Electrochemistry Microtex
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Electrochemistry Definition

Electrochemical power sources or batteries are studied under the inter-disciplinary subject of Electrochemistry dealing with the reactions occurring at the interface of electronic conductors (active materials) and ionic conductors (electrolyte), production of electrical energy from chemical cells (or conversion of chemical energy into electrical energy)  and its reverse reaction where electrolytic cells are employed for chemical transformations. 

Electrochemical Power Sources (Batteries)

The energy conversion processes in batteries are based on the oxidation-reduction reactions (redox reactions).  The cells are classified into electrolytic cells and galvanic cells. Examples for electrolytic cells are cells used for extraction of metals like aluminium, magnesium etc. and batteries when being charged. Galvanic cells or batteries are capable of delivering current to us as opposed to electrolytic cells, into which we have to pass current for the reaction to occur.

Oxidation simply means the removal of electron/electrons (from anodes during a discharge reaction) and reduction is the process of addition of these electrons to the other electrode (cathode) through an external circuit, an ionically conducting electrolyte being the medium of ion transfer inside the cell.  During cell discharge, electrons pass from the anode (negative plate) to the cathode (positive plate) through an external circuit and ions flow inside the cell to convert chemical energy into electrical energy.

Typical examples for anode are:

Li → Li+ + e    

Pb → Pb2+ + 2e

Zn → Zn2+ + 2e

 

Examples of cathodes are:

PbO2 ⇄ Pb2+ +2e (Lead-acid battery)

LiFePO4 (Li-iron sulphate battery)

NiOOH + 2e  ⇄ Ni(OH)2 (Ni-cadmium battery)

Cl2 + 2e ⇄ 2Cl (Zinc-chlorine battery)

Br2 + 2e ⇄ 2Br (Zinc-Bromine battery)

Primary and secondary cells - electrochemistry

A cell is an independent unit of a galvanic system. When more than one cell is connected in either series or parallel fashion, this arrangement is called a battery. Essential components of a cell are Positive electrode or plate (cathode), negative electrode or plate (anode), electrolyte and other inactive components like container, separator, small parts like bus bars, pillar posts, terminal posts, etc.

Galvanic cells are classified into primary and secondary (or rechargeable or storage) cells.  In the primary cells, the reactions cannot be reversed once the discharge has come to an end due to exhaustion of the active materials, whereas in the secondary cells the active materials can be brought back to the previous status by passing current into the cell in the opposite direction.

Familiar examples of the primary cells are cells used in wristwatches, the electric torches and many controls like TV remotes and AC remotes. The ubiquitous lead-acid battery used for starting automobiles and home inverters/UPS and Ni-Cd, Ni-MH and Li-ion cells are examples for secondary batteries. Fuel cells differ from (primary) batteries in the sense that the reactive constituents are fed from outside, as against the availability of the same inside the batteries.

Potentials of electrodes (half cells) and voltage of a cell and Mass-independent entity of galvanic cells:

The potential (voltage) of an electrode is a fundamental electrochemical property and its value does not depend on the quantity of the electrode material. Thermodynamically it is an intensive property as against the capacity (which is an extensive property) of an electrode which depends on the mass of the active material it contains.

The voltage of a cell is the combination of two electrode potential or voltage values of the anode (negative electrode or plate) and the cathode (positive electrode or plate). The potential values of negative electrodes are always negative (lying below zero volts in the EMF series, See standards textbooks or handbooks). The zero volts refers to the standard electrode potential of hydrogen electrode (SHE).

The negative electrode materials are invariably metals or alloys, with a few exceptions like carbon and hydrogen, which are the negative active material in Ni-MH and Ni-H2 cells. The cathodes have positive potentials and they are mostly oxides, halides, sulphides etc., with the exception of oxygen which acts as cathode active material in metal-air cells. There should be an electrolyte to conduct ions inside the cell.

The voltage is the driving force for the current. It is a combination (algebraic difference) of the two values of the positive and the negative potentials. The voltage can be likened to the height of a water tank or the level of water in the tank and the current to the diameter of the pipe coming out from the tank. The higher the water level in the tank, the faster the water will come out. Similarly, the higher the diameter of the pipe, the more will be the volume of water that comes out.

How to determine the voltage of a cell?

The cell voltage can be determined from the two electrode potential values or it can be computed by using Gibbs equation and Standard Gibbs free energies of formation (ΔfG˚). The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of 1mole   of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at 1 bar of pressure and the specified temperature, usually 298.15 K or 25 °C).

Gibbs free energy (G)

In thermodynamics, Gibbs free energy is a measure of the work that can be extracted from a system and in the case of batteries, work is done by liberating ions at one electrode (anode) followed by the movement to the other (cathode). The change in energy is mainly equal to the work done, and in the case of galvanic cell, the electrical work is done through the motion of ions due to the chemical interaction between the reactants to give rise to the products.  Hence, the energy is given in terms of ΔG, the change in the Gibb’s free energy, which represents the maximum amount of chemical energy that may be obtained during the energy conversion processes.

Whenever a reaction occurs, there is a change in the free energy of the system:

∆G = – nFE°

where F = constant known as the Faraday (96,485 C or 26.8 Ah)

n = number of electrons involved in stoichiometric reaction

E° = standard potential, V.

The values of ∆G can be computed from the other three values, n, F and E.

The cell voltage of a galvanic cell can be computed from the expression

ΔG° = ΣΔG°f products – ΣΔG°f reactants    

The standard molar free energies of formation  can be obtained from standard text books [Hans Bode, Lead-Acid Batteries, John Wiley, New York, 1977, p.366].

PbO2 + Pb + 2H2SO4 ⇄ 2PbSO4 + 2H2O

ΔG° = ΣΔG°f products – ΣΔG°f reactants    

∆Gº      = [2(193.89) + 2(56.69)] [(52.34) + 0 – 2(177.34)] 

= 94.14 kcal / mole

= 94.14 kcal / mole × 4.184 kJ / mole

= 393.88 kJ / mole

Eº         = ΔGº/nF

= (393.88 × 1000)  / 2 × 96485

= 2.04 V

The corresponding increase in free energy is equal to the electrical work done on the system. Hence,

−ΔG = nFE or ΔG = −nFE and ΔGº   = −nFEº.

Cell voltage from electrode potentials

The combination of the two electrode potentials will give the cell voltage:

Ecell = Ecathode or positive electrode – E anode or negative electrode

Or E cell  = EPP  – ENP

 

According International Union of Pure and Applied Chemistry (IUPAC) conventions of 1953 and 1968, a galvanic cell is written in such a way that the right hand electrode (RHE) is the positive electrode where reduction occurs and the left hand electrode is the negative electrode, where oxidation occurs and the electrons flow from left to right [McNicol B.D; Rand, D.A.J in McNicol B.D; Rand, D.A.J (ed.) Power Sources for Electric Vehicles, Chapter 4, Elsevier, Amsterdam, 1984]. The RHE is the cathode and the LHE is the anode

 

Ecell = ERHE − ELHE

The values for electrode potentials can be had from Textbooks and Handbooks.

Cell voltage from electrode potentials for lead-acid cell

Ecell      = Ecathode or positive electrode – E anode or negative electrode

LHE Pb½H2SO4½H2SO4½PbO2 RHE

RHE is cathode E°Rev = 1.69 V for Pb4+ + 2e ⇄  Pb2+  and

LHE  anode E°Rev  = −0.358 V for Pbº − 2e _ Pb2+

Ecell      = 1.69 – (-0.358) = 2.048 V.

Cell voltage from electrode potentials for Ni-Cd Cell

RHE Cd|KOH|KOH|NiOOH LHE

LHE E°Rev  = 0.49 for NiOOH +2e ⇄Ni(OH)

RHE E°Rev  = – 0.828 V for Cd ⇄ Cd2+ +2e

Ecell                =0.49 V (0.828) = 1.318 V

 The E°Rev  of nickel electrode under standard conditions is 0.49 V. E°Rev  of the MH electrode depends on the partial pressure of the hydride-forming materials, according to  

2MH ⇄ 2M + H2

The preferred partial hydrogen pressure of MH electrode is of the order of 0.01 bar, E°Rev  ranges generally between –0.930 and –0.860 V.    So

Ecell                =0.49 V (0.89) = 1.3 V.

Cell voltage from electrode potentials for Li-ion cell of the LCO Chemistry

RHE C | LiPF6 in DMC +DEC +PC | LiCoO2 LHE

RHE E°Rev = 0.1 V (vs Li metal) for LiC6 ⇄ xLi+ + xe +  C6

LHE E°Rev = 3.8 V (vs Li metal)  for Li1-xCoO2 + xe  Discharge → LiCoO2

The total reaction is    C6 +LiCoO2 ⇄LixC6 + Li1-xCoO2     

Ecell           = 3.8 – (0.1) = 3.7 V.

Cell voltage from electrode potentials for Li-ion cell of the LiFePO4 chemistry

RHE C | LiPF6 or LiODFB in (EC+EMC+DEC) | LiFePO4 LHE

RHE E°Rev  = 0.1 V (vs Li metal) for LiC6 ⇄ xLi+ + xe +  C6

LHE E°Rev = 3.5 V (vs Li metal)  for FePO4  + xe + xLi = Discharge →   xLiFePO4 + (1-x) FePO4

LIODFB = Lithium difluoro(oxalato)borate

The total reaction LiFePO4 + 6C →LiC6 + FePO4

Ecell           = 3.3 – (0.1) = 3.2 V

Mass-dependent quantities of galvanic cells: Current, power and energy

Power is given in the unit of watts and time factor is not involved in power.

P = W = V*A

Energy refers to the power spent over a period of time and so the unit involves hours.

Energy                   1 W.Second = 1 Joule

Energy                   = Wh = W*h = V*A*h = 3600 joules.

1 kWh                    = 1000 Wh.

Capacity is the quantity of electricity (Ah) that a battery can deliver.

If any two of the terms in Wh or of kWh are given, the other can be calculated (Wh = VAh).

850 Wh of a 12 V battery can deliver 850 Wh/12 V = 71 Ah.  The duration over which this 71 Ah can be drawn depends not only on the current, but also on the type of chemistry. For example, a Li-ion battery, can deliver 70 A for 1 hour. But the lead-acid battery, on the on other hand, can stand up to 1 hour if the discharge current is 35 A. But, a VRLA battery can deliver 70A only for about a little less than 40 minutes.

The wattage delivered by a Li-ion cell at 70 A = 70 A*3.6 V= 252 W.

But the wattage delivered by a lead-acid cell at 70 A = 70 A* 1.9 V= 133 W.

One can see that the Li-ion cell can deliver more wattage on per-cell basis for the same current.

 

Similarly the energy delivered by a Li-ion cell at 70 A = 70 A*3.6 V *1h= 252 Wh.

But the energy delivered by a VR  lead-acid cell at 70 A = 70 A* 1.9 V * 0.66 h= 88 Wh.

We can see that the Li-ion cell can deliver more energy on per-cell basis for the same current

 

Specific capacity is Ah per unit weight (Ah/kg or mAh/g).

Specific energy is the Wh per unit weight (Wh/kg).

Energy density is the Wh per unit volume (Wh/litre).

 

Note:

The term gravimetric energy density has been replaced by specific energy and volumetric energy density by energy density

Electrochemistry - Theoretical Specific capacity and Theoretical Specific Energy of electrode active materials

The unit of electricity is coulomb, which is 1 ampere second (A.s). The Faraday constant (F) refers to the amount of charge carried by 1 mole of electrons. Since 1 electron has a charge of 1.602 x 1019 coulombs (C), one mole of electrons should have a charge of 96485 C/mole.

1 F = 1(6.02214 *1023) * (1.60218*10-19 C) = 96485 C (i.e. 96485 C/mole).

6.02214 *1023 is Avogadro number (Avogadro constant), which is defined as the number of atoms, moles or ions in one mole of that substance. It is useful in relating the mass of a substance to the number of particles in the substance. Thus, 0.2 mole of any substance will contain 0.2 *Avogadro number of particles.  The charge on an electron based on modern experiments is 1.60217653 x         10-19 coulombs per electron. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadro’s number of 6.02214154 x 1023 particles per mole [https://www.scientificamerican.com/article/how-was-avogadros-number/].

 

1 F 96485 C/mole = 96485 A.s/60*60 s = 26.8014 Ah/mole

Specific capacity and specific energy for lead-acid cell

The molecular weight or the atomic weight in grams divided by the number of electrons participating in the reaction gives the gram equivalent of the respective material. One gram equivalent will deliver 96,485 coulombs (most authors round it off to 96,500 C) which is equal to 26.8014 Ah.

207.2 g of lead metal can be equated with 2F electricity = 2× 26.8014 Ah = 53.603 Ah. (Reaction:   Pb →Pb2+ + 2e).

Therefore the amount of negative active material (NAM) in a lead-acid cell required for 1 Ah (which is known as capacity-density) = 207.2 / 53.603 = 3.866 g /Ah [Bode, Hans, Lead-Acid Batteries, John Wiley, New York, 1977, p.292.].

The reciprocal of the capacity density is called the specific capacity 

Specific capacity = nF / Molecular weight or atomic weight. (n= Number of electrons participating in the reaction).

The specific capacity of negative active material

The specific capacity of negative active material (NAM), Pb = 56.3/207.2 = 0.259 mAh /g = 259 Ah/kg. This value multiplied by the cell equilibrium potential is Theoretical Specific energyTheoretical Specific energy of NAM lead = 259*2.04 V = 528.36 Wh/kg

The specific capacity of positive active material (PAM)

Similarly, the amount of positive active material in a lead-acid cell required for 1 Ah (which is known as capacity density) = 239.2 / 53.603 = 4.46 g /Ah.

The specific capacity of positive active material (PAM), PbO2 = 56.3/239 = 0.224 mAh /g = 224 Ah/kg. Theoretical Specific energy of PAM lead dioxide = 224*2.04 V = 456.96 Wh/kg.

Lithium Ion cell

Specific capacity and specific energy for Li-ion cell carbon anode

Specific capacity of LiC6          = xF/n*Molecular Weight

= 1 * 26.8/ 1*72 mAh/g (Stoichiometrically 72 g of C is required for 1

mole of Li storage to form LiC6.  Since Li is available from LCO cathode, its mass is not taken into account of total anode mass. Only carbon is taken into consideration. X = 1; 100 % intercalation of Li+)

                                                = 0.372 Ah/g

                                                =  372 mAh/g = 372 Ah/kg

Specific energy  LiC6                = 372*3.7 V

= 1376 Wh/kg

Specific capacity and specific energy for LiCoO2 (LCO)

Specific capacity LiCoO2

= 0.5  Li+ + 0.5 e + Li0.5 CoO2 (x= 0.5, 50 % intercalation of Li+)

= xF/n*Mol Wt

                                                =0.5*26.8/ 1 * 98       Li= 6.94    Co = 58.93  2 O= 32

                                                      = 13.4 / 98 Ah/g  = 0.1368 Ah/kg

                                                =  137 mAh/g =  137 Ah/kg.

Specific energy of LiCoO2              = 137*3.7 V = 507 Wh/kg  (x= 0.5, 50 % intercalation of Li+)

If the value x is taken as 1, the specific capacity will be doubled, 137*2= 274 mAh/g = 274 Ah/kg

Specific energy of LiCoO2        = 274 *3.7 V (x= 1.  Full (100 %) intercalation of Li+)

= 1013 Wh/kg 

Specific capacity and specific energy for LiFePO4

 Specific capacity of LiFePO4   

=  xF/n*Mol Wt

=  26.8/157.75 = 169.9  mAh/g = 170 mAh/g = 170 Ah/kg

Specific energy of LiFePO4 = 170*3.2 V = 544 Wh/kg

Electrochemistry - Theoretical Specific Energy of a cell

The maximum specific energy derivable from an electrochemical power source is given by:

Theoretical Specific Energy = 26.8015× (nE/Σmoles)   Wh/kg   where n and E have their usual notations; n, the number of electrons participating in the reaction and E, the cell voltage.

Note

  1. Smoles refers to the summation of all  the reactants and one need not worry about the products
  2. Since the unit is given in Wh / kg (also written as Wh kg -1), the total weight is to be given in units of kg.

The specific energy lead-acid cell

A familiar example will be taken for calculation of theoretical specific energy.

First we have to write down the reaction and calculate the molar values of the reactants.  We need not worry about the products.  For lead-acid battery, the reaction is:

PbO2 + Pb + 2H2SO4   ⇄ 2PbSO4 + 2H2O                                  Eº = 2.04 V.

Σmoles    =  239 +207+ 2*98 in g

               = 0.642 kg

Theoretical Specific Energy       = 26.8× (nE/Σmoles)  Wh/kg 

                                                       = 26.8*(2*2.04/0.642) Wh/kg

                                                       = 26.8015*(6.3551) Wh/kg

                                                       = 170.3 Wh/kg.

According to Tobias Placke [J Solid State Electrochem (2017) 21:19391964], the specific energy can also be calculated as given below for lead-acid cell:

Specific energy of a cell = 

Specific energy in electrochemistry                                   

                                    =1[1/(224*2.04) + 1/(259*2.04) + 1/(273*2.04)]

                                    = 1[(1/457) + (1/528) + (1/557)]

                                    = 1/(0.002188 + 0.001893 + 0.001796)

                                    = 1/0.005877

                                    = 170 Wh/kg

The specific energy of Ni-Cd cell

2NiOOH + Cd ⇄ 2Ni(OH)2 + Cd(OH)2                                                  Eº = 1.33 V

Theoretical Specific Energy       = 26.8× (nE/Σmoles)  Wh/kg 

                                                = 26.8*(2*1.33/0.296) Wh/kg

                                                = 26.8015*(8.9865) Wh/kg

= 240.8  Wh/kg

The aqueous KOH electrolyte in these alkaline cells does not participate in the  cell reaction and

hence not taken into consideration while calculating the specific energy values. But, some authors

would like to include the weight of water in the calculation.

Then the figure for specific energy would come down to 214.8 Wh/kg if the Σmoles is substituted by

0.332. The result will be 214.8 Wh/ kg.

The specific energy of LiFePO4 cell

(x=1. 100 % intercalation)

 = 26.8015× (nE/Σmoles)   Wh/kg 

 = 26.8 [(1*3.2)/(72+157.75)     LiFePO4  +  6C + zero Li

= 26.8[(1*3.2)/(229.75)]     = 26.8*0.013928

=  0.37329        Wh/g

=  373               Wh/kg

The specific energy of LCO cell

(x=1;   100% intercalation)

 = 26.8015×   Wh/kg        169.87

= 26.8 [(1*3.7)/(72+97.87)]     LiCoO2 + 6C + zero Li

= 26.8 *[(3.7)/(169.87)]    

= 26.8 *0.02178

= 0.58377 Wh/g

= 584 Wh/kg

If x = 0.5 (50 % intercalation of Li ions), we have to replace 26.8 by half of this value, i.e., 13.4.  The result would be 584/2 = 292 Wh/kg.

Practical (Actual) Specific Energy of a cell/battery

https://pushevs.com/2015/11/04/gs-yuasa-improved-cells-lev50-vs-lev50n/ 

Real time Specific energy of a battery    =  (Mean voltage * Ah) / (Mass of the battery)

                                                            = (3.7 V*50 Ah1) / 1.7 kg (Yuasa LEV50 single cell)

                                                            = 185 /1.7

                                                            = 108.8 Wh/kg

                                                            = (14.8*50)/ 7.5 (Yuasa LEV50-4 battery)

                                                            = 98.7  Wh/kg

Real time energy density of a battery      = Wh/Volume = 17.1*4.4*11.5 = 865 cc

                                                            = 185/0.865 = 214 Wh / litre

                                                            = Wh/Volume = 17.5*19.4*11.6 = 3938 cc = 3.94 litre

                                                            = 14.8*50 / 3.94 = 187 Wh / litre

There is about 10 % reduction in specific energy when conversion occurs from cell to battery (Low kWh) and about 13 % reduction in energy density when conversion occurs from cell to battery (Low kWh)

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