Northern Arizona University
CHM 152, General Chemistry II
Sections 2 & 3, Spring 2004
Quiz #11
Dr. Brandon Cruickshank
April 26, 2004
Due by 5:00 p.m. on Friday, April 30.
Name _______________________
1.
In the following nuclear equation, identify the missing product.
43
20 Ca
a)
46
22Ti
b)
46
21 Sc
c)
+ α → X + 11 H
44
22Ti
d)
e) none of these
42
18 Ar
Remember that an alpha particle has an atomic number of 2 and a mass number of 4.
2.
232
90Th
−
is a radioactive nuclide. When one of these atoms decays, a series of α and β particle emissions occur,
taking the atom through many transformations to end up as an atom of
emitted in converting
232
90Th
into
b) 8
a) 6
208
82 Pb .
How many α particles are
208
82 Pb ?
c) 2
d) 24
e) 4
A beta particle will not change the mass number. The difference in mass numbers is 232 − 208 = 24. An
alpha particle has a mass number of 4, so we need 6 alpha particles to equal a mass of 24.
3.
14
A sample of wood from an Egyptian mummy case gives a C count of 9.4 cpm/g C (counts per minute per
14
gram of carbon). How old is the wood? (The initial decay rate of C is 15.3 cpm/g C, and its half-life is 5730
yrs.)
a) 6400 yr
t1 =
2
0.693
k
b) 4570 yr
c) 4030 yr
d) 3420 yr
e) none of these
−4
k = 1.21 × 10 1/yr
 [A]t 
ln 
 = − kt
 [A]0 
 9.4 
−4
ln 
 = − (1.21 × 10 1/yr)t
 15.3 
t = 4030 yr
4.
If a tree dies and the trunk remains undisturbed for 13,750 years, what percentage of original
14
present? (Half-life of C = 5730 yrs)
a) 5.20%
b) 18.9 %
c) 2.20%
d) 45.0%
14
C is still
e) none of these
0.693
k
t1 =
2
−4
k = 1.21 × 10 1/yr
ln[A]t = −kt + ln[A]0
−4
ln[A]t = −(1.21 × 10 1/yr)(13,750 yr) + ln(100)
ln[A]t = 2.94
2.94
[A]t = e
= 18.9%
5.
How many electrons are transferred in the following reaction?
−
−
+
2 ClO3 + 12 H + 10 I → 5 I2 + Cl2 + 6 H2O
a) 12
b) 5
c) 2
d) 30
e) 10
The above equation is balanced. Choose either half-reaction to determine the number of electrons
transferred. The iodide/iodine half-reaction is the easier one, so let's use that one.
−
−
10 I → 5 I2 + 10 e
6.
When the equation for the following reaction in basic solution is balanced, what is the sum of the coefficients?
−
−
−
MnO4 (aq) + CN (aq) → MnO2 (s) + CNO (aq)
b) 8
a) 13
−
−
c) 10
d) 20
e) 11
−
+
3(CN + H2O → CNO + 2H + 2e )
−
−
+
2(3e + 4H + MnO4 → MnO2 + 2H2O)
−
−
−
+
3CN + 2H + 2MnO4 → 3CNO + 2MnO2 + H2O
−
This is the balanced equation in acidic solution. Add 2OH to each side of the equation to balance in basic
solution.
−
−
−
3CN + 2H2O + 2MnO4 → 3CNO + 2MnO2 + H2O + 2OH
−
Canceling one water molecule from each side of the equation gives:
−
−
−
3CN + H2O + 2MnO4 → 3CNO + 2MnO2 + 2OH
−
The coefficients in the balanced equation add to 13.
7.
What is ∆G° at 25°C for the reaction below?
2 Au
2
a) −5.07 × 10 kJ
3
d) +1.01 × 10 kJ
D
∆G° = −nF Ecell
3+
2+
+ 3 Ni → 3 Ni
2
b) +5.07 × 10 kJ
3
e) −1.01 × 10 kJ
+ 2 Au
2
c) −7.24 × 10 kJ
D
First, calculate Ecell
, then calculate ∆G°.
Au
3+
−
D
Ered
= 1.50 V
+ 3e → Au
2+
Ni → Ni
−
D
Eoxi
= 0.25 V
+ 2e
D
D
D
Ecell
= Eoxi
+ Ered
= 1.50 V + 0.25 V = 1.75 V
D
∆G° = −nF Ecell
∆G° = −(6 mol)(96,500 J/V⋅mol)(1.75 V)
6
3
∆G° = −1.01 × 10 J = −1.01 × 10 kJ
8.
For a reaction in a galvanic cell both ∆H° and ∆S° are positive. Which of the following is true?
a)
o
Ecell
will increase with an increase in temperature.
b)
D
Ecell
will decrease with an increase in temperature.
c)
D
Ecell
will not change when the temperature increases.
d) ∆G° > 0 for all temperatures.
e) None of the above statements is true.
This reaction has a favorable ∆S and an unfavorable ∆H. It will be temperature dependent.
Endothermic reactions are favored by an increase in temperature. This should make sense because
endothermic reactions absorb heat.
9.
Given the following standard reduction potentials in acidic solution, the strongest reducing agent would be?
−
3+
+ 3 e ↔ Al (s)
Al
−
AgBr (s) + e ↔ Ag (s) + Br
Sn
−
4+
+ 2 e ↔ Sn
−
3+
−
2+
+0.13
2+
+0.77
+ e ↔ Fe
Fe
3+
a) Fe
E°(V)
−1.66
+0.07
b) Fe
2+
c) Br
−
3+
d) Al
e) Al
The strongest reducing agent will be the substance that is most easily oxidized. Al(s) is the most easily
oxidized with an oxidation potential of +1.66 V.
10. Copper will spontaneously reduce which of the following?
2+
a) Fe
and Ag
+
b) Fe
2+
c) Ag
+
3+
d) Al
2+
e) Fe
3+
and Al
+
Looking at Table 19.1 of the text, you can use the diagonal rule. Cu is diagonally across and below Ag .
You could also calculate the cell potential.
+
−
D
Ered
= 0.80 V
Ag + e → Ag
Cu → Cu
2+
−
+ 2e
D
Eoxi
= − 0.34 V
D
D
D
Ecell
= Eoxi
+ Ered
= 0.80 V + (−0.34 V) = 0.46 V
A positive cell potential indicates that the reaction is spontaneous.
11. Calculate the equilibrium constant at 25°C for the reaction:
Sn
a) 1.9 × 10
D
Ecell
=
−4
b) 2.1
2+
2+
+ Ni → Sn + Ni
c) 8.6
d) 1.5 × 10
13
e) 5.2 × 10
3
0.0257 V
ln K
n
First, calculate the cell potential, then calculate the equilibrium constant.
Sn
2+
−
+ 2e → Sn
2+
Ni → Ni
−
+ 2e
D
Ered
= − 0.14 V
D
Eoxi
= 0.25 V
D
D
D
Ecell
= Eoxi
+ Ered
= − 0.14 V + 0.25 V = 0.11 V
0.0257 V
ln K
n
0.0257 V
0.11 V =
ln K
2
lnK = 8.56
8.56
3
K = e
= 5.2 × 10
D
Ecell
=
12. Consider an electrochemical cell based on the spontaneous reaction
−
2 AgCl (s) + Zn (s) → 2 Ag (s) + 2 Cl + Zn
2+
If the zinc ion concentration is kept constant at 1 M, and the chloride ion concentration is decreased from 1 M to
0.001 M, the expected change in cell voltage is:
c) decrease 0.06 V
b) increase by 0.18 V
e) increase by 0.35 V
a) increase by 0.06 V
d) decrease by 0.18 V
0.0257 V
ln Q . This is the Nernst equation which will allow you to calculate a non-standard cell
n
potential. Since the problem is only asking for the change in cell voltage, we only need to calculate
0.0257 V
−
ln Q
n
E = ED −
(
0.0257 V
0.0257 V
ln Q = −
ln [Cl− ]2 [Zn 2+ ]
n
2
0.0257 V
2
ln [0.001 M ] [1 M ] = +0.18 V
change in cell voltage = −
2
change in cell voltage = −
(
)
)