Chapter 18: Electrochemistry
2. Refer to Sections 4.4 and 18.1.
Remember that the oxidation is always shown on the left side of the cell notation. For part
(c), note that the Pt serves as an inert electrode and is not part of the chemical equation.
a.
Ag(s) → Ag+(aq) + eSn4+(aq) + 2e- → Sn2+(aq)
oxidation half-reaction
reduction half-reaction
Multiply the oxidation half-reaction by two, then add the two half-reactions together.
2Ag(s) → 2Ag+(aq) +2eSn4+(aq) + 2e- → Sn2+(aq)
2Ag(s) + Sn4+(aq) + 2e- → 2Ag+(aq) +2e- + Sn2+(aq)
2Ag(s) + Sn4+(aq) → 2Ag+(aq) + Sn2+(aq)
b.
Al(s) → Al3+(aq) + 3eCu2+(aq) + 2e- → Cu(s)
oxidation half-reaction
reduction half-reaction
Multiply the oxidation half-reaction by two and the reduction half-reaction by three, then
add the two half-reactions together.
2Al(s) → 2Al3+(aq) + 6e3Cu2+(aq) + 6e- → 3Cu(s)
2Al(s) + 3Cu2+(aq) + 6e- → 2Al3+(aq) + 6e- + 3Cu(s)
2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)
c.
Note: The cell notation should be:
Pt | Fe2+, Fe3+ || MnO4-, H+ | Mn2+ | Pt
Fe2+(aq) → Fe3+(aq) + eMnO4-(aq) + 5e- → Mn2+(aq)
MnO4-(aq) + 5e- + 8H+(aq) → Mn2+(aq)
MnO4-(aq) + 5e- + 8H+(aq) → Mn2+(aq) + 4H2O(l)
oxidation half-reaction
reduction half-reaction
Balance electrons by multiplying the oxidation equation by 5, then add the two halfreactions together.
5Fe2+(aq) → 5Fe3+(aq) + 5eMnO4-(aq) + 5e- + 8H+(aq) → Mn2+(aq) + 4H2O(l)
5Fe2+(aq) + MnO4-(aq) + 5e- + 8H+(aq) → 5Fe3+(aq) + 5e- + Mn2+(aq) + 4H2O(l)
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
18. Refer to Section 18.2, Table 18.1, and Example 18.3.
Write the half-reactions and the associated redox potentials. When combining the half-reactions,
add the potentials to calculate E°. Bear in mind that multiplying a half-reaction by a factor
does not change that E° value.
a.
2[Cr2+(aq) → Cr3+(aq) + e-]
Sn4+(aq) + 2e- → Sn2+(aq)
Sn4+(aq) + 2e- + 2Cr2+(aq) → 2Cr3+(aq) + 2e- + Sn2+(aq)
Sn4+(aq) + 2Cr2+(aq) → 2Cr3+(aq) + Sn2+(aq)
b.
Mn2+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + 2eH2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)
Mn2+(aq) + 2H2O(l) + H2O2(aq) + 2H+(aq) + 2e→ 2H2O(l) + MnO2(s) + 4H+(aq) + 2e2+
Mn (aq) + H2O2(aq) → MnO2(s) + 2H+(aq)
o
E ox
= 0.408 V
o
E red
= 0.154 V
E° = 0.562 V
o
E ox
= -1.229 V
o
E red
= 1.763 V
E° = 0.534 V
38. Refer to Section 18.3, Table 18.1, and Example 18.5.
Note that in the reaction given, OH- is a product, thus the solution is basic.
a. 3[S2-(aq) → S(s) + 2e-]
o
E ox
= 0.445 V
o
E red
= -0.140 V
2[NO3-(aq) + 2H2O(l) + 3e- → NO(g) + 4OH-(aq)]
23S (aq) + 2NO3 (aq) + 4H2O(l) + 6e → 3S(s) + 6e + 2 NO(g) + 8OH (aq)
3S2-(aq) + 2NO3-(aq) + 4H2O(l) → 3S(s) + 2 NO(g) + 8OH-(aq) E° = 0.305 V
b. ∆G° = -nFE°
∆G° = -(6 mol.)(9.648 x 104 J/mol.·V)( 0.305 V)
∆G° = -1.77 x 105 J = -1.77 x 102 kJ
c.
E° =
0.0257 V
0.0257 V
ln K ⇒ 0.305 V =
ln K
n
6
lnK = 71.2
K = e71.2 = 8.4 x 1030
Note: the answer is different if you use the equation
∆G° = -RTlnK:
-1.77 x 105 J = -(8.31 J/ mol.·K)(298K)lnK
71.4 = lnK
K = e71.4 = 1.0 x 1031
62. Refer to Sections 18.5, 18.6, and Example 18.8.
Use the equation on page 538 for the reaction of a lead storage battery to determine the moles of
e- per mole of Pb.
453.6 g
1 mol.
2 mol. e −
a. 12.0 lb Ca ×
×
×
= 272 mol. e −
1 lb
40.078 g Ca 1 mol. Ca
272 mol. e − ×
3.2 V =
9.648 ×10 4 C
= 2.62 ×107 C
1 mol. e −
J
2.62 × 107 C
J = (3.2 V)(2.62 × 107 C) = 8.4 × 107 J
b. $ 0.09 = 1 kWh = 3.600 x 106 J
8.4 × 107 J ×
1 kWh
$ 0.09
×
= $ 2.1
6
3.600 × 10 J 1 kWh