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Optional Homework 5 – for practice only!!!!!
1.) Show the oxidation numbers for all species in the chemical reactions below. Indicate which
species is oxidized, which species is reduced, which species is the oxidizing agent, and which
species is the reducing agent for each reaction by writing the half-reactions: They are not
balanced – please do no balance them!
a. PbS(s) + O2 (g) → PbO (s) + SO2 (g)
Pb+2 S-2
O2 0
Pb+2 O-2 S+4 O-2
4 e-1
S-2 → S+4 + 6e-1
+ O20 → 2O-2
O2 is reduced, it is the oxidizing agent
S-2 is oxidized, it is the reducing agent
b. K2Cr2O7 (aq) + HI (aq) → KI (aq) + CrI3 (aq) + I2 (s) + H2O (l)
K+1 I-1 Cr+3 I-1 I20 H+1 O-2
K+1 Cr+6 O-2 H+1 I-1
3 e-1
2I-1 → I20 + 2e-1
+ Cr+6 → Cr+3
Cr+6 is reduced, it is the oxidizing agent
I-1 is oxidized, it is the reducing agent
2.) Consider the following balanced redox reaction:
2CrO2-1(aq) + 2H2O(l) + 6ClO-1(aq) → 2CrO4-2(aq) + 3Cl2 (g) + 4OH-1 (aq)
Cr+3 O-2
H+1 O-2
Cl+1
Cr+6 O-2
Cl0
O-2 H+1
a.) Which species is being oxidized?
Cr+3 or CrO2-1
b.) Which species is being reduced?
Cl+1 or ClO-1
c.) Which species is the oxidizing agent?
Cl+1 or ClO-1
d.) Which species is the reducing agent?
Cr+3 or CrO2-1
e.) Write the oxidation half-reaction
Since the equation is ALREADY balanced, we just need to show the half reaction that
contains the species being oxidized. You can write it the simplistic way, or carry the
entire species down. But it is NOT necessary to balance the half reaction!
Cr+3 → Cr+6
Or
CrO2-1 → CrO4-2
f.) Write the reduction half-reaction
Cl+1 → Cl0
Or
ClO-1 → Cl2
3.) A Hydrogen electron makes a transition from n=6 to n=3. Calculate the energy, frequency, and
wavelength of the photon. In what region of the spectrum is this photon?
This energy equation gives you the absolute value of the energy, regardless if
the electron jumped levels or fell from levels. Therefore, we MUST state, with
a sign or words, whether the energy was absorbed (+E when an electon jumps)
or released (-E when an electron falls) as part of the final answer
 1
1 
E = -2.179 x 10-18 J  2 − 2 
 nH
nL 
1 
 1
− 2
2
3 
6
E = -2.179 x 10-18 J 
1 
 1
− 
 36 9 
E = -2.179 x 10-18 J 
E = -2.179 x 10-18 J (-0.083)
E = 1.809 x 10-19 J released or -1.809 x 10-19 J
Now we need to calculate the wavelength, and we SHOULD never use calculated values to
determine other values. If we use the energy to calculate the wavelength, and our energy was
WRONG, our wavelength value would also be incorrect!
1
= 1.0968 x 107 m-1
λ
 1
1

−
n 2 n 2
2
 1




It really doesn’t matter who is n1 and who is n2 as long as you remember that all wavelengths
are positive numbers – so if you plug in your n values backwards – just take the absolute value
of your answer!!
1
1 
 1
= 1.0968 x 107 m-1  2 − 2 
λ
6 
3
1
= 1.0968 x 107 m-1
λ
1 1 
 − 
 9 36 
1
= 1.0968 x 107 m-1 (0.083)
λ
1
= 9.1034 x 105 m-1
λ
λ=
1
= 1.0980 x 10-6 m (or 1098 nm)
9.1034 x 10 5 m - 1
1.0980 x 10-6 m
Since there is NO other way but to use a calculated value to determine frequency, we MUST use
either wavelength or E. Each should give the same answer. In fact, this is a way to check you
math. If you calculate the frequency based on the wavelength and you get a different number
than the frequency calculated from the energy, somewhere you have made a math mistake! (the
numbers will not be exact due to rounding, but they should be close!!)
E = hυ
υ = E/h
-19
υ = 1.809 x 10 /6.6.26 x 10-34 Js
c = λυ
υ = c/λ
υ = 2.9979 x 108m/s /1.0980 x 10-6 m
υ = 2.7301540 x 1014 sec-1
υ = 2.7303279 x 1014 sec-1
Remember a Hertz (Hz is a reciprocal second)
υ from E = 2.730 x 1014 Hz
υ from wavelength = 2.730 x 1014 Hz
Same number!! 
4.) Calculate the de Broglie wavelengths for a pea (m= 0.522 grams), a carrot (m=8.75 grams), and a
potato (m= 227 grams) thrown by Randy Johnson thus moving at (a speed) v = 92 miles/hr.
h
λ=
mv
92
miles
1hr
1 min 5280ft 12in 2.54cm
1m
= 41 m/sec
x
x
x
x
x
x
hr
60 min 60 sec 1mile
1ft
1in
100cm
0.522 grams x 1kg = 0.000522kg
1000g
kgm 2
1J =
s2
8.75 grams x
6.626 x 10 - 34
1 kg
= 0.00875kg
1000 grams
kgm 2
227 grams x
s
2
s
λ=
= 3.10 x 10-32 meters
0.000522 kg x 41m/sec
1 kg
= 0.277 kg
1000 grams
6.626 x 10 - 34
kgm 2
s
2
s
λ=
= 1.85 x 10-33 meters
0.00875 kg x 41m/sec
6.626 x 10 - 34
kgm 2
s
2
s
λ=
= 5.83 x 10-35 meters
0.277 kg x 41m/sec