Electrolytic reactions are
non-spontaneous reactions that are run by the surrounding doing work on the
system (DG>0 and DG= minimum work needed to make the non-spontaneous
reaction run forward. The backward spontaneous
reaction has a certain potential that is greater than zero. The forward reaction has a potential less
than zero. As an example say the forward
reaction has a potential, E = -2 V. By
using an externally supplied voltage greater than +2 V, we can get the
non-spontaneous reaction to go forward.
The reaction is run in electrolytic
cells. These reactions go the
opposite direction as a voltaic reaction.
Using the table of standard
reduction potentials, we said that the half-reaction with the greater reduction
potential will go from left to right, and substances with lower reductions
potentials will be oxidized (their reduction half-rxn will go backward, from
right to left).
i.e., Higher Reduction Potential ®
Lower Reduction Potential ¬
EMF = Ered, cathode – Ered, anode >
0 (Thus DG<0, and the reaction
is product favored)
For an electrolytic reaction this reaction
will be reversed
Higher Reduction Potential ¬
Lower Reduction Potential ®
EMF < 0 (Thus DG>0. This
reactant favored reaction will be run by the surrounding working on the
system.)
Electrolysis of NaCl – Based on the table of reduction potentials, we
predict that chlorine will react with sodium metal. The chlorine will be reduced and the sodium
will be oxidized in this spontaneous reaction making sodium chloride.
|
|
|
Eºred (V) |
|
Cl2 (g) + 2e- ® |
2 Cl- (aq) |
1.36 V |
|
Na+ (aq) + e- ® |
Na (s) |
-2.71 V |
Net Reaction: Cl2 (g) + 2Na (s) + 2 e-
® 2 NaCl (s) + 2 e-
Using an electrolytic cell we
can run the reverse reaction, reacting sodium chloride to make sodium metal and
chlorine gas. Solid sodium chloride does
not conduct electricity since the ions are fixed in their crystal lattice. So we need liquid sodium chloride to conduct
an electric current. CaCl2 is
typically added to lower the melting point from 802°C to around 600ºC (remember how we added a solute to
lower the freezing point of water).
2 Cl- (L) ® Cl2 (g) + 2 e- Ered° ~ +1.36 V *Oxidation (Anode)
Na+ (L)+ e- ® Na(L) Ered° ~ -2.71 V
*Reduction (Cathode)
Net Rxn 2 Na+ + 2 Cl- + 2 e- ® 2 Na(L) + Cl2
(g)
E°= -2.71 – (-1.36) V = -4.07 V *
DGº = nFEº
= 785 kJ, So we need to do over 785 kJ of work to run
this non-spontaneous reaction.
*We should treat these
potentials as estimates since they are taken from half-reactions in liquid
water under standard conditions, and we are far from standard conditions.
As a voltaic cell, the
reaction of sodium and chlorine producing sodium chloride would produce 4.07 V
(at standard conditions). To run the
reaction producing sodium and chlorine, we would have to apply a DC potential
greater than this in the reverse direction.
If we have an adjustable voltage supply and a meter measuring current
(flow of electrical charge such as electrons) as we slowly increased the
voltage, we would have zero current until we got to -4.07 V. When we got to -4.07 V, we would suddenly get
a flow of current and the reaction would turn on. We could not increase the voltage greater
than -4.07 V.
The reduction of water to
hydrogen and hydroxide has a higher reduction potential than Al3+. If we tried to use electrolysis to reduce
aluminum, water would be reduced instead because it is easier. Since we cannot use an aqueous solution to
reduce aluminum, we could use the molten refined product from bauxite, the
aluminum ore. Refining bauxite, leaves
Al2O3. The problem
with electrolysis of Al2O3 is its melting point is >
2000° C which is hard to handle. In 1886 Charles Hall of the
4 X Al+3
+ 3 e- ® Al (L) (Cathode,
reduction)
3X C (s) + 2 O2- ® CO2 +
4 e- (Anode, oxidation)
4 Al+3 + 3 C + 6 O2- + 12 e- ® 4 Al (L) + 3 CO2 + 12 e-
Graphite electrodes are used
for the anode, and the oxygen from the molten Al2O3
reacts with the graphite to produce carbon dioxide.
~2% of all
Electrolysis is used to
produce Al, Mg, Na, Cl2, NaOH, and purify Cu (with Ag, Au, and Pt
byproducts!).
Ionic substances have high
melting points, so could we do electrolysis in an aqueous solution at a more
reasonable temperature? With water
present, we have it as a possible reactant along with H+,
|
|
|
Eºred / V |
|
F2 +
2 e- ® |
2 F- (aq) |
2.87 |
|
Cl2 +
2 e- ® |
2 Cl- (aq) |
1.36 |
|
O2 +
4 H+ + 4 e- ® |
2 H2O (L) |
1.23 |
|
Cu2+ + 2 e- ® |
Cu (s) |
.34 |
|
Ni2+ +
2 e- ® |
Ni (s) |
-.28 |
|
2 H2O + 2 e- ® |
H2 + 2 |
-.83 |
|
Al3+ + 3
e- ® |
Al (s) |
-1.66 |
|
Na+ +
e- ® |
Na (s) |
-2.71 |
Consider electrolysis of an
aqueous NaF solution.
At the anode the possible
oxidation reactions are. (Note, I am writing the reduction half-reaction backward)
2 F- ® F2 + 2 e- Ered°= 2.87 V
2 H2O ® O2 + 4 H+ + 4 e- Ered°=1.23 V Easier to oxidize
The second half reaction is
more favorable (from thermodynamics) because the greater the reduction
potential, the less likely the reduced form (F- and H2O)
will be oxidized.
At the cathode the possible
reduction reactions are
2 H2O + 2 e- ® H2 (g) + 2
Na+ + e- ®
Na -2.71
V
Water is easier to reduce
than sodium, and this is what is observed.
So looking at the table
above, half reactions above the top blue line in italic print at 1.23 V are substances harder to oxidize than
water, and so it is more likely thermodynamically that water will be
oxidized. Half reactions below the lower
blue line in italic print at -0.83 V
are harder to reduce than water, and so are less likely thermodynamically to be
reduced. This is why we could not
electrolytically reduce aluminum in water, we would just make hydrogen instead.
So, for electrolysis of an
aqueous NaF solution, the net reaction will be
4 H2O +
4 e- ® 2H2 (g) + 4 OH- -.83 V (Cathode reduction)
2 H2O ® O2 + 4 H+
+ 4 e- 1.23 V (Anode
oxidation)
6 H2O + 4 e- ® 2 H2+O2+
4H+ + 4OH- + 4e-
E°(cell) = -.83 - 1.23 = -2.06 V
The hydrogen ion and hydroxide products can combine to produce 4 waters
to have a net reaction of 2 moles of water producing 2 moles of hydrogen and 1
mole of oxygen. The volume of oxygen
will be half of the hydrogen produced.
This supports Avogadro’s hypothesis about equal volumes of gas having
equal number of particles, and it supports the formula of water having twice as
many hydrogen particles as oxygen particles.
So we will need an external D.C.
power supply will a voltage greater than 2.06 V to run this non-spontaneous
reaction.
If we used NaCl
instead of NaF, we would still reduce water to make hydrogen but at the anode
the choice is
2 Cl- ® Cl2 + 2 e- Eºred = +1.36 V
2 H2O ® O2 + 4
H+ + 4 e- Eºred
= +1.23 V
The second reaction is
favored thermodynamically, but at high Cl- concentrations the first
reaction happens (the Cl – oxidation is favored kinetically). So the first half reaction
has a lower activation energy but the change in Gibbs free energy is not as
large a negative number as the second reaction. So the first half-reaction is not as
exergonic as the second. So the net
reaction observed is
2 Cl- ® Cl2 + 2 e- 1.36 V (Anode
oxidation)
4 H2O + 4 e- ® 2H2 (g) + 4
4 Cl- + 4 H2O + 4e- ® 2 H2 +
4 OH- + Cl2(g) Net
Ionic Equation
E°(cell) = -.83V – 1.36V = -2.19 V
The molecular equation would be
4 NaCl + 4 H2O + 4e- ® 2 H2 +
4 NaOH + Cl2(g)
This is the net reaction using
sodium chloride and water, both ready available, to produce the important
commercial products hydrogen, chlorine, and sodium hydroxide.
Aqueous Electrolysis –
1.
Neglecting exceptions like the chlorine mentioned above, metals cations and
other substances with a reduction potential greater than -0.8 V, (the reduction
potential for reduction of water to H2 and OH-) will be
reduced during electrolysis in water.
This includes metals such as Zn, Fe, Ni, and Ag, but does not include
Al, Mg, and Na.
2. Species with an reduction potential less than 1.2 V,
(the reduction potential for the oxidation of water to O2 and H+)
will be oxidized in aqueous electrolysis.
An reduction potential less than 1.2 V is equivalent to saying an
oxidation potential greater than -1.2 V.
The net of the above statements is species dissolved in
water between with reduction potentials between 1.2 V and -0.8 V can be reduced
and oxidized in water. If either species
potential exceeds these limits, water will react instead.
Electrolysis with Active Electrodes - Electroplating
e.g. Ni has a reduction potential between 1.2 and -0.8 V,
so we can oxidize Ni at one electrode and reduce the Ni+2 at the
other. So two Ni electrodes in a NiSO4
solution will react with one reducing Ni+2 and the other
oxidizing Ni metal. This can be used to
put a protective or decorative coating of, for example Ni, Au, or Ag on a
cheaper, more easily oxidized metal.
Typically, electroplating is used to produce thin films from 0.03-.05 mm
thick (30-50 mm).
Ni+2 + 2 e- ® Ni E°= -.28 V (Cathode,
reduction)
Ni ® Ni+2 + 2 e- E°= -.28 V (Anode,
oxidation)
No net reaction and E°=0
1 mole of Ni will be
deposited when 2 moles of electrons flow through the circuit.
Electroplating is used to purify
copper by oxidizing if from an impure anode and reducing it on a cathode (page
928-929 from 9th Edition, BLB).
Less active metals such as gold and platinum drop off of the impure
anode and are collected as valuable sludge in the bottom of the electrolysis
cell. More active metals such as lead
and zinc are oxidized from the anode but are not reduced on the cathode, and so
they stay in solution.
Electroplating of Ag and Au
requires Ag(CN)2- and Au(CN)2+
in order to get a nice smooth metal film.
NiSO4 and H2CrO4 are used to plate Ni
and Cr. Note the electroplating of Ag
and Au has killed people due to accidents producing HCN (g).
Electric current is measured
in amps, 1 amp = 1A = 1 Coulomb of
charge / sec
or 1A s = 1 C
Faraday’s Constant 1 F = 96,500 C / mol of charge converts moles of electrons to Coulombs (A s)
Example 1. Say I run 1.0 amp
through a Ni platting bath for 30.00 minutes (1800. sec). What mass of Ni plates out? Two electrons are needed to reduce Ni2+ cations
to nickel metal.
1800. sec X 1.0
A X 1 C X 1 mol e- X 1
mol Ni X 58.69 g Ni = .547
g Ni
A s 96,500 C 2 mol e- mol Ni
Example 2. How many minutes
will it take 10.0 A to deposit 3.00 g of Au from Au(CN)2+? (7.3 minutes)
Example3 . Often you want some thickness. A problem might give you a current to use and
a thickness and surface area, and ask the time needed to deposit that thickness. Given a current and a thickness and surface
area, calculate a time you would need to calculate a volume to be deposited and
given a density, you could calculate a mass of metal and then calculate
everything else.
Electrical Work
Remember DG = w, so if DG<0, then work is done on the surroundings by a
spontaneous reaction, and if DG>0,
then the surrounding do work on the system.
We have seen
DG
= -nFE, so if E>0, the process is spontaneous and the system does work on
the surroundings.
Power is defined as energy
per unit time, and the unit of power is a Watt (W), so 1W = 1J/s. Electrical work is usually expressed as watt
seconds = J. A kilowatt hour is 1000
watts X 1 hour (3600s) = 3.6 X 106 J. Also remember that a
1V = J/C and 1C =
1As
So VC = VAs = J.
A recent power bill from MID
said I’m using about 14 KWh/day and I’m being charged $.0649 $0.0881/
KWh. So we can calculate the cost of
electrolysis.
Sample Exercise 20.18 (page
818 (page 792 from BLB 8th Edition), I don’t like the way they work
it so I will do it my equivalent way)
Calculate the KWh and the
cost of producing 1000 kg of Al from Al+3 electrolysis using a EMF
of 4.50 V.
1000 kg X 1000g Al X 1
mol Al X 3 mol e- X 96500
As = 1.07 X 1010 As
1 kg 26.98 g Al 1 mol Al 1 mol e-
1.07 X 1010 As X
4.50 V X 1 KWh =
1.32 X 104 KWh
3.6 X 106 J
The cost would be $0.0881 $.0649 X 1.32 X 104 KWh =
$ 1163. 857
1
KWh
As BLB notes, this
electrolysis process is only 40 % efficient, so the process would cost 2.5
times more (100/40). There are also the
added cost of mining the Al ore, initial processing, shipping, clean up,
heating the ore to melt it, and then this electrolysis. Remember above I said that recycling aluminum
saves about 91% of the production cost.
Aluminum metal has a lower melting point and we don’t need to reduce
it. Impurities are added to produce
certain properties, so recycling may involve purification or the aluminum or a
limitation on the uses of the metal. For
example recycled aluminum cans might be limited to use as aluminum cans.