256 ❯ STEP 4. Review the Knowledge You Need to Score High
- C—The impure silver must be oxidized so it will
go into solution. Oxidation occurs at the anode.
Reduction is required to convert the silver ions to
pure silver. Reduction occurs at the cathode. The
cathode must be pure silver; otherwise, it could
be contaminated with the cathode material. - B—If the voltage was not equal to E°, then the
cell was not standard. Standard cells have 1 M
concentrations and operate at 25°C with a par-
tial pressure of each gas equal to 1 atm. No gases
are involved in this reaction, so the cell must be
operating at a different temperature or a differ-
ent concentration (or both). - A—The calculation is
×
2.00 C
s3,600 s
h
(3.00 h)1 F
96,500 C
1 mol Hg
2 F200.6 g Hg
1 mol HgYou can estimate the answer by replacing 96,500
with 100,000; 3,600 × 3 with 10,000; and
200.6 with 200, which gives
2.00 C
s10,000 s
h1 F
10 0,000 C
1 mol Hg
2 F
200 g Hg
1 mol Hg1
10
200 gHg
1 20 g
×
=
≈
(which is slightly low owing to the rounding).
- A—The gases produced are hydrogen (at the
cathode) and chlorine (at the anode). Hydrogen is
odorless, whereas chlorine has a distinctive odor. - D—The bromate ion, BrO 3 - , is gaining elec-
trons, so it is being reduced. Reduction always
occurs at the cathode. - B—The half-reactions giving the overall reac-
tion must be as follows:
3 [Zn^2 +(aq) + 2 e- → Zn(s)] E° = –0.76 V (given)
2 [M(s) → M^3 +(aq) + 3 e-] E° =?
2 M(s) + 3 Zn^2 +(aq) → 2 M^3 +(aq) + 2 Zn(s)
E° = 0.90 V (given)
Thus, –0.76 +? = 0.90, giving? = 1.66 V. The
half-reaction under consideration is the reverse
of the one used in this combination, so the sign
of the calculated voltage must be reversed. Do
not make the mistake of multiplying the volt-
ages when the half-reactions were multiplied to
equalize the electrons.- B—This is a reduction half-reaction. Reductions
take place at the cathode. - C—The only gases in the table are hydrogen,
sulfur dioxide, and chlorine. Of the three gases,
the only odorless gas is hydrogen. The formation
of hydrogen is through reduction; therefore, it is
necessary to determine the oxidation half-reac-
tion. Sulfur dioxide forms through reduction, so
it cannot be the gas. The gas with the distinc-
tive odor must be chlorine, which is produced
through oxidation. - B—Based on the reduction potentials in the
table, tin is a sufficiently strong reducing agent
to reduce the dithionate ion to sulfur dioxide, a
gas with a distinctive odor. Tin cannot reduce
water to produce hydrogen, and tin cannot oxi-
dize chloride ion to produce chlorine. None of
the other substances is a gas. - C—Using the cell potentials, the copper(II) ions
will undergo reduction (use the half–reaction
from the table), and the tin metal will undergo
oxidation (reverse the half–reaction from the
table). Combine the resultant half–reactions and
cancel electrons to get this answer. A is not a net
ionic equation. B is the reverse reaction. D is non-
spontaneous (according to the cell potentials).