CHEM 101
Chemistry II
Problem Set VIII
Answers
1) We are given the data that a solution of an unknown salt of rhenium is electrolyzed
with 95.0% efficiency, and we get 19.51 g of rhenium. We are also told that we apply
2.75 amps for 2.15 hours. We are asked a) how many faradays are used to deposit the
rhenium; and b) what is the charge on the rhenium? We'll approach these in order.
a) The number of coulombs (and thereby, faradays) can be determined from the
amperage and the time. ?C = (2.75 A)(2.15 h)(3600 sec/h) = 21285 C, which, given that 1
F = 96500 C, yields 0.221 F, of which 95%, or 0.210 F is used to deposit the rhenium.
b) Rhenium has an atomic mass of 186.2 g/mol, so 19.51 g of rhenium
corresponds to 0.105 mole rhenium. We used 0.210 Faradays, which is 0.210 moles of
electrons to deposit 0.105 moles of rhenium metal. Taking the ratio of these, we get:
0.210mol
= 2 mole electrons/mole rhenium which leads us to conclude that the charge
0.105mol
on rhenium is +2 (i.e. we have Re2+)
2) We are given the following cell, and asked to get the ∆G°.
Rb|Rb+(aq)||Pb2+|Pb.
The balanced equation for this cell is:
2Rb + Pb2+ → 2Rb+ + Pb
∆G°
0.0 -24.3
-282.2 0.0
(kJ/mol)
which data leads to a ∆G°rxn = (2(-282.2) + 0.0) - (0.0 + (-24.3)) = -540.1 kJ.
This ∆G° value, given an n = 2, leads, via ∆G° = -nFE°cell to: E°cell =
− 540100 J
= 2.80 V.
− (2)(96500C )
Given that E°Pb/Pb2+ = -0.13, and that E°cell = E°cath - E°an leads to:
2.80= -0.13 - (x), and x = E°Rb+/Rb = -2.93 V.
3) We are given the following cell, which we are told is spontaneous for all
concentrations under standard conditions. This means that all concentrations, pressures,
etc…are held at standard conditions.
O3(g) + 2H+(aq) + 2Co2+ → O2(g) + H2O(l) + 2Co3+(aq)
So we have for our two half-cells:
E° = +2.07 V
O3(g) + 2H+(aq) + 2 e- → O2(g) + H2O(l)
E° = +1.82 V
Co3+(aq) + e- → Co2+(aq)
Therefore, we can calculate our E°cell = 2.07 - 1.82 = +0.25 V, so yes, we are
spontaneous.
Now, we have a buffer system comprised of HClO/ClO- (Ka = 3.5 x 10-8, so pKa =
7.46). Using the concentrations we are given, and the Henderson-Hasselbalch equation,
we can get the pH of the buffer, and from that, the [H+], which we will need. The
[
]
 ClO − 
 (0.10 ) 
 = 7.46 + log 
 =
Henderson-Hasselbalch equation is: pH = pKa + log 
 (0.15) 
 [HClO ] 
7.28 which gives us a [H+] = 10-7.28 = 5.25 x 10-8 M.
Now, here is the important thing to notice: every other condition is still standard!
The only thing we have changed in the [H+]. Therefore, using the Nernst Equation, we
get:
[
]
2
Co 3+ PO 2
RT 
ln
Ecell = E°cell nF  Co 2+ 2 PO 3 H +
[
= 0.25 -
2.57 x10 −2
2
]




[ ]
2

(1M )2 (1atm )
ln 
 (1M )2 (1atm ) 5.25 x10 −8

(
)
2

 = 0.25 – 0.431 = -0.181 V,


and the reaction is non-spontaneous.
4) We are told that this cell is: Hg2+2(aq) + H2(g) → 2Hg(l) + 2H+(aq) in which the
concentration of mercuric ion is controlled by the Ksp of Hg2Cl2, which we are asked to
find. We need to obtain the E°cell and, using the Nernst Equation, obtain the [Hg22+] and
from this, the Ksp of Hg2Cl2. We know the E°S.H.E. = 0.0 V (definition) and we can find
E°Hg+/Hg = 0.80 V. This yields the E°cell = 0.80 - 0.00 = 0.80 V. We are also given the Ecell
[ H ] = [10. ] = 1 .Lastly, n =2.
P [ Hg ] 10
. [ Hg ] [ Hg ]
+ 2
= 0.268 V, and we can determine Q =
H2
2+
2
2
2+
2
2+
2
 1
Substituting, we get: 0.268 V = 0.80 V -(1.285 x 10-2)ln 
 Hg 2 +

2
[
 1
0.532 V = -(1.285 x 10-2) ln 
 Hg 2 +

2
[
]


 or ln  1
 Hg 2 +



2
[
]
]

 which further leads to 


 = 41.401 which gives us (careful!)


[Hg22+] = 1.047 x 10-18. With a [Cl-] = 1.0 M, we have, finally, a Ksp = (1.023 x 109
)(1.0)2 = 1.047 x 10-18.
5) The electrode potentials given in the appendix are referenced to the standard hydrogen
electrode at 0.00 V (definition). Since most commercial cells are aqueous (meaning, as I
am sure you remember, that they involve water), the electrolysis of water becomes an
issue. Put another way: if we were to try to construct an aqueous cell with a cell potential
large than that for the electrolysis of water (E°red = -0.828 V; E°oxid = -1.229 V), we
would simple electrolyze water. (E°cell = 1.229 – (-0.828) = 2.057 V)