Name: _________________________________________
Date:____________________________
Optional Homework 7 – for practice only!!!
1.) Determine Eocell for the following reaction under standard conditions:
Fe+3 (aq) + 1e-1  Fe+2 (aq)
Cu+2 + 2e-1  Cu(s)
Fe+3 (aq) + 1e-1  Fe+2 (aq)
Cu+2 + 2e-1  Cu(s)
Ered = 0.77 V
Ered = 0.34 V
Since the reduction for Fe+3 is the larger, more positive E value, we will keep this
reaction as the reduction. This means that we will flip the second reaction and
change the sign of E.
Fe+3 (aq) + 1e-1  Fe+2 (aq)
Cu (s)  Cu+2 + 2e-1
Ered = 0.77 V
Eox = - 0.34 V
We must also make sure that the number of electrons lost = number of electrons
gained, so we will multiply the first reaction by 2 but we DO NOT change the
value of E!!
2Fe+3 (aq) + 2e-1  2Fe+2 (aq)
Cu (s)  Cu+2 + 2e- 1
Ered = 0.77 V
Eox = - 0.34 V
2Fe+3 + Cu (s) → 2Fe+2 + Cu+2 Eocell =
0.43 V
2.) What if Fe+3 was at a concentration of 0.20 M, Cu+2 was at a concentration of 0.50 M and Fe+2 was at
a concentration of 0.10 M. What would the value of Ecell be at 298 K?
Fe+3 = [0.20]
Cu+2 = [0.50]
Fe+2 = [0.10]
2Fe+3 + Cu (s) → 2Fe+2 + Cu+2 Eocell =
Q=
0.43 V
[Cu 2 ][Fe 2 ]2 [0.50][0.10] 2
=
= 0.13
[Fe 3 ]2
[0.20]2
Ecell = Ecello –
0.0592V
log Q
n
Ecell = 0.43 V -
0.0592V
log 0.13
2
Ecell = 0.43 V – (-0.026V) = 0.456 V
= 0.46 V
3.) The alloy SmCo5 forms a permanent magnet because both samarium and cobalt have unpaired
electrons. How many unpaired electrons are in cobalt? In samarium? Show the unpaired electrons
(only the unpaired electrons) in orbital boxes.
Co: 1s22s22p63s23p64s23d7
Cobalt has 3 unpaired electrons in the 3d orbital
Sm: 1s22s22p63s23p64s23d104p65s24d105p66s24f6
Samarium has 6 unpaired electrons
In the 4f orbital
4.) Lead can displace silver from solution:
Pb(s) + 2Ag+1 (aq) → Pb+2 (aq) + 2Ag (s)
As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore.
Calculate K and Go at 25oC for this reaction. (Hint: determine Eocell!)
Pb → Pb+2 + 2e-1 Ereduction = -0.13 so Eoxidation = 0.13 V
2Ag+1 + 2e-1 → 2Ag Ered = 0.80 V
Eocell = 0.13 V + 0.80 V =
Eocell =
0.93 V
0.0592V
log K
2
0.93 V = 0.0296 log K
31.42 = log K
1031.42 = K =
2.6 x 1031
∆Go = -RTlnK
∆Go = -(8.314 J/molK)(298K)ln(2.6 x 1031)
∆Go = -1.8 x 105 J/mol = -1.8
x 102 kJ/mol
5.) What is the maximum number of unpaired f electrons that an atom or ion can possess. Explain how
your choice was made.
The maximum number of unpaired f electrons is 7. A filled f orbital holds 14 electrons. Each
electron fills an orbital singly at first before they pair up. Using the orbital box diagram, the
most unpaired electrons occurs when each orbital is holding 1 electron.
6.) Define/explain the lanthanide contraction:
The lanthanide contraction results when more protons are added to the elements
but the 4f group electrons slide underneath the 6 s electrons.
7.) Explain how it affects atomic size down a group
This results in the size of the atoms not changing even though we are going down
a group. Normally size would increase, but these species in the 5d row are almost
the exact same size as the species in the 4d row!
8.)
Explain how it affects the densities of the transition metals
Density = mass/volume. Since the size isn’t changing, the volume isn’t changing
but the mass is definitely getting larger! For molybdenum to tungsten, the size
stays nearly the same, but the mass almost doubles. If the mass increases, but we
are dividing by the same number (the volume) then we see large increases in
density for our 5d elements! In fact, this row contains the densest elements!
9.) Using the ideas of Lewis and resonance structures, DRAW the resonance structures of the following
species (the number in parentheses is how many structures there are!) Off to the side, please
include your calculations for the number of valence electrons
a. NO2-1 (2)
N: 5
O: 6 x 2 = 12
17 + 1 = 18 e- total
O N O
a. SO2 (3!)
S: 6
O N O
O: 6 x 2 = 12
18 e- total
O
S
O
O
S
O
O
S
O
b. NO3-1 3)
N: 5
O: 3 x 6 = 18
23 + 1 = 24 e- total
O
O
O
O
N
N
N
O
O
O
O
O