Chem 6c Midterm 1

Only available on StudyMode
  • Download(s) : 23
  • Published : April 2, 2013
Open Document
Text Preview
Chem. 6C Midterm 1
Version A
October 19, 2007
Name__________________________________________
Student Number _________________________________

All work must be shown on the exam for partial credit. Points will be taken off for incorrect or no units. Non graphing calculators and one hand written 3” × 5” note card are allowed.

Problem 1
(of 15 possible)

Problem 2
(of 26 possible)

Problem 3
(of 20 possible)

Problem 4
(of 15 possible)

Problem 5
(of 6 possible)

Problem 6
(of 28 possible)

Problem 7
(of 10 possible)

Problem 8
(of 21 possible)

Problem 9
(of 9 possible)

Midterm Total
(of 150 possible)

I would like my grade to be posted on line by my student number ___________________________________

1

1) Answer the questions below using the following data
Half Reaction
Eº (V)
Ag2+(aq) + e- Ag+(aq)
1.99
1.50
Au3+(aq) + 3e- Au(s)
Fe3+(aq) + e- Fe2+(aq)
0.77
+
2H (aq) + 2e
H2(g)
0.00
Pb2+(aq) + 2e- Pb(s)
-0.126
Ni2+(aq) + 2e- Ni(s)
-0.23
-1.66
Al3+(aq) + 3e- Al(s)
Ca2+(aq) + 2e- Ca(s)
-2.76

(3 pts) Which is the strongest reducing agent?

Ca(s)

(3 pts) Which is the strongest oxidizing agent?

Ag2+(aq)

(3 pts) Will Al(s) dissolve in 1 mol·L-1 HCl?

Yes

(3 pts) Can Pb(s) reduce Ag2+(aq)?

Yes

(3 pts) In the electrochemical cell Ni(s)|Ni2+(aq)||Au3+(aq)|Au(s). What is the emf when the cell is at equilibrium?
At equilibrium the emf = 0 V

2

2) Answer the following questions using the given galvanic cell

MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) +4H2O(l)
Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l)

Eº = 1.51 V
Eº = 1.33 V

(3 pts) When current is allowed to flow what species is oxidized? Cr3+
(2 pts) What is the value of Eºcell?
1.51 V – 1.33 V = 0.18 V
(3 pts) What is the oxidation state of Cr in Cr2O72-?
6+
(5 pts) What is the balanced equation of this cell?
6(MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) +4H2O(l))
5(2Cr3+(aq) + 7H2O(l) Cr2O72-(aq) + 14H+(aq) + 6e-)
6MnO4-(aq) +10Cr3+(aq) + 11H2O(l) 6Mn2+(aq) + 5Cr2O72-(aq) + 22H+(aq) (4 pts) What is the value of Q, reaction quotient, for this cell reaction? Q = ([Mn2+]6[Cr2O72-]5[H+]22)/([MnO4-]6[Cr3+]10)
= ((0.20)6(0.30)5(0.010)22)/((0.10)6(0.40)10) = 1.48×10-41
(5 pts) What is the emf at 25ºC as read on the digital voltmeter? E = Eº - (RT/nF)lnQ
= 0.18V-((8.3125 Jmol·K-1)(298.15 K))/((30)(96485 C·mol-1))ln(1.48×10-41) =0.26 V
(4 pts) What is the value of the equilibrium constant at 25ºC for the net spontaneous cell reaction?
lnK = nFEº/RT
K = enFEº/RT = e((30)(96485 C·mol-1)(0.18 V))/((8.3145J·mol-1·K-1)(298.15K)) = 1.89×1091 3) The following data was obtained experimentally at 25ºC

3

[A]
(mol·L-1)
0.0001
0.0003
0.0002
0.0004

[B]
(mol·L-1)
0.0200
0.0200
0.0100
0.0300

[C]
(mol·L-1)
0.0200
0.0200
0.0400
0.0100

Initial Rate
(mol·L-1·s-1)
1.66×10-7
4.99×10-7
6.66×10-7
1.87×10-7

(15 pts)What is the rate law?
Rate=k[A]x[B]y[C]z
1.66×10-7 = k[0.0001]x[0.0200]y[0.0200]z
4.99×10-7 = k[0.0003]x[0.0200]y[0.0200]z
0.33 = (0.33)x
ln(0.33) = xln(0.33)

x = 1.00

1.66×10-7 = k[0.0001][0.0200]y[0.0200]z
6.66×10-7 = k[0.0002][0.0100]y[0.0400]z
0.25 = (0.50)(2.00)y(0.50)z

0.5 = (2.00)y(0.50)z

ln(0.50) = yln(2.00) + zln(0.50)
y = (ln(0.50) - zln(0.50))/ln(2.00) = -1.00 + z
1.66×10-7 = k[0.0001][0.0200]y[0.0200]z
1.87×10-7 = k[0.0004][0.0300]y[0.0100]z
0.89 = (0.25)(0.67)y(2.00)z
3.56= (0.67)y(2.00)z
ln(3.56) = yln(0.67) + zln(2.00) = (-1.00 + z)ln(0.67) + zln(2.00) 1.27 = 0.40 + 0.29z

z=3

y = -1.00 + z = -1.00 + 3

y=2

Rate = k[A][B]2[C]3
(5 pts) What is the rate constant?
1.66×10-7 = k[0.0001][0.0200]2[0.0200]3
k = 5.2×105 L5·mol-5·s-1

4

4) (15 pts) Balance the following equation:
Zn(s) + NO3-(aq) Zn(OH)42-(aq) + NH3(s) (basic solution)
NO3- NH3
Zn Zn(OH)42Zn Zn(OH)42NO3- NH3
2Zn + H2O Zn(OH)4
NO3- NH3 + 3H2O
Zn + 4H2O + 4OH- Zn(OH)42- + 4H2O NO3- + 9H2O NH3 + 3H2O + 9OHZn + 4OH- Zn(OH)42- + 2eNO3- +...
tracking img