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3. Electrochemical Technology
3.1 General Considerations
The design, characterization and operation of electrolytic devices and processes is
the province of electrochemical engineering. The electrochemical engineer must
consider
aspects
of
fundamental
electrochemistry,
industrial
scale
electrochemistry and the essential link between them, namely electrochemical
technology.
Fig. 3.1 Scale and Endeavour's of electrochemical engineering
In concept, an analytical approach to the design and characterization of an
electrochemical reactor might consider a complete treatment of the following
aspects:
1. Mass balance (including the electrical charge balance) for all reactants and
products.
2. Heat balance across all reactor compartments and components.
3. Voltage balance across the reactor.
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Table 3.1 Classification of electrochemical cells
Electrochemical reactor: The term 'electrochemical reactor' is used to mean a
purpose-built electrochemical cell (or number of cells in a common package)
which performs a useful duty.
The electrodes and the separator are the only components in an electrolytic cell
which are not to be found in other chemical reactors. The detailed design and
characterization of such reactors may involve complex, interactive procedures
and understanding of factors such as potential and current distribution,
electrolyte flow patterns, electrode thermodynamics and kinetics, mass and heat
transfer besides the cost and the reactor components. These factors are common to
all chemical engineering processes with the important exception of potential
distribution.
Consider the general electrochemical reaction:
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where 1 mole of the reactant A receives n moles of electrons to yield nP moles of
product P.
3.1.1 Fractional conversion XA
This is the fraction of reactant which is consumed by the electrochemical reaction.
For a batch process:
Eq------3.1
where mo is the initial, molar amount of reactant and mt is the molar amount at time
t. For a continuous flow through reactor, the fractional conversion is related to the
inlet and outlet concentration of reactant:
Eq----3.2
If the electrolyte volume V is constant
Eq-------3.3
Equations may then be written in terms of the reactant concentration.
For a batch process:
Eq--------3.4
and for a continuous, flow through reactor:
Eq-------3.5
Since electrolysis is a heterogeneous process, the fractional conversion depends on
the ratio of the active electrode area to the cell volume and to the flow rate of the
electrolyte. A high conversion per pass is desirable if, for example, the starting
material is not to be recycled [e.g. an effluent treatment) or the product must be
extracted during each cycle, e.g. when the product is unstable. This is obtainable
with most cell designs only when a slow flow rate is used which leads to a long
residence time and poor mass transport conditions. Hence, a cell with a high
conversion per pass at a high flow rate is often a desirable goal and certainly
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provides a driving force for designing cells with a high surface area per unit
volume.
3.1.2 Materia1 yield
Material yield θp (which is more precisely called the overall operational yield) is
the maximum molar amount of desired product obtained from 1 mole of reactant,
taking the reaction stoichiometry into account
Eq----------3.6
The material yield determines the annual consumption of raw material for the
desired tonnage of product.
In addition, however, the material yield is important because for θp < 100% it will
be necessary either to accept the lower price normally associated with an impure
material or to introduce additional unit processes for purifying the product and
handling the by-product.
3.1.3 Overall conversion-related yield
This is the ratio of the material yield to the fractional conversion:
Eq-----3.7
3.1.4 Current efficiency
The current efficiency φ is the yield based on the electrical charge passed during
electrolysis, i.e. it is a yield based upon the electron as a reactant:
From Faraday's laws of electrolysis:
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Eq------3.8
we may define two convenient versions of the current efficiency, depending
upon the period during which the charge is measured:
1. the overall current efficiency
Eq-----3.9
Eq-----3.10
2. the interval current efficiency
A value φ of below 100%, indicates either:
(1) that the back-reaction occurs to some extent in the cell; or
(2) (more likely) that by-products are being formed.
3.1.5 The overall selectivity
SP is the ratio of desired product to total products:
Consider the reaction schemes 1 and 2 below.
In both cases:
Eq------3.11
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3.1.6 (Electrical) Energy consumption (Es)
The electrolytic energy costs of an electrochemical process are related closely to the
energy efficiency. The energy consumption may be referred to the amount of
substance on a molar mass or volume basis.
Eq-----3.12
The more common:
Eq-----3.13
while for gaseous or liquid products,
Eq-----3.14
where Vm is the molar volume. If the Faraday constant has the units of A s mol-1
and Ecell in V then Es will take the units of J mol-1 as in the 1st equation, J kg -1 as
in the 2nd equation and J m-3 as in the 3rd equation.
It is more common to express Es in terms of kWh mol-1 or kWh kg-1 or kWh m-3 .
This is achieved readily upon division of the right-hand side of equations by 3.6 x
106. (kW-hr = 3600 kJ)
It can be seen that the energy consumption does not depend directly on current
density but only on the cell voltage and the current efficiency. Hence, the energy
consumption can be minimized only by selecting the electrolysis conditions so that
the current used solely for the reaction and making the cell voltage as low as
practicable.
3.1.7 Cell Voltage
The cell voltage is a complex quantity made up of a number of terms, i.e.:
Eq------3.15
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are the equilibrium potentials for the anode and the cathode reactions
respectively, so that
may be calculated from the free energy change for
the overall cell reaction by:
eq-----3.16
In the case of self-driving cells (e.g. batteries and fuel cells) Ecell is positive and the
energy consumption is negative, i.e. the cell produces electrical energy.
are the anode and cathode overpotentials respectively. For a simple
electron transfer reaction, the overpolential is given by the expression:
Eq-----3.17
So, the overpotential will depend on the transfer coefficient and on the exchange
current density. The design of cells with minimum resistances is the first problem
of electrochemical engineering.
Fig. 3.2 Schematic voltage component in a divided cell, illustrated by a plot of potential vs.
distance x in the interelectrode direction.
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1. the internal cell resistance Rcell is decreased by making the interelectrode gap
smaller and using a highly conducting electrolyte (molten salts or water with a high
concentration of electrolyte).
2. any separators will cause substantial increases in cell resistance although they
may be essential for a good current and material yield.
3. In electrolytic processes the overpolential and iR terms represent energy
inefficiencies and, hence. will make the cell voltage a larger negative value.
4. The final term in Ecell equation recognizes that there will be a potential drop in
the electrodes themselves and the busbars which carry the current, in the various
connectors and in the other parts of the electrical circuit.
3.1.8 Mass transfer coefficient
For a process under mass transport control (via convective diffusion) the limiting
current iL expresses the duty of the reactor:
Eq-----3.18
So, the definition of the mass transport coefficient:
Eq-----3.19
3.1.9 Space-time and space velocity
These parameters describe the investment costs for an electrochemical process. The
space-time τST is defined by:
Eq------3.20
as the ratio of reactor volume to volumetric flow rate. If V is the effective volume
of electrolyte in the reactor, then τST is equivalent to the mean residence time of
electrolyte in the reactor. If VR has units of m3 and Q is in m3 s-1, τST will be in
seconds.The space-velocity S is defined as the ratio of volumetric flowrate to
reactor volume:
Eq--------3.21
and is therefore the reciprocal of space-time:
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Eq-------3.22
The space-velocity describes the investment costs per unit volume of
electrolyte.The space-velocity is effectively the volume of electrolyte which may be
processed in unit reactor volume in unit time, e.g. the units of m3 m-3 h-1 are
obtained if the right-hand side of equation is multiplied by 3600.
3.1.10 Space-Time yield
The space-time yield ρST is one of the most valuable statements of reactor
performance. It expresses the mass of product per unit time w/t which can be
obtained in a unit cell volume VR i.e.:
Eq------3.23
It can be seen that the space-time yield is related to the space-velocity as follows:
Eq-------3.24
where ΔC is the concentration change and M is the molar mass of material. From
Faraday's laws of electrolysis;
Eq--------3.25
Substitution of eq. 3.25 into eq. 3.23 gives:
Eq-------3.26
The electrode area per unit volume As called the specific electrode area
so that A/VR = As is the electrode area per unit reactor volume;
Eq------3.27
and it can be seen that the space-time yield is directly proportional to the useful
current density ( =I φ ) and
to As . If I is in A m- 2, M in kg mol -1 I , As in m-1 and F in A s mol-1, ρST takes the
units of kg m- 3 s-1
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3.2 Electrolysis Applications
Michael Faraday observed that when 96500 coulombs of electrical charge placed
through an electrolytic solution that one equivalent of substance is plated out onto an
electrode. If the current passing through the cell is double then 2 equivalents would
be deposited. He concluded that the quantity of substance deposited on the electrode
is directly proportional to the current passing through the cell.
The Gram Equivalent Weight (GEW) of Oxidizing and Reducing Agents
Gram Equivalent weight or mass is the grams of substance per equivalent. This can
be determined by dividing the atomic mass or molecular mass of the substance by
the number of moles of electrons gained or lost under balanced conditions.
GEW = atomic or molecular weight / moles of electrons gained or lost per mole of
substance being oxidized or reduced. For example:
We know that in a copper plating solution:
Cu+2 + 2e- Æ Cu
So it takes 2 moles of electrons per mole of copper so the gram equivalent weight is:
GEW = atomic weight of Cu / 2 = 63.5 / 2 = 31.75 grams/equivalent
What would be the gram equivalent weight of KMnO4 as an oxidizing agent in acidic
solution:
1.Look up the half reaction in the Standard Reduction Potential Table.
MnO4-2+ 8H+ + 5 e- Æ Mn+2 + 4H2O
2. Note the moles of electrons shown in the half reaction. According to the half
reaction there are 5 moles of electrons per mole of the Permanganate
3. Determine the Molecular weight of the substance
For KMnO4 that would be 39.1 + 54.9 + 4(16) = 158 g/mole
4. Plug the molecular weight and moles of electrons into GEW formula
GEW = Molecular Weight of KMnO4 / 5 = 158 / 5 = 31.6 grams / equivalent
Exercise
Determine the gram equivalent mass for the following using the Standard Reduction
Potential Table.
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1. K2Cr2O7 as an oxidizing agent in acidic solution. (ans. GEW=49 g/eq)
2. NaClO as an oxidizing agent in basic solution. (ans. GEW=37.25 g/eq)
3. SnCl2 as a reducing agent. (ans. GEW=94.85 g/eq)
3.2.1 Stoichiometry of Electrolysis
According to the Faraday's Laws:
weight deposited at an electrode = (current/amps) (time/sec) (GEW of substance
deposited) / 96500
1.Electrolytic stoichiometry
For the electrolysis of sodium Chloride in water, we would need to consider not only
the Na+ and Cl– as possible reactants but also the water. At the cathode of an aqueous
NaCl cell, we would get the reduction of water take place.
2H2O + 2e– → H2 + 2OH–
E°red = –0.83 V
rather than the reduction of the sodium
Na+ + e– → Na
E°red = –2.71 V
Notice that the reduction potential for the water reduction is a lot more positive than
that for the sodium.
Looking at the two possible reactions at the anode of the NaCl (aq) cell, we see that
the Cl– and the H2O are both candidates for oxidation. The two possible halfreactions are:
2H2O → O2 + 4H+ + 4e–
E°ox = –E°red = –1.23
2Cl– → Cl2 + 2e–
E°ox = –E°red = –1.36
Since the water has a slightly more positive oxidation potential than the chlorine
reaction, it should be oxidized more readily and since the two possible reactions are
quite close in potentials, any overpotential can quickly be sufficient to cause the
chlorine oxidation to occur.
Example-1
In an aqueous solution of CuSO4, Copper is deposited at the cathode according to the
following half reaction:
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Cu+2 + 2e- Æ Cu(s)
Water is oxidized at the anode producing Oxygen gas according to the following half
reaction:
2H2O Æ O2 + 4H+ + 4 eIf 0.404 grams of Cu was deposited at the cathode as current was passed through the
cell in 5 hours, calculate the following:
1. Current in amps that must be passed for the indicated time.
2. Mass of Oxygen gas deposited at the anode
3. Volume of Oxygen collected at STP (R = .0821 liter-atm/mole-K)
Solution
1. Determining the Current in amps
a.Convert the 5 hours to seconds
5 hours × 3600 sec = 18,000 sec
b. Calculate the gram eq weight of Cu
GEW = Atomic weight of Cu / 2 moles of electrons/mole Cu
= 63.5/ 2 = 31.75 g/eq
c. Using the Faraday formula above to determine the current.
weight deposited at an electrode = (Current in amps) (time in sec) (GEW of
substance deposited) / 96500
0.404 grams = (x)(18,000)(31.75)/ 96,500
x = (.404)(96,500) / (18,000)(31.75) = 0.068 amps = 68 milliamps
2. Determining the mass of Oxygen deposited at the Anode
a. Determine the gram equivalent weight of O2 according to the half reaction:
2H2O Æ O2 + 4H+ + 4 eGEW of O2 = molecular weight of O2 / 4
GEW = 2(16)/4 = 8 grams/eq
b. Plug in the time in sec, the amps, and the GEW of O2 into the Faraday formula
mass = 0.101 grams O2
3. Determining the volume of O2 at STP
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a. Convert grams from Part 2 to moles of O2 .
0.101 grams × 1 mole / 32 grams = 3.17 × 10-3 moles O2
b. Recognizing that STP is a Temp of 273 K and 1 atm Pressure plug the moles,
temperature, and pressure into the Ideal Gas Law Equation and solve for Volume in
liters.
PV = nRT
V = n R T/P = (3.17 × 10-3 moles) (.0821 l-atm/mol-K)(273 K) / 1 atm
V = 0.071 liters = 71 ml of O2 will be deposited
Exercise
When an aqueous solution of KI is electrolyzed the following half reactions occur:
Anode:
-
2I Æ I2 + 2e-
Cathode: 2H2O(l) + 2e- Æ H2(g) + 2OH-(aq)
If a current of 8.52 × 10-3 amps is passed through the cell for 10 minutes calculate
the following
(R=.08206 l.atm/mol.K):
a. mass of I2 produced at the cathode. (ans. 6.72 mg)
b. mass of H2 gas deposited at the anode. (ans 0.053 mg)
c. Volume in liters of H2 deposited at STP(0°C, 1 bar) (ans.0.594)
2. Fused Salt stoichiometry
Consider the reaction in the molten NaCl electrolysis cell.
2Cl– → Cl2(g) + 2e–.
2 moles of electrons per mole of
Cl2(g)
Na+ + e– → Na(l).
1 mole of electrons per mole of Na
metal
Recall that the charge on the electron is Q = nF where F is the Faraday constant
(96500 C/mol). Thus to produce one mole (23 g) of sodium, we need Q = 1×96485 C
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= 96500 C. To produce one mole (70.9 g) of chlorine gas, we need 2×96500 =
1,930,000 C.
Current is defined as the amount of charge passing a point in a circuit in one second.
I = Q/s
The units are
1 Ampere = 1 Coulomb/second (1A = 1C/s).
Example-2
A current of 50.0 A passing through an NaCl(l) electrolysis cell in 1 hour, how much
sodium and chlorine will we produce.?
50.0 A × 3600 s = 180,000 C
n = Q/F = 180,000/96,485 = 1.87 mol e–
It takes lots of current to produce very little sodium and chlorine.
Example-3
What volume of Cl2(g) at STP (0°C, 1 bar) will be produced by a current of 20.0 A
in 2.00 h in the same cell as used in the previous example?
3. The Hall process is a process for the production of aluminum on an industrial
scale. A molten mixture of aluminum ore (Al2O3) and cryolite (Na+)3(AlF6–)(l). The
cryolite is the solvent and it is used because it has a lower melting point than pure
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Al2O3. A carbon electrode acts as the anode and aluminum forms the cathode such
that the following reactions occur:
Cathode . . .
AlF63- + 3 e = Al + 6 F-
Anode
2Al2OF62- + C(s) + 12 F- = + 4 AlF63- + CO2 + 4 e
...
Overall cell reaction 2Al2O3 + 3 C = 4 Al + 3 CO2
The overall reaction is simple despite the complicated mechanism in the electrolysis.
In the examples below, a simple formulation for the cathode and anode reaction is
used to illustrate the shoichiometry of electrolysis.
Example-4
What mass of aluminum will be produced in 1.00 h by electrolysis of Molten AlCl3
using a current of 10.0 A? what is the energy consumption if the applied potential is
6V and 85% current efficiency.
Q = I×t = 10.0 A × 3600 s = 3.60×104 C.
n = Q/F = 3.60×104 C / 96500C/mol e– = 0.373 mol e–
⎡1mol Al ⎤
⋅ ⎥ = 0.124 mol Al metal
⎣ 3 mol e ⎦
0.373 mol e– × ⎢
0.124 mol × 27 gm/mol = 3.357 gm Al metal
Energy consumption =
− nFE cell
φ
=
− 0.373 × 96500 × −6
= 254 kJ/mol = 0.0706 kW-h
0.85
Example-5
How many Faradays and how many coulombs must be passed through a molten
mixture of Al2O3 and Na3AlF6 to produce 1 Kg of Al metal? What is the current
efficiency if the actual required electric charge would be 1.33 × 107 coulomb.
Solution
The Hall process can be oversimplified by these reactions,
Al3+ + 3 e = Al . . . Cathode
C(s) + 2 O2- = CO2 + 4 e . . . Anode
To produce 1 kg = 1000g of Al metal
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1000 gm
= 37.037 mole of Al metal
gm
27
mol
Theoretically 3F needed to deposit 1mol Al metal
37.037 × 3F/mol = 37.037× 3× 96500 = 1.07 × 107 coulomb
or using the faraday's formula:
Wt = I × t × GEW × (1/96500)
1000 gm = Q × (27/3) × (1/96500)
Q = 1.07 × 107 C Producing Al is an expensive process.
The current Efficiency φ =
mnF
Q
φ =1.07×107/1.33 × 107 = 80%
Problems
1. Thirty minutes (30 m) of electrolysis of a solution of CuSO4 produced 3.175 g Cu
at the cathode. How many Faradays and how many Coulombs passed through the
cell? What is the current? ( Ans. Q= 9650 C, I= 5.36 A)
2. An electrolysis cell with Fe(NO3)3 solution is operated for 2.0 hrs at a constant
current of 0.10 A, how much Fe metal is plated out if the efficiency is 90%.(At.wt.
Fe=55.8). (Ans. 0.125 gm)
3. An electrolysis cell contains MSO4 solution is operated for 1.0 hr at constant
current of 0.200 A. If the current efficiency is 95%, and 0.399 g of M plates out,
what is the atomic weight of the element M? (Ans. Cadmium)
4. If a current of 1.00 A is used in the electrolysis of H2O, how many seconds will it
take to produce 22.4 ml H2 at STP? (Ans. 193 sec)