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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

2. the interval current efficiency Eq-----3.10

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 ( =Iφ ) 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- 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- 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 half-

reactions 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- Cu(s)

Water is oxidized at the anode producing Oxygen gas according to the following half

reaction:

2H2O O2 + 4H+ + 4 e-

If 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 O2 + 4H+ + 4 e-

GEW 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- I2 + 2e-

Cathode: 2H2O(l) + 2e- 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 AlF6

3- + 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–

0.373 mol e– × ⎥⎦

⎤⎢⎣

⎡⋅emol

Almol31

= 0.124 mol Al metal

0.124 mol × 27 gm/mol = 3.357 gm Al metal

Energy consumption = φ

cellnFE− = 85.0

696500373.0 −××− = 254 kJ/mol = 0.0706 kW-h

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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037.3727

1000=

molgmgm mole of Al metal

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 Q

mnF=φ

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)