1. No category

Electrochemistry
1) a. The cell will not produce a current because the circuit is not complete. A salt
bridge is needed. This provides a means for ions to flow thus completing the
circuit and maintaining charge neutrality in the cells.
b.
Cu2+(aq) + 2 e−  Cu(s)
E°½ = 0.337 V
3+
−
Al(s)  Al (aq) + 3 e
E°½ = 1.66V
c. The copper electrode is the cathode. The aluminum electrode is the anode.
d. Current flows from the anode (where electrons are produced) to the cathode
where electrons are consumed. Electrons flow from the Al electrode to the Cu
one.
e. The anions flow from the salt bridge to the anode. The cations flow from the
salt bridge to the cathode compartment.
f. 3 Cu2+(aq) + 2 Al(s)  3 Cu(s) + 2 Al3+(aq)
There are 6 electrons transferred.
g. E°cell = 1.997 V
2) A
3) A
4) a. 0.294 V
b. 1.051 V
c. 1.211 V
5) 1.43 V
6) a. 3 I3− (aq) + 2 Au(s) + 8 Br−(aq)  9 I− (aq) + 2 AuBr4−(aq)
E°cell = 1.378
V
b. Ni(s) + Sn2+(aq)  Ni2+(aq) + Sn(s)
E°cell = 0.144 V
7) B
8) C
9) A
10) a. Ag+(aq) + 2 e−  Ag(s)
E° = 0.799 V
Cr(s)  Cr3+(aq) + 3 e−
E° = 0.74 V
+
3 Ag (aq) + Cr(s)  3 Ag(s) + Cr3+(aq)
b. The silver electrode is the cathode (where reduction occurs). The chromium
electrode is the anode (where oxidation occurs).
c. Ag will be deposited on the cathode so the silver anode will increase in mass.
The chromium anode will dissolve, form ions, and decrease in mass.
d. E°cell = 0.799 V + 0.74 V = 1.539 V
11) a. Al
b. NO3(aq)
c. I2(s)
12) B
13) B
14) D
15) a. Mn is +7, Br is +5, Cr is +6
b. Reduction potentials:
I. MnO4(aq) + 8 H+(aq) + 5 e−  Mn2+(aq) + 4 H2O() E°red = 1.51 V
II. BrO3(aq) + 12 H+(aq) + 10 e−  Br2() + 6 H2O() E°red = 1.52 V
III. Cr2O72(aq) + 14 H+(aq) + 6 e−  2 Cr3+(aq) + 7 H2O()
E°red = 1.33 V
c. Atoms in high oxidations states are easily reduced.
16) ΔG = 2.70 kJ mol1
Keq = 2.98
17) ΔG = 24.3 kJ mol1
Keq = 5.46  105
18) D
19) B
20) B
21) a. Ecell will increase.
b. No change in Ecell.
c. Ecell will decrease.
22) a. E°cell=1.076V
b. E cell = 1.068 V
c. E cell = E°cell = 1.076 V
d. E cell = 0.829 V
23) A
24) E
25) B
26) A
27) B
28) A
29) C
30) B
31) 239 g
32) a. Li(s) is the anode; Ag(s) is the cathode.
b. Ecell at 37 C will be 3.5 V because all of the reactants and products are solids
(Q = 1).
33) a. The oxidation reactions and potentials for Mg and Fe are:
Mg(s)  Mg2+(aq) + 2 e−
E° = 2.37 V
Fe(s)  Fe2+(aq) + 2 e−
E° = 0.440 V
This works because Mg has a higher oxidation potential. The overall reaction and
cell potential would be:
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s) E° = 1.93 V
b. The oxidation reactions and potentials for Mg and Pb are:
Mg(s)  Mg2+(aq) + 2 e−
E° = 2.37 V
Pb(s)  Pb2+(aq) + 2 e−
E° = 0.126 V
This works because Mg has a higher oxidation potential. The overall reaction and
cell potential would be:
Mg(s) + Pb2+(aq)  Mg2+(aq) + Pb(s) E° = 2.24 V
34) No, it will not provide a long-term solution. Steel will promote the further corrosion
of brass because Zn has a higher oxidation cell potential than Fe.
35) B