Solutions To Problem Set 2
Solutions To Problem Set 2
Solutions To Problem Set 2
You should read Chapters 4, 12, and 13. You should review carefully:
Ch.4: 4.4
Ch.12: 12.1-12.3
Ch.13: 13.2, 13.3, 13.6 and 13.7
1. Oxtoby 4.39: Write a balanced equation to represent the reaction between each of the
following pairs of substances in the presence of water:
(a) hydrogen bromide and calcium hydroxide
2. Tell which of the following are oxidation-reduction reactions. For the redox reactions,
identify the reducing agent, oxidizing agent, the substance being reduced, and the
substance being oxidized.
(a) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Is oxidation-reduction reaction;
Reducing agent: Zn, reducing H+ to H0 (in the form of H2)
Oxidizing agent: H+, oxidizing Zn0 to Zn2+
3. Rust stains can be removed by washing a surface with a dilute solution of oxalic acid
(H2C2O4). The reaction is:
Fe2O3(s) + 6H2C2O4(aq) 2Fe(C2O4)33-(aq) + 3H2O(l) + 6H+(aq)
(a) Is this an oxidation-reduction reaction?
No
(b) What mass of rust can be removed by 1.0 L of a 0.14 M solution of oxalic acid?
(6 mol oxalic acid)/(1 mol rust) = (0.14 mol oxalic acid)/(X mol rust)
=> X mol rust = 0.023
4. Oxtoby 12.19: Acidified water can be electrolyzed to produced hydrogen and oxygen.
Write equations for the half-reactions that occur at the anode and at the cathode, and
verify that they combine correctly to give the overall decomposition of water to its
elements.
2H2O 2H2 + O2
5. Oxtoby 12.34: A galvanic cell uses a cadmium cathode immersed in a CdSO4 solution and
a zinc anode immersed in a ZnSO4 solution.
(a) Write a balanced equation for the cell reaction.
(b) A steady current of 1.45 A is observed to flow for a period of 2.60 h. How much
charge passes through the circuit during this time? How many moles of electrons
is this charge equivalent to?
(0.141 mol e-) (1 mol Cd/2 mol e-) (112.411 g/1 mol Cd) = 7.91 g increase
in Cd
1:1 molar ratio between the oxygen and copper generated on an atom to atom basis
implies the same amount of electrons required to convert both ions. Since oxygen
in solution has oxidation number of -2, the oxidation number of Cu ions in solution
must be +2.
7. Oxtoby 12.78: Tarnish forms on silver spoons when airborne sulfur compounds react with
the silver to produce a dark coating of silver sulfide (Ag2S). Tarnish can be removed by
putting the corroded spoons in an aluminum container (a pie plate works) containing a
warm solution of baking soda (sodium hydrogen carbonate, NaHCO3). The NaHCO3(aq)
does not react but serves only as an electrolyte in an electrochemical cell in which the
aluminum container is the anode. Write a balanced chemical equation for the tarnish-
removing reaction.
Since the aluminum container is the anode, aluminum is oxidized and silver is
reduced.
3Ag2S(s) + 2Al(s) 6Ag(s) + 2Al3+(aq) + 3S2-(aq)
8. Oxtoby 13.10: A galvanic cell is constructed by connecting a standard Pt|Fe2+, Fe3+ half-
cell to a standard Cd2+|Cd cathode half-cell.
(a) Write balanced chemical equations for the half-reactions at the anode and the
cathode and for the overall cell reaction.
9. Oxtoby 13.20: Many bleaches, including chlorine and its oxides, oxidize dyes in cloth,
destroying their color. Predict which of the following is the strongest bleach at a given
concentration and pH 0: NaClO3(aq), NaClO(aq), Cl2(aq). How does the strongest
chlorine-containing bleach compare in strength with ozone (O3(g))?
Considering the standard reduction potenetials for the anions, NaClO is strongest
bleach, followed by NaClO3 and Cl2. NaClO, however, is not as strong as O3
because its reduction potential is lower.
10. Oxtoby 13.24: Extending Figure 13-5, the following additional reduction potentials are
measured at pH 14:
Since two moles of electrons are release1d per one mole of HgO that is reduced,
0.0046 mol electrons are released.
12. Oxtoby 13.73: Estimate the cost of the electrical energy needed to produce 2.1 x 1010 kg (a
year’s supply for the world) of aluminum from Al2O3(s) if electrical energy costs 10 cents
per kilowatt-hour (1kW-h = 3.6 MJ = 3.6 x 106 J) and if the cell voltage is 5 V.
First, let’s calculate the number of aluminum atoms in a year’s supply of Al.
1000 g 1 mol Al
(2.1× 1010 kg) × ( )×( ) = 7.78 × 1011 mol Al
1 kg 26.982 g
The production of aluminum from alumina is a three electron process:
Al3+ + 3e − → Al 0
With this information, we can calculate the number of amount of charge transfer
required to produce the Al.
3 e- 96485.342 C
(7.78 × 1011 mol Al) × ( )×( −
) = 2.25 × 1017 C
mol Al mol e
We know that the electrical work performed is equal to the product of the charge
transferred and the cell voltage. Welec = Q ⋅ ∆E
(2.25 × 1017 C) × (5 V) = 1.13 × 1018 J
The electric company charges for this work. Let’s calculate our total bill.
1 kW - h $ 0 .1
(1.13 × 1018 J) × ( 6
)×( ) = $3.13 × 1010
3.6 × 10 J kW − h