Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr
(www.ionicviper.org) on April 5, 2008. Copyright Margret J. Geselbracht 2008. This work is licensed
under the Creative Commons Attribution Non-commercial Share Alike License. To view a copy of this
license visit http://creativecommons.org/about/license/.
H2 STORAGE IN Pd METAL
1. Hydrogen is a cleaner, greener alternative to fossil fuels. However, not
many of us would care to drive a car with a big tank of hydrogen on the
back (ever heard of the Hindenburg?). Metal hydrides are a safe
alternative for storage of hydrogen. Certain metals act like a hydrogen
sponge; hydrogen is able to neatly fit into all of the empty spaces in the
metallic structure. To think about the hydrogen storage capacity of
different metals, think about the percentage of “empty space” in the
structure.
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Commons license
(a) Palladium (Pd) adopts the fcc structure with a unit cell length of 3.8908 Å. Calculate the
percentage of the fcc structure that is empty space in metallic Pd. Note: You will need the radius
of a palladium atom in the fcc structure; you can calculate this based on the fact that the atoms
are in direct contact with each other across the diagonal of any face in the unit cell.
Length of face diagonal:
a2 + a2 = (4r)2
(3.8908)2 + (3.8908)2 = 16r2
r of Pd atom = 1.3756 Å
Volume of unit cell = a3 = (3.8908 Å)3 = 58.900 Å3
Volume of Pd atom = (4/3) π r3 = (4/3) π (1.3756)3 = 10.9035 Å3
fcc unit cell contains 4 Pd atoms so volume of unit cell taken up by Pd atoms
= 4 (10.9035 Å3) = 43.6139 Å3
% of empty space = 100 – (43.6139 Å3) / (58.900 Å3) = 25.953
(b) Although expensive, palladium has great promise as a hydrogen sponge due to the high solubility
of hydrogen in Pd metal. Hydrogen atoms occupy the octahedral holes. If 70% of the octahedral
holes are filled by hydrogen atoms and the lattice does not expand upon hydrogenation, how
many grams of hydrogen will be contained in one cubic centimeter of the palladium hydride?
How does this number compare to the density of liquid hydrogen (70.8 kg/m 3)?
Density is simply mass per unit volume. So, if we determine the mass of hydrogen in
grams in one unit cell and divide that by the volume of the unit cell in cubic centimeters,
we will have the density of hydrogen in that structure.
Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr
(www.ionicviper.org) on April 5, 2008. Copyright Margret J. Geselbracht 2008. This work is licensed
under the Creative Commons Attribution Non-commercial Share Alike License. To view a copy of this
license visit http://creativecommons.org/about/license/.
There are 4 Pd atoms in the fcc unit cell and for every Pd atom, there is one possible
octahedral hole. Hydrogen only fills 70% of these. So, the mass of hydrogen within one
unit cell is
1 mol

  4.687  1024 g
4 H atoms 0.71.008 g mol -1 
23
6.022  10 atoms 
Now find the volume of one unit cell in cm3:
3.8908 Å = 3.8908 x 10-10 m = 3.8908 x 10-8 cm
volume of unit cell = (3.8908 x 10-8 cm)3 = 5.8900 x 10-23 cm3
So, the density is simply mass per unit volume:
24

 4.687  10 g   0.07957 g cm -3
5.8900  10 -23 cm 3 
How does this compare to liquid hydrogen? You need to do some unit conversions:

1000 g 
1 m 3 


70.8 kg m  1 kg 1  10 6 cm 3   0.078 g cm -3
-3
So, the density of hydrogen in palladium hydride is slightly greater than the density of liquid
hydrogen! I would much rather carry around some palladium hydride than a tank of liquid
hydrogen!