then E° from ∆G° = -nFE°;
or we can use the equation derived in Exercise 17.105. For this reaction, n = 2 (from
oxidation states).
[(253 K)(263.5 J/K) + 315.9•103 J]
E°-20 = (T ∆S° - ∆H°)/nF = ———————————————
(2 mol e-)(96,485 C/mol e-)
= 1.98 J/C = 1.98 V
b. E°-20 = E°-20 – RT/nF ln Q = 1.98 V = RT/nF (1/[H+]2[HSO4-]2
E°-20 = 1.98 V – (8.3145 J/K•mol)(253 K)/(1 mol e-)(96,485 C/mol e-)
ln(1/(4.5)2(4.5)2 = 1.98 V + 0.066 V = 2.05 V
c. From Exercise 17.56, E = 2.12 V at 25 °C. As the temperature decreases, the cell
potential decreases. Also, oil more viscous at lower temperature, which adds to the
difficulty of starting an engine on a cold day. The combination of these two factors
results in batteries failing more often on cold days than on warm days.