Chemistry 2000 (Spring 2008)
Problem Set #6: Redox Reactions and Electrochemistry
Solutions
Answers to Questions in Petrucci (only those w/out answers at the back of the book)
5.77
(a)
3 × [ IBr(aq) + 3 H2O(l) → IO3-(aq) + Br-(aq) + 6 H+(aq) + e- ]
2 × [ BrO3-(aq) + 6 H+(aq) + 6 e- → Br-(aq) + 3 H2O(l) ]
3 IBr(aq) + 3 H2O(l) + 2 BrO3-(aq) → 3 IO3-(aq) + 5 Br-(aq) + 6 H+(aq)
(b)
3 × [ Sn(s) → Sn2+(aq) + 2 e- ]
1 × [ C2H5NO3(aq) + 6 H+(aq) + 6 e- → C2H5OH(aq) + NH2OH(aq) + H2O(l) ]
3 Sn(s) + C2H5NO3(aq) + 6 H+(aq) → 3 Sn2+(aq) + C2H5OH(aq) + NH2OH(aq) + H2O(l)
(c)
3 × [ As2S3(s) + 8 H2O(l) → 2 H3AsO4(aq) + 3 S(s) + 10 H+(aq) + 10 e- ]
10 × [ NO3-(aq) + 4 H+(aq) + 3 e- → NO(g) + 2 H2O(l) ]
3 As2S3(s) + 4 H2O(l) + 10 NO3-(aq) + 10 H+(aq) → 6 H3AsO4-(aq) + 9 S(s) + 10 NO(g)
(d)
1 × [ I2(aq) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 10 e- ]
5 × [ H5IO6(aq) + H+(aq) + 2 e- → IO3-(aq) + 3 H2O(l) ]
I2(aq) + 5 H5IO6(aq) → 7 IO3-(aq) + 9 H2O(l) + 7 H+(aq)
(e)
4 × [ 2 S2F2(g) + 6 H2O(l) → H2S4O6(aq) + 4 HF(aq) + 6 H+(aq) + 6 e- ]
3 × [ 4 S2F2(g) + 8 H+(aq) + 8 e- → S8(s) + 8 HF(aq) ]
20 S2F2(g) + 24 H2O(l) → 4 H2S4O6(aq) + 3 S8(s) + 20 HF(aq)
5.78
(a)
(b)
(c)
(d)
(e)
(f)
2 × [ Fe2S3(s) + 6 OH-(aq) → 2 Fe(OH)3(s) + 3 S(s) + 6 e- ]
3 × [ O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq) ]
2 Fe2S3(s) + 3 O2(g) + 6 H2O(l) → 4 Fe(OH)3(s) + 6 S(s)
3 × [ 4 OH-(aq) → O2(g) + 2 H2O(l) + 4 e- ]
4 × [ O2-(aq) + 2 H2O(l) + 3 e- → 4 OH-(aq) ]
4 O2-(aq) + 2 H2O(l) → 3 O2(g) + 4 OH-(aq)
2 × [ CrI3(s) + 32 OH-(aq) → CrO42-(aq) + 3 IO4-(aq) + 16 H2O(l) + 27 e- ]
27 × [ H2O2(aq) + 2 e- → 2 OH-(aq) ]
2 CrI3(s) + 10 OH-(aq) + 27 H2O2(aq) → 2 CrO42-(aq) + 6 IO4-(aq) + 32 H2O(l)
4 × [ Ag(s) + 2 CN-(aq) → [Ag(CN)2]-(aq) + 1 e- ]
1 × [ O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq) ]
4 Ag(s) + 8 CN-(aq) + O2(g) + 2 H2O(l) → 4 [Ag(CN)2]-(aq) + 4 OH-(aq)
1 × [ B2Cl4(aq) + 8 OH-(aq) → 2 BO2-(aq) + 4 Cl-(aq) + 4 H2O(l) + 2 e- ]
1 × [ 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) ]
B2Cl4(aq) + 6 OH-(aq) → 2 BO2-(aq) + 4 Cl-(aq) + 2 H2O(l) + H2(g)
3 × [ C2H5OH(aq) + 5 OH-(aq) → C2H3O2-(aq) + 4 H2O(l) + 4 e- ]
4 × [ MnO4-(aq) + 2 H2O(l) + 3 e- → MnO2(s) + 4 OH-(aq) ]
3 C2H5OH(aq) + 4 MnO4-(aq) → 3 C2H3O2-(aq) + 4 MnO2(s) + OH-(aq) + 4 H2O(l)
20.16 Al and H2C2O4
In order to reduce Eu3+ to Eu2+, a stronger reducing agent than Eu2+ is required (so that E˚
for the cell will be positive). Looking at the standard reduction potentials, we can see that
Al and H2C2O4 are stronger reducing agents than Eu2+ (because the standard reduction
potentials to make them are more negative therefore the “standard oxidation potentials”
would be more positive, giving a positive cell potential).
20.22
(a)
(b)
(c)
(d)
Fe(s) | Fe2+(aq) || Cl-(aq) | Cl2(g) | Pt(s)
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
Pt(s) | Cu+(aq) , Cu2+(aq) || Cu+(aq) | Cu(s)
Mg(s) | Mg2+(aq) || Br-(aq) | Br2(l) | Pt(s)
20.38 2 Cl-(aq) → Cl2(g) + 2 ePbO2(s) + 4 H+(aq) + 2 e- → Pb2+(aq) + 2 H2O(l)
PbO2(s) + 4 H+(aq) + 2 Cl-(aq) → Pb2+(aq) + 2 H2O(l) + Cl2(g)
(a)
Ecell = +0.189 V
forward reaction is spontaneous
(b)
Ecell = +0.106 V
forward reaction is spontaneous
(c)
Ecell = -0.405 V
reverse reaction is spontaneous
E˚cell = +1.798 V
E˚cell = +1.563 V
E˚cell = +0.361 V
E˚cell = +3.421 V
E˚cell = +0.097 V
20.72 [Fe2+] = 0.37 M
E˚cell = +0.016 V
K = 3.3
(because ∆rG˚ = -νeFE˚ = -RTlnK)
2
at equilibrium (aFe = 1)
K = (aFe2+)(aCr2+)
(aFe) (aCr3+)2
Set up ICE table to find [Fe2+]. (Recall ICE tables from high school equilibrium
calculations. If not, see equilibrium section in text.)
20.73 [Fe2+] = 0.0014 M
E˚cell = +0.029 V
K = 3.1
(because ∆rG˚ = -νeFE˚ = -RTlnK)
at equilibrium (aAg = 1)
K = (aFe3+)(aAg)
(aFe2+) (aAg+)
Set up ICE table to find [Fe2+]. (Recall ICE tables from high school equilibrium
calculations. If not, see equilibrium section in text.) Simplify calculations by assuming that
change in concentration is much less than 2.0 M (so [Ag+] does not significantly change).
Always confirm assumptions once final answer is obtained.
20.82
(a)
E˚cell = +0.985 V
(b)
This electrode doesn’t involve a gas, so there are none of the practical difficulties associated
with handling gases.
(c)
Ksp = 1.7 × 10-10
20.83
(a)
Einitial = -0.400V
(b)
Ecell = -0.323 V
(c)
Ecell = +0.0215 V
(non-standard conditions so use Nernst equation)
(method of successive approximations gives Ecell = -0.306 V)
(answer should be between +0.021 V and +0.022 V)
Additional Practice Problems
1.
Determine the oxidation states for each atom in the following sulfur oxide anions (only one
resonance structure is given).
(a)
peroxodisulfate: S2O82(b) dithionite S2O42(c) thiosulfate S2O32-2
..
.. O
..
-2
..
.. O
+6
S
.. O ..
-2
2.
(a)
(b)
(c)
(d)
(e)
-1
..
..-1
O
.. O
..
..O-2 ..
S
+6
.. O..
-2
-2
..
.
O
...
-2
.. O-2 .. ..O
..
+3 .
. S S .. +3
.. O .. .. O..
.. ..
-2
-2
.. S0 ..
-2
-2
..
..
+4
.
.. O
.. S O
.. .
.. O ..
-2
(f)
Balance the following redox reactions in acidic aqueous solution.
3 UO22+ + Te + 4 H+ → 3 U4+ + TeO42- + 2 H2O
2 PbSO4 + 2 H2O → Pb + PbO2 + 2 SO42- + 4 H+
4 AsH3 + 24 Ag+ + 6 H2O → As4O6 + 24 Ag + 24 H+
2 MnO4– + 5 HCN + 5 I– + 11 H+ → 2 Mn2+ + 5 ICN + 8 H2O
H5IO6 + 7 I– + 7 H+ → 4 I2 + 6 H2O
This is an example of a comproportionation reaction, a reaction in which the same species
(I2) is produced by both the oxidation and reduction half reactions.
3 UO2+ + Cr2O72– + 8 H+ → 3 UO22+ + 2 Cr3+ + 4 H2O
3.
(a)
(b)
Balance the following redox reactions in basic aqueous solution.
3 OCl- + I- → IO3- + 3 ClP4 + 2 H2O + 4 OH- → 2 HPO32- + 2 PH3
(c)
(d)
(e)
(f)
2 Co + 3 OCl + 3 H2O → 2 Co(OH)3 + 3 Cl–
2 As + 6 OH– → 2 AsO33– + 3 H2
[PbO2(OH)]– → Pb + O2 + OH4 Au + 8 CN– + O2 + 2 H2O → 4 [Au(CN)2]– + 4OH(Note: while gold is not oxidized by oxygen in aqueous solution, addition of cyanide makes
this possible. This reaction is heavily used in gold mining operations, especially if the gold
concentration of the ores is low. The “cyanide ponds” that are used for this extraction
process are serious environmental hazards given the huge quantities of such a toxic
substance!)
–
4.
The cell described below develops a reversible emf of -0.97 V at 25 ˚C (assume exact
temperature so infinite sig. fig.).
Pt (s) | H 2(g) (1 bar ) | H +(aq ) (pH = 5.0) || V(2+
aq ) (0.0010 mol /L ) | V(s)
(a)
The half-reactions are
+
H 2(g) → 2H(aq)
+ 2e - E 0 = 0V
2+
V(aq)
+ 2e - → V(s)
E V0 2+ / V
Overall:
2+
+
H 2(g) + V(aq)
→ 2H(aq)
+ V(s) E 0 = E V0 2+ / V
with νe = 2.
The emf generated by the cell under the given conditions is E = -0.97 V. We can calculate
E0 using the Nernst equation:
RT
E = E0 −
lnQ
ν eF
2
⎛
⎞
RT
RT ⎜ (aH + ) ⎟
0
lnQ = E +
ln
E =E+
ν eF
ν e F ⎜ (aH 2 )(aV 2+ )⎟
⎝
⎠
2
8.314472 J K -1mol-1 )(298.15 K) ⎛ (10−5.0 ) ⎞
(
⎟
= −0.97V +
ln⎜
⎜ (1)(0.0010)⎟
2(96485.342 C/mol)
⎝
⎠
= −1.18 V
Since E 0 = E V0 2+ / V , E V0 2+ / V = −1.18 V .
(b)
For the overall reaction from part (a),
∆ rGm0 = −ν e FE 0
= −2(96485.342 C/mol)(−1.18 V) = 227 kJ/mol
However, we can also write ∆ rGm0 in terms of the standard free energies of formation of the
reactants and products:
+
∆ rGm0 = 2∆ f G 0 (H(aq)
)+ ∆ f G 0 (V(s))− ∆ f G 0 (H 2(g))+ ∆ f G 0 (V(aq)2+)
[
]
2+
= −∆ f G 0 (V(aq)
)
2+
so ∆ f G 0 (V(aq)
)= −227 kJ/mol.
5.
Using the data appended to this problem, calculate the equilibrium constant for the reaction
below at 25 ˚C (exact temperature).
2+
Hg 2+
2(aq ) ⇔ Hg(l) +Hg (aq )
Which of the two aqueous ions will be more abundant at equilibrium? (Metallic mercury is
insoluble in water, so it precipitates out if formed. If necessary, assume that the aqueous
layer is in direct contact with metallic mercury initially.)
Data:
Hg(2+aq ) +2e− → Hg(l)
E 0 = +0.851V
−
Hg 2+
2(aq ) +2e → 2Hg(l)
E 0 = +0.7973V
If we turn the first half-reaction around, we have
Hg(l) → Hg(2+aq ) +2e−
E 0 = −0.851V
−
E 0 = +0.7973V
Hg 2+
2(aq ) +2e → 2Hg(l)
The overall reaction is
2+
E 0 = −0.054 V
Hg 2+
2(aq) → Hg(l) + Hg(aq)
with νe = 2. We can do the calculation in one or two steps. I’ll do it in one step here:
⎛ −∆ G 0 ⎞
⎛ ν FE 0 ⎞
K = exp⎜ r m ⎟ = exp⎜ e
⎟
⎝ RT ⎠
⎝ RT ⎠
⎛ 2 96485.342 C/mol −0.054 V ⎞
(
)(
)⎟
= exp⎜⎜
-1
-1
⎟
⎝ (8.314472 J K mol )(298.15 K)⎠
= 0.015
This value of the equilibrium constant implies that the reaction is reactant-favored, i.e. there
2+
at equilibrium.
will be more Hg 2+
2 than Hg
6.
Mercury(II) sulfide is only sparingly soluble in water. The sulfide ion is a stronger base than
hydroxide, so the solubility equilibrium is:
2+
HgS(s) + H 2O(l) ⇔ Hg(aq)
+ HS−(aq) + OH−(aq) .
The equilibrium constant for this reaction is 2×10-53 at 25 ˚C (exact temperature).
(a)
Let’s start by setting up the equilibrium expression:
K = aHg 2+ (aHS - )(aOH - )
( )
Since the equilibrium constant is really tiny, this equilibrium will have a negligible effect on
the hydroxide concentration in solution. Therefore aOH - = 10−7 . We will make one Hg2= for
each HS-, so the activities of these two ions will be equal. Thus we have
2
K = aHg2+ (aOH- )
(
)
aHg2+ = K aOH2 × 10 −53
= 1× 10 −23
−7
10
The concentration is proportional to the activity by a factor of 1 mol/L, so the solubility of
HgS is 1×10-23 mol/L. In other words, in 1.0 L of water, we should have 1×10-23 mol of the
Hg2+ and HS- ions. But wait! That many moles is
(1×10-23 mol)(6.022142×1023 mol-1) = 9 molecules!
In other words, we only expect to find nine molecules of HgS dissolved in 1 L of water. This
=
(b)
7.
(a)
equilibrium constant was clearly not measured by measuring a concentration of dissolved
material!
We might try something analogous to the AgCl solubility measurement. We could make up
an Hg|HgS electrode. (This is clearly not quite the same as an Ag,AgCl electrode since Hg is
a liquid, but in principle we could make something like this.) The likely half-cell reaction
would be
HgS(s) + H 2O(l) + 2e - → Hg(l) + HS(aq)
+ OH(aq)
Having made this electrode, we could mesure its half-cell potential and calculate the
corresponding standard reduction potential. Then, combining this standard reduction
potential with the standard reduction potential of Hg2+ to Hg (which is already available), we
could calculate the equilibrium constant for the solubility equilibrium.
Suppose that we operate an ethanol fuel cell in Lethbridge at 25 ˚C (exact temperature).
The partial pressure of O2 is 0.19 bar and the partial pressure of CO2 is about 0.034 bar. The
fuel cell operates on liquid ethanol.
The simplest way to do this problem is probably to work out the free energy change, then go
from there to an emf. We can’t do anything without a balanced reaction, of course:
C 2H 5OH (l) + 3O 2(g) → 2CO 2(g) + 3H 2O (l)
∆ rGm0 = 2∆ f G 0 (CO 2(g) )+ 3∆ f G 0 (H 2O(l) )− ∆ f G 0 (C 2H 5OH(l) )
= 2(−394.4) + 3(−237.1) − (−174.8) kJ/mol = -1325.3 kJ/mol
∆ rGm = ∆ rGm0 + RT lnQ
2⎞
⎛
aCO 2 ) ⎟
(
0
⎜
= ∆ rGm + RT ln
⎜ a 3⎟
⎝ ( O2 ) ⎠
⎛ (0.034 )2 ⎞
⎟
= −1325.3 kJ/mol + (8.314472 ×10 kJ K mol )(298.15 K)ln⎜⎜
3 ⎟
⎝ (0.19) ⎠
= −1329.7 kJ/mol
We’re going to use the equation ∆rGm = −νeFE to calculate E. Our next problem is to figure
out νe. There are three ways to do this:
The hard way: Use the standard method for balancing redox reactions in solution to get the
half-reactions. We can then read νe off from our balanced half-reactions.
The easier way: Look up the oxygen half-reaction from the table of standard reduction
potentials at the back of the book. In this case, it doesn’t matter if we look up the reaction
for acidic or basic reaction conditions because the aqueous intermediates cancel out anyway.
Note that there are 4 electrons for every O2. For our balanced reaction then, νe = 3(4) = 12.
The easiest way: In fuel cells that use oxygen as the oxidant, it’s always 4 electrons per
oxygen molecule, so just remember this ratio.
−∆ rGm
E=
ν eF
−3
1329.7 ×10 3 J/mol
=
= 1.1485 V
12(96485.342 C/mol)
-1
-1
(b)
If we wanted to make a 240V battery, assuming that the operating voltage is similar to the
emf (which, remember, is measured under reversible conditions, and so is generally higher
than the operating voltage), then we would need 240/1.1485 = 209 cells.