Final Review
first some electrochem/redox
The oxidation state (number) of Cl in HOCl is:
a +2
b -1
c0
d +1 review rules for assigning ON's
+1
e +3
In the reaction: H2O(l) + 3ClO- (aq)
the number of electrons transferred is:
a 1
b 2
pick ClOc 3
or
d 4
2ClO- + 2ee 5
-2
ClO2- (aq) + Cl2(g) + 2OH-(aq)
ClO2- + 2eCl2
36-1
A Voltaic cell consists of two half cells: lead metal immersed in 1M Pb(NO3 )2 (aq)
and silver immersed in 1M Ag(NO)3 (aq). What is the spontaneous reaction that
occurs in this cell?
a Pb(s) + 2Ag+ (aq)
b 2Ag(s) + Pb 2+ (aq)
c Pb2+(aq) + 2Ag+ (aq)
d Pb(s) + 2 Ag(s)
e no reaction occurs
Pb2+(aq)
Pb(s) +
Pb(s) +
Pb2+(aq)
+ 2Ag(s)
2 Ag+(aq)
Ag(s)
+ 2 Ag +(aq)
Pb2+ + 2e
Pb -0.126
Ag+ + e
Ag +0.799
2Ag+ + Pb
2Ag + Pb2+ +0.925
For the reaction 2I-(aq) + Cl2(g)
I2(s) + 2Cl-(aq)
)G0 is - 1.6 x 102 kJ under standard conditions at T = 298. E0 for this
reaction is
a -0.83
b 0.83
c -1.66
d 1.66
e 0.42
)G0 = -nFE0
-1.6 x 105 J = -(2) (96485) E0
E0 = +0.82(9)
36-2
Under standard conditions the following redox reactions are observed to occur
in aqueous solution
A+ + B
A+ + C
2B+ + D
A + B+
no reaction
2B + D 2+
The order of reactivity (most easily oxidized to least easily oxidized is:
a C > A > D >B
b D > B >A>C
one with most +ve is cathode
SRP's
c C >A> B >D
this lets you order the E0's
d B > D >A>C
e cannot be determined
so A > B
C >A>B > D
C>A
B> D
}
most easily oxidized
36-3
Given 2U3+(aq) + Cd 2+(aq)
Cd(s) + 2U 4+(aq)
The potential of this cell reaction at 298 K when
[U 3+] (aq) = 0.10M, [U 4+ (aq)] = 2.0 x 10 -3 M, [Cd 2+aq)] = 0.20 M is:
a + 0.20
b + 0.12
c +0.29
d + 0.25
e +0.37
Cd2+ + 2e
U 4+ + e
Cd -0.4
U 3+ -0.52
E = E0 -0.0591/2 log { 2 x 10-3)2/ (0. 1)2 x 0.2
= 0.12 + 0.08
The alkaline Leclanche Cell is based on the reaction:
Zn(s) + 2MnO2(s)
ZnO2(s) + Mn2O3(s)
As the cell discharges:
a metallic zinc is plated out
b zinc oxide is consumed
c. the voltage increases
d. the voltage decreases
e. the voltage does not change
36-4
C. [10 points]
A proposed mechanism for the gas phase reaction of chlorine with chloroform is:
k1
º
Cl2(g) k
2Cl(g) : fast equilibrium
-1
k2
Cl(g) + CHCl3(g)
HCl + CCl3(g) : Slow
CCl3(g) + Cl (g)
k3
CCl4(g) : Fast
a) Write the overall reaction
Cl2(g) + CHCl3(g)
HCl(g) +
CCl4(g)
36-5
b) Find an expression for the rate law.
Rate = k2 [Cl(g)] [ CHCl3(g)]
k1[Cl2(g)] = k-1 [Cl(g)]2
[Cl] = (k1/k-1)1/2[Cl2]1/2
Rate = k2 (k1/k-1)1/2[Cl2]1/2 [ CHCl3(g)]
= k obs[Cl2 (g)]1/2 [ CHCl3(g)]
Note step one : in the exam it may be written as:
Cl2(g)
º
2Cl(g) : fast equilibrium K1
equilibrium constant
in which case [Cl(g)] = K 1/2 [Cl2]1/2
36-6
The reaction 2N2O5(g)
4NO2(g) + O2(g)
is first order in [N2O5] and first order overall. Which of the following linear
plots would be obtained for the reaction? t represents time
a) -d[N2O5]/dt
b) ln [N2O5]
c) ln [N2O5]
t
t
t
e) [N2O5]
d) 1/[N2O5]
t
t
36-7
D [10 points]
The overall reaction for the lead storage battery is:
Pb (s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)
2PbSO4(s) + 2H2O(l)
E0 = 2.04 V at 298 K
a) Calculate E at 298 for this battery when [H+] = [HSO4-] = 4.5M
n=2
E= E0 - 0.0591/2 log (1/ {4.52 x 4.52})
= 2.04 + 0.77 = 2.117 = 2.12 V
b) For the cell reaction in the lead storage battery, )H0 = -315.9 kJ and
)S0 = 263.5 JK-1. Calculate E0 at -20 oC
)G0 = ) H0 - T )S0
= -315.9 - (253) (263.5 x 10-3 )
= -382.6 kJ
)G0 = -nFE0
E0 = 382600 / 2 x 96485 = 1.98V
watch the units!!
)G0 in Joules
36-8
c) Calculate E at -20 oC when [H+] = [HSO4-] = 4.5M
E = E0 -RT/nF ln Q
= 1.98 - (8.314 x 253)/ (2 x 96485)ln (1/(4.52 x4.52)
= 1.98 + 0.066
= 2.046V = 2.05 V
watch it!!! T = 253 not 298
so cannot use 0.0591
36-9
The reaction A
B is first order in A and first order overall with
a rate constant = 4.00 x 10-4 s-1 at 323K. The percentage of A that will react in a
1 hour period is.
a) 0.42 b) 14.4 c) 23.7 d) 69.4 e) 76.3
ln [A]0/[A] = kt = 4.00 x 10-4 x 60 x 60
so [A]0/[A] = 4.2
let [A]0 = 100 ; [A] = 23.7 so 76.3 % has reacted
A current of 375 amperes is passed through a molten mixture containing Al3+
ions for 300 s. The mass of Aluminum deposited is:
a 31.5g
mass = M It/Fn = 27 x 375 x 300/ 96485 x 3 = 10.5
b 23.2 g
c 3.15 g
d 10.5g
e 1.05 g
36-10
For the reaction Sn2+(aq) + Ni(s)
Sn(s) + Ni 2+ (aq)
Keq is 5.0 x 103 under standard conditions. E0 is ...
a 0.055
b -0.055
c -0.11
d 0.11
e 0.42
E0 = RT/nF ln K
= (8.314 x 298/2 x 96485) ln 5 x 103
= 0.109
Given the following standard reduction potentials
Ag + (aq) + e
Ag (s)
0.8
Fe 3+ + e
Fe 2+(aq)
0.77
Cu 2+ + 2e
Cu(s)
0.34
The strongest reducing agent is:
easiest to reduce has most +ve E
a Cu(s)
so hardest to reduce is Cu 2+
b Cu 2+ (aq)
- means Cu(s) is easiest to oxidize
c Ag + (aq)
and is best reducing agent
d Ag (s)
36-11
e Fe 2+ (aq)
It is planned to deposit gold via the process
Au(s) + 2CN-
Au(CN)2- + e
If an electric current of 30mA is passed for 55.0 minutes what
mass of gold is deposited
m = MIT/Fn = 196.9 x 30 x 10-3 x 55.0 x 60/ 96485
= .20g
In the above question how many moles of electrons
flowed?
charge passed = 30 x 10 -3 x 55.0 x 60 = 99 C
96485 C = 1 mole of electrons
so 99C = 99/96485 moles = 1.02 x 10 -3
36-12
2NO(g) + Cl2(g)
2NOCl (g) at -10 oC
a) If at a particular time [NOCl] increases at the rate of 0.03 mol L-1 s-1
What is the rate of disappearance of Cl 2(g) at the same time?
Rate = -1/2 d[NO]/dt = -d[Cl2]/dt = 1/2 d[NOCl]/dt
so: -d[Cl2]/dt = 1/2 (0.030)
d[Cl2]/dt = -0.015 mol L-1 s-1
b) If [NO]init
[Cl2]init
0.10
0.10
0.10
0.20
0.20
0.20
What is the rate law?
Rate init (mol L-1 min-1)
0.18
0.35
1.45
Rate = k[NO]2[Cl2]
36-13
c) Calculate the Rate constant
take any one of the values
Rate = .35 mol L-1 min -1 = k [.10 mol L-1]2[.20 mol L-1]
k = 1.8 x 102 mol -2 L2 min -1
note units of k for third order reaction
36-14
OK That’s it!!!
Goodbye and Good Luck with the Final Exam
chemistry made simple!
36-15