CE60017 - Sustainable Energy Technologies

Question 1

What’s a redox reaction?

Answer 1

A reaction involving electron transfer, with a species being oxidised and another being reduced.

Question 2

Define oxidation and reduction:

Answer 2

Oxidation: loss of e- and increase in oxidation state

Reduction: gain of e- and a decrease in oxidation state

Question 3

What are the equations for:

Li-ion batteries
Electrolysers
Fuel cells
CO2 reduction

Answer 3

Li-ion batteries
LiCoO2 + C6 → CoO2 + LiC6

Electrolysers
2H2O → 2H2 + O2

Fuel cells
2H2 + O2 → 2H2O

CO2 reduction
CO2+H2 → HCOOH

Question 4

Define anode and cathode:

Answer 4

During discharge, the positive is the cathode and the negative is the anode.
During charging, the positive is the anode and negative is the cathode.

Anode - where oxidation occurs

Cathode - where reduction occurs

Question 5

What’s the difference between galvanic and electrolytic cells?

Answer 5

Galvanic:
Spontaneous reactions
dG < 0
Electrons are spontaneously formed at anode
Electrons are supplied by the reaction

Electrolytic:
Non spontaneous
dG > 0
Electrons supplied to cathode to drive reaction
Electrons supplied by an external battery

Question 6

What is absolute potential?

Answer 6

The amount of electric potential energy carried by a unitary point charge located at a specific point.

Or

The Work needed to move a unit positive charge from infinity to a specific point

The absolute potential cannot be measured - we always use relative potentials, i.e. the difference between the potential at two electrodes

The most common “reference potential” is the Standard Hydrogen Electrode (SHE), which is the potential of the redox couple H+/H2

So, clearly, the potential at which hydrogen can be reduced or oxidized is 0V.

Question 7

How is cell potential calculated?

Answer 7

V.cell = V (reduction at cathode) - V (reduction at anode)

A reaction is spontaneous if
Vcell > 0

Question 8

Pros and cons of batteries:

Answer 8

Pros:
High Round-trip efficiency (i.e. Ratio between the energy you get out of a battery and the energy you put into it)
Easily scalable
Relatively high energy density

Cons:
Expensive
Energy and power density are intrinsically coupled
Self-discharge
Efficiency loss over time
Requirement of critical metals

Question 9

What’s a half cell and full cell?

Answer 9

A half-cell is a single electrode in an electrochemical cell, while a full cell is a complete electrochemical cell that consists of two half-cells connected by a salt bridge.
The electrode potential of a half-cell is determined by the energy required to move ions from the half-cell to the solution, and vice versa.

Looking at a half cell…
Testing the electrode of interest against the metal (for example lithium)
Since Li/Li+ has a potential of 0, it allows you to look at only the potential of lithium insertion at the electrode of interest
You have an abundance of Li available, which means you are not limited by it and can look at limitations at the electrode of interest only.

Looking at full cell…
Testing the battery with the cathode and anode of interest
The voltage we measured (cell potential) is affected by both anode and cathode.
The effect of anode and cathode cannot be deconvoluted
It measures the performance of a real battery.

Useful to find out how different parts of battery perform.

Question 10

Properties of coin cells:

Answer 10

Small area,
One cathode, one anode,
Single-sided electrodes,
Large void space requires more electrolyte,
Low currents.

Question 11

Properties of pouch cells:

Answer 11

Larger area,
Multiple stacked electrodes,
Often double-sided electrodes,
Minimal void space requires less electrolyte,
Higher currents.

Question 12

What is the difference between battery capacity (or charge), power, and energy?

Answer 12

Battery capacity (or charge): Total charge that can be stored in a battery (Ah)

Power: the power that can be delivered is the product of the current (measured in A) and the cell potential (measured in V)
Power is measured in W = V x A

Energy: the total energy stored in a battery is the integral of the supplied power over time
Energy if measured in Wh= V x Ah

1Wh=3600 J

Question 13

What is DoD and SoD?

Answer 13

Depth of discharge (DoD) = the amount of charge (capacity) extracted compared to the total amount (at the same discharge rate) – expressed in fraction or percentage

State of charge (SoC) = the amount of charge (capacity) still available to extract compared to the total amount (at the same discharge rate)

								SoC = 1-DoD

Question 14

How is state of charge (SoC) found?

Answer 14

SoC = 1-DoD

Depth of discharge (DoD) = the amount of charge (capacity) extracted compared to the total amount (at the same discharge rate) – expressed in fraction or percentage

State of charge (SoC) = the amount of charge (capacity) still available to extract compared to the total amount (at the same discharge rate)

Question 15

How is theoretical charge (in mAh/g) calculated?

Answer 15

Q = nF/3.6M

Where:
n - number of electrons
F - Faraday constant (96485 C/mol)
M - molecular mass
3.6 is the conversion factor, 1 C = 0.28 mAh

For example, if CoO2 is used as an electrode for Li –ion batteries, the reaction is
LiCoO2→Li+ +e− + CoO2
Each LiCoO2 (M=98g/mol) can store 1 electron (n=1) so the theoretical capacity is:

Q=1*96485/3.6/98= 274 mAh/g

By convention, when calculating the theoretical capacity of the cathode, we include the weight of litihium, when doing it with the anode we don’t.

Question 16

How is power calculated?

Answer 16

P = IV

(Then energy (E) = capacity (Ah) * cell potential (V))

Question 17

How is gravimetric energy density calculated?

Answer 17

Gravimetric energy density = cell potential (V) * gravimetric capacity (Ah/g)

The gravimetric capacity can be maximized by finding lighter materials, that can store a charge with a smaller weight.

The cell potential can be maximized by choosing cathode and anode materials with the biggest potential difference possible.

Question 18

What is cell potential?

Answer 18

The cell potential is determined by the difference in potential between the anode and the cathode.

To maximize the cell potential, we should choose the anode with the lowest possible voltage and the cathode with the highest possible one.

Question 19

How are batteries tested?

Answer 19

Electron flow is provided and the electric potential is measured.

Batteries are tested galvanostatically, i.e. at constant current (i.e. With a constant flow of electrons) and the potential is measured.

In an ideal battery:
- The cell potential is equal to the standard cell potential (4.1 in this case)
- The cell potential is constant over time
- The charge potential is equal to the discharge potential (4.1V in this case)
- What current we apply while measuring the potential (i.e. How fast we charge/discharge) doesn’t affect the total charge.

Once the battery has been fully charged or discharged, the potentiostat will keep changing the potential to maintain the current, this current will not be the effect of charging discharging the battery anymore, but will come from degradation of the battery. For this reason, we need to stop the experiment before this happens, using a CUT-OFF potential.

Question 20

How does the charge-discharge curve behave for a real battery?

Answer 20

The cell potential is lower than the theoretical value during discharge.
You get less energy out of the battery.
The difference between theoretical and real is called overpotential.

The cell potential changes with the state of charge (or time)
What current we apply while measuring the potential (i.e. How fast we charge/discharge) has an impact on the total charge.

Question 21

What is overpotential?

Answer 21

The difference between the theoretical cell potential and the experimental one.

Question 22

Why are there differences between electrode potential / what causes overpotential?

Answer 22

Polarisation losses V=E-iR
(Polarization losses are proportional to the current we draw (i), and are caused by the resistance of the electrolyte to the flow of ions.
The proportionality constant is R, also known as ohmic resistance)

Electrode / Activation overpotential
(The activation overpotential is characteristic of an electrode and is due to kinetics limitation to the charge transfer process
(i.e. Kinetics limitation for Li+ to become LiC6))

Electrode overpotential - concentration overpotential
(The concentration overpotential is due to depletion of the reactant next to the electrode surface.
It is essentially a measure of the «extra driving force» needed to transport the reactant (Li+) to the surface))

Question 23

What is the C-rate?

Answer 23

How fast we can charge/discharge a battery.

C-rate = 1 / time to charge or discharge

The higher the C-rate (i.e. the faster we discharge) the shorter the duration of discharge will be.

For an ideal battery, although the time is different, the capacity (i.e. the product of current and time) should be the same
In reality, as we will see, the capacity decreases the faster we charge/discharge.

Question 24

How is coulombic efficiency found?

Answer 24

Q (charge) / Q (discharge) *100

Question 25

What is coulombic efficiency?

Answer 25

A measure of the capacity that you can get out of a battery, compared to the capacity that you put in during charge
An important consideration when balancing electrode capacities, and for considering overall specific energy or capacity of the cell.

Question 26

What is voltaic efficiency?

Answer 26

The ratio between the average discharge voltage to the average charge voltage (resistance of the cell, polarisation losses)

= E (discharge) / E (charge) * 100

Question 27

How is battery efficiency found?

Answer 27

= QE (discharge) / QE (charge) *100
= voltaic efficiency * coulombic efficiency

It is the ratio of the energy retrieved from thebattery, to the energy provided to thebattery, when coming back to the same SOC state = Columbic efficiency X Voltaic Efficiency

Question 28

How is it known if a reaction will be spontaneous?

Answer 28

A reaction is spontaneous if it results in an increase in free energy.

This is marked by a negative dG value in the Gibbs Free Energy Equation.

Since ΔG=-nFVcell…
A reaction is spontaneous if (cell potential) Vcell > 0

Question 29

What does the reduction potential of a half reaction tell us?

Answer 29

The tendency for a material to be reduced.

The potential for Fe 3+ to Fe 2+ is more positive than Fe 2+ to Fe. This means that Fe 3+ has a greater tendency to gain electrons and be reduced to Fe 2+ compared to Fe 2+ to be reduced further to Fe.
Under standard conditions, the reduction of Fe 3+ is favourable.

Metal ions at the top of the series are good at picking up electrons. They are good oxidising agents. (for example Cu2+ is good at picking up electrons and becoming Cu) The oxidising ability of the metal ions increases as you go up the series.
Metals at the bottom of the series are good at giving away electrons. They are good reducing agents. (for example Fe is good at giving away electrons and becoming Fe2+)
The reducing ability of the metal increases as you go up the series.

Question 30

Define:

Specific energy
Energy density (or volumetric energy density)
Specific power
Specific volume
Gravimetric Battery Capacity

Answer 30

Specific energy (or gravimetric energy density)= the stored energy per unit mass ( Wh/Kg)

Energy density (or volumetric energy density) = the stored energy per unit volume ( Wh/l)

Specific power = power per unit mass ( W/Kg)

Specific volume =power per unit volume ( W/L)

Gravimetric Battery Capacity = Ah/g or mAh/g

Question 31

How can the gravimetric capacity and cell potential of a battery be maximised?

Answer 31

The gravimetric capacity can be maximized by finding lighter materials, that can store a charge with a smaller weight.

The cell potential can be maximized by choosing cathode and anode materials with the biggest potential difference possible.

Question 32

What is open circuit potential?

Answer 32

The potential difference between the electrodes when not drawing any current, this should be equal to the theoretical cell potential.
In reality, the potential we measure is different, it has an overpotential.

Question 33

Why do we use Li-ion batteries?

Answer 33

• Li is lightweight
• High power density
• Good energy density compared to other batteries
• Great cyclability
• More environmentally friendly than lead acid

All the elements on the first group have a single electron on the outer electron sphere, therefore:
- Removing the electron is “easy” and reversible
- They have an oxidation state of +1, meaning they can store one charge and being transported across a battery as +1 ion.

This means Li, Na, K are all good candidates
Mg and Ca even more so, as they can store 2 charges per atom
Out of all of them, Li is the lightest!

All other battery chemistry will intrinsically be heavier and therefore have a lower energy density.

Question 34

What are the main components of a battery?

Answer 34

1. Current collectors (conductive materials e.g. Cu or Al connected to the electrodes (anode and cathode) that allow the flow of electrons to the external circuit)
2. Electrodes (anode often made of graphite and cathode made of a metal oxide)
3. Electrolyte (medium allowing ion flow - can be solid, liquid, or a gel)
4. Separator (permeable membrane that prevents direct contact between the anode and cathode while allowing the flow of ions)
5. Container (holds all the components together and provides structural support and protection for the battery)
6. Terminal (points of connection where the battery is connected to an external circuit to deliver or receive electrical energy)

Question 35

Why do we need separators in batteries?

Answer 35

To separate the anode from the cathode to keep them electrically separated.

Must be porous to allow Li ions to pass back and forth between the anode and the cathode

Must be insulating to electrons.

Can be made from:
- Thin polymer films
- Ceramic
- Ceramic/polymer blends

Question 36

What are current collectors typically made from?

Answer 36

Copper or carbon-coated Aluminium

Choice based on costs and compatibility.

Question 37

What are binders in batteries?

Answer 37

Binders are materials that hold together the components within the battery cell, maintaining the structural integrity of the electrodes and ensuring proper electrical contact between the active materials and current collectors.

Binders are typically polymers that possess adhesive properties and chemical stability within the electrolyte environment of the battery. They help to immobilize the active materials (such as lithium cobalt oxide, graphite, etc.) onto the current collectors, preventing them from detaching or undergoing undesirable reactions during charge/discharge cycles.

Common binder materials used in lithium-ion batteries include polyvinylidene fluoride (PVDF) and its copolymers, such as PVDF-HFP (hexafluoropropylene).

Question 38

What do we look for in an electrolyte?

Answer 38

The main requirement of an electrolyte is that it needs to be stable at the potential of the cathode and anode.

Question 39

What does SEI stand for, regarding batteries?

Answer 39

Solid Electrolyte Interface

The SEI is the layer of electrolyte degradation products that is formed at the surface of the electrode
The SEI can be electrically conductive and/or ionically conductive.

The Solid Electrolyte Interface (SEI) is a protective layer that forms on the surface of electrodes in lithium-ion batteries, consisting of electrolyte decomposition products. It prevents further electrolyte decomposition and enables stable lithium-ion transport while allowing efficient battery operation.

Question 40

What happens if the solid electrolyte interface (SEI) is electrically conductive?

Answer 40

Electrons will be transported to the new interface, where the electrolyte will keep degrading indefinitely.

Question 41

What happens if the solid electrolyte interface (SEI) is not electrically or ionically conductive?

Answer 41

If a totally insulating layer is formed the degradation reaction stops, but so does the desired lithium reaction, so the battery is dead.

Question 42

What happens if the solid electrolyte interface (SEI) is ionically conductive, but not electrically conductive?

Answer 42

This is the ideal scenario.
Since the electrons cannot be transport through the SEI, they can’t reach the electrolyte and further electrolyte degradation is not possible. At the same time, Li-ions can travel through the SEI to meet the electrons at the electrode surface and react with the electrode, so the battery is still working.

Question 43

What is the ideal scenario for an battery SEI?

Answer 43

The solid electrolyte interface being ionically conductive but not electrically conductive is ideal.

Since the electrons cannot be transport through the SEI, they can’t reach the electrolyte and further electrolyte degradation is not possible.
At the same time, Li-ions can travel through the SEI to meet the electrons at the electrode surface and react with the electrode, so the battery is still working.

Question 44

How is the SEI formed?

Answer 44

The Solid Electrolyte Interface (SEI) forms on the surface of lithium-ion battery electrodes through a series of electrochemical reactions between the electrolyte and the electrode materials during the initial charging and discharging cycles.
These reactions result in the deposition and rearrangement of electrolyte decomposition products, creating a thin, passivating layer that protects the electrode surface from further degradation and promotes stable battery performance.

The SEI is typically formed in the first cycle of the battery by decomposition of electrolyte at the electrode surface. The formation of the SEI prevents further degradation, but changes the energy levels.

It consists of many organic and inorganic compounds with thickness of around 20 nm.

Composition and structure depends on cathode, anode, electrolyte and cycling conditions used.

Low Coulombic efficiency due to irreversible Li consumption. Efficiently prevents further chemical reactions between electrolyte and anode.

Question 45

What is LUMO and HOMO?

Answer 45

Lowest unoccupied molecular orbitals (LUMO) and highest occupied molecular orbitals (HOMO) of the electrolyte.

If μA of anode > LUMO of electrolyte
The electrolyte is reduced at the anode

If μC of cathode < HOMO of electrolyte
The electrolyte is oxidized at the cathode

Electrolytes have a stable operating voltage ‘window’ in which they can be cycled without degrading.
This ‘window’ is the energy gap between the lowest unoccupied molecular orbitals (LUMO) and highest occupied molecular orbitals (HOMO) of the electrolyte.

(μ is the Fermi level - A and C refer to anode and cathode)

Question 46

Describe the ideal solid electrolyte interface (SEI):

Answer 46

1. High ionic conductivity to reduce Li ion diffusion resistance.
2. Appropriate thickness to allow easy diffusion of Li ions into the electrode, but prevent further electrolyte decomposition.
3. Robust mechanical performance to accommodate non-uniform volume fluctuation, and withstand repeated Li deposition and plating processes.
4. Excellent stability with respect to structure, shape, morphology and chemistry during long-term cycling performance.

Question 47

Describe properties of the anode:

Answer 47

Graphite is the most common anode for Li-ion batteries

Lithium in stored via intercalation between the graphene sheets

It can store one lithium atom per 6 carbons: LiC6, with a theoretical capacity of Q=370 mAh g-1

Question 48

What is the issue with having metallic Li anodes?

Answer 48

Metallic Li anode in the rechargeable battery is plated/stripped repeatedly upon cycling.

The initially present nucleation sites and positions contribute an imperative role in the following deposition nature/behaviour of Li.

Protrusions with large curvature possess a significantly higher electric field at their tip sites that are prone to attract more Li ions for deposition, creating more local inhomogeneity.

With repeated cycling, these finally grow into dendritic morphologies.

On repeated recharge, dendrites grew across the electrolyte from the anode to the cathode, leading to dangerous short-circuits in the cell in the presence of the flammable organic liquid electrolyte.

Question 49

How can metallic Li anodes be protected?

Answer 49

The following approaches have been investigated to control the different reactions occurring on the lithium surface to suppress the formation of dendritic growths.

Electrolyte additives: salts, reactive organic compounds
Artificial SEI
Solid Electrolytes
Structuring the electrode architecture
Polymer coatings
Coatings with carbon materials

Question 50

What does structuring the electrode architecture involve?

Answer 50

Providing a conductive 3D host structure which can stabilise Li metal deposition by:
- Creating a larger, favourable surface area for Li nucleation and growth,
- Providing porosity to accommodate Li metal growth,
- Regulating localised current density in the electrode.

It is done to protect the Li anode and prevent dendritic growths.

Carbon is often chosen as it is cheap, lightweight, safe, and conductive.

Question 51

How are electrolyte additives (salts, reactive organic compounds) used to protect metallic Li anodes (from dendritic growths)?

Answer 51

Electrolyte additives can react with the metallic Li anode to form a more robust and passivating SEI film than that formed by Li salts in pure organic solvents, inhibiting further reaction between surface of Li and electrolyte. A more chemical and stable SEI can be formed.

Salt anions in nonaqueous electrolytes strongly affect the surface chemistry of metallic Li anodes.

The inner layer of SEI close to the surface of Li metal consists of inorganic salts. The composition is determined by the reduction of the anions present in solution.

LiPF6, LiSO3CF3, LiBF4 and LiTFSI salts are typically the more reactive salts with Li surface and consequently adversely influence the cycling efficiency of Li than LiClO4. The thicker interphase layers formed may lead to higher resistance, and decreased cell performance.

As the availability of salts/solvents is limited, additives that possess a higher reduction voltage compared to electrolyte salts and solvents can be employed, to contribute to stable SEI formation.

These additives are normally added in small amounts (ppm level) to the electrolyte.

Question 52

Compare solid and liquid electrolytes:

Answer 52

Liquid:
- Poor chemical stability
- Leakage
- Flammable
- Dendrites form easily
- High ionic conductivity

Solid:
- Safe
- Chemically stable
- No leakage
- Prevent dendrite formation
- Poor ionic conductivity
Higher ionic conductivities, but brittle, and poor interfacial contact with electrodes.

Question 53

What are the 2 main types of solid electrolyte?

Answer 53

Ceramic
Polymer
Can combine the two.

Can be true solid polymer, or gel polymer electrolytes.

PEO based.
Poly-ionic liquids.
Biopolymers.

Good interfacial adhesion with electrodes.
Poor thermal stability.

Question 54

How does the formation of an artificial SEI help protect metallic Li anodes?

Answer 54

Essentially the formation of an artificial SEI:
By coating with a protective layer of e.g. carbon or CN.
Introducing a thin layer of SE on the electrode.
By inducing artificial SEI formation first before assembling the battery.

This prevents the formation of dendrites.

Question 55

Why do Li-S systems have such high capacity?
What is the redox eq for this?

Answer 55

A single S atom can store 2 Li atoms, and sulphur is quite lightweight.

S + 2Li+ + 2e- ⇌ Li2S
(E° ≈ 2.15 V vs Li/Li+)

Overall conversion occurs via
formation of polysulfides Li2Sx, 8<x<1)

Question 56

What are the advantages of Li-S batteries?

Answer 56

High theoretical capacity (1675 mAh g-1),
High theoretical specific energy (2500 Wh kg-1),
Current best is 500 Wh kg-1, compared to Li-ion 150-250 Wh kg-1,
Relatively lightweight,
Sulfur is cheap

(Likely to enter the market where mass is the critical factor above all else, e.g. unmanned aerial vehicles, space and automotive (buses and trucks, rather than consumer vehicles where volume is a constraint).

Question 57

What are disadvantages of Li-S systems?

Answer 57

Many reaction steps needed to form Li2S
(S8 → Li2S8 → Li2S6 → Li2S4 → Li2S3 → Li2S2 → Li2S and many more)

All reaction intermediates are soluble, which means they can enter the electrolyte.

Polysulfide shuttle effect decreases capacity and rechargeability

Large volume expansion from S to Li2S,
Slow kinetics from Li2S2 to Li2S

Uncontrolled Li2S precipitation

Large amount of electrolyte needed

Not compatible with conventional organic carbonate electrolytes

Extremely low electrical conductivity of sulfur (5 × 10-30 S cm-1 at 25°C)

Formation of lithium dendrites

Question 58

How can issues with Li-S systems be managed?

Answer 58

1. Using separators

A functional separator interlayer can:
- Physically block or trap polysulfides to minimise the shuttle effect by tailoring nanostructure,
- Chemically trap polysulfides by introducing functional groups or dopants,
- Contain catalysts to accelerate polysulfide conversion to reduce shuttling.

2. Using carbon-sulphur composite cathodes
   Carbon provides:
   - Conductivity,
   - Structural scaffold,
   - Trapping sites for S and polysulfides,
   - Catalytic/functional surface.
   - As cathode conversion reactions involve dissolution and precipitation of active material species, electrolyte plays a much more active role than just providing ionic transport between electrodes.

Question 59

How is energy density (Wh/kg) of a Li-S system calculated?

Answer 59

E = Q.sVM.s / M.tot

Question 60

How can specific energy in Li-S batteries be maximised?

Answer 60

Maximise sulfur loading:
Higher sulfur loading increases the areal capacity of the cathode (mA h cm2),
Increased fraction of active material offsets the ‘dead weight’ of inactive components,
However, the coating and drying processes are problematic for thick electrodes, and thicker electrodes result in increased ionic and electronic resistance, and therefore lower rate capability.

Reduce electrolyte loading:
Electrolyte contributes the highest weight and volume fraction of the cell components and has a huge impact on specific energy at cell level.

Reduce/eliminate weight of other inactive components (e.g. current collector, binder).

Question 61

How is sulphur utilisation found?

Answer 61

Sulfur use = m (s used) )/ m (s total) =Q (practical) / Q (theoretical)

Sulfur utilisation can be affected by many factors including porosity, surface area, morphology, electrolyte loading.

Achieving high sulfur loadings is not necessarily always desirable, if full conversion to Li2S does not occur:
- Unutilised sulfur can cover and passivate the carbon surface, blocking charge transfer reactions,
- Unused sulfur adds mass to the cell without contributing capacity, decreasing energy density.

Question 62

Why is water a bad electrolyte for Li batteries?

Answer 62

• Highly reactive with Li
• Water can corrode the electrodes and other components of the battery. Lithium reacts with water to form lithium hydroxide and hydrogen gas.
• Pure water has poor conductivity for ions compared to other electrolytes commonly used in Li-ion batteries, such as lithium salts dissolved in organic solvents like ethylene carbonate (EC) or dimethyl carbonate (DMC).

Question 63

Why may electrolyte additives be used?

Answer 63

Flame retardant

SEI formation improvement and improved recyclability by suppressing anodic decomposition

Cathode protection

Improve thermal stability

Improve wettability

Improve corrosion resistance

Question 64

How may carbon materials be used as supports for sulphur cathodes in rechargeable batteries?

Answer 64

Physical Confinement:
1. Hierarchical Porosity: Refers to the structure of the carbon material, which contains pores at different length scales (macro, meso, and micropores). This structure helps accommodate the volume changes that occur during the charge-discharge cycles of the battery.

2. Yolk-Shell Structure/Encapsulation: This involves enclosing the sulfur within a carbon shell (yolk-shell structure), which helps to prevent the dissolution of sulfur into the electrolyte and improves the stability of the electrode.
3. Capillary Absorption of Sulfur Species Through Porosity: The porous nature of the carbon material allows for the absorption and retention of sulfur species, which is beneficial for the battery’s performance.
4. Polymeric Coating: Coating the carbon material with a polymer can further enhance the stability and performance of the electrode by providing additional protection and preventing the dissolution of sulfur.

Chemical Confinement:
1. Heteroatom Doping (N, O): Introducing heteroatoms such as nitrogen (N) and oxygen (O) into the carbon structure can modify its properties, such as increasing its conductivity and enhancing its interaction with sulfur species.

2. Catalyst Decoration/Metal Additives: Adding catalysts or metal additives to the carbon material can improve the kinetics of the electrochemical reactions, leading to better battery performance.
3. Directed Nucleation: Controlling the nucleation of sulfur species on the carbon surface can help prevent the formation of undesirable products and improve the stability of the electrode.

Question 65

What are the 3 main points of the energy trilemma?

Answer 65

Sustainability
Energy equity
Energy security

The 2 main options to increase sustainability are battery recycling and different battery chemistry.

Question 65

Why use Na ion batteries?

Answer 65

Na itself is cheaper and more abundant

We can use Al instead of Cu as current collector (more abundant, cheaper, and lighter, though less conductive)
[Potential to provide energy at 23/kWh/annum, less than half the cost of LIBs and a quarter that of lead-acid systems of the same size.]

The electrode materials can be cheaper and more sustainable

It is similar to Li-ion batteries (Same manufacturing and technology for development)

Not as good as lithium in terms of performance, but second lightest atom in the 1st group and Sodium’s standard redox potential is only 340 mV higher than Li+/Li.

Question 66

Describe Na ion battery performance:

Answer 66

Sodium has:
Higher standard redox potential
Higher ionic radius
Is heavier than lithium?

Cell potential is lower to overall power will be lower.
Larger ionic radius makes it harder for diffusion.
Regarding insertion into electrode material, it will be harder as the ions are bigger and bulkier.

[all compared to Li]

Sodium’s higher standard redox potential (Na -2.71 V vs Li -3.04 V) and larger ionic radius (Na+ 1.02 Å vs Li+ 0.76 Å) generally mean that the capacity and voltage achieved will be lower, therefore projected energy density will be lower.

Question 67

Pros and Cons of Na ion batteries (compared to Li):

Answer 67

Advantages of Na-ion:
Sodium is more abundant and cheaper than lithium
Can use hard carbon anions, instead of graphite used in Li-ion.
Do not require cobalt in the cathode
Can replace Cu with Al as current collector
Overall cheaper

Disadvantages of Na-ion:
Na-ion batteries have a shorter cycle life than Li-ion
Lower gravimetric energy density
Lower volumetric energy density
Graphite is a widely-used anode material for lithium-ion batteries. However, Na-ions do not intercalate into graphite due to the larger ion size.
NIB electrodes show slightly lower specific capacities than those for LIB, resulting in lower energy density. Therefore, the cost reduction is not significant in terms of the cost per energy.

Question 68

What are Na ion batteries good for?

Answer 68

Large-scale stationary energy storage and low-speed electric vehicles, despite lower energy density (than Li).
This is due to the lower cost and greater abundance.

To be competitive with Li-ion, improvements must be made in terms of cycle life and they must be easy to manufacture at larger scale.

Anode materials with high specific capacities and appropriately low redox potentials to improve the energy density of NIBs could replace LIBs.

Question 69

What may tin, Sn, be used for in batteries?

Answer 69

Metallic Sn is one of the most intensively investigated anode material for room-temperature NIBs, due to its ability to alloy up to 3.75 Na per Sn, corresponding to the high theoretical capacity of 847 mA h g−1.

Sodiation (replacement of metal ions (typically lithium) with those of sodium) of Sn is a multi-step alloying process.

However Sn prone to huge volumetric expansion (420%) which causes cracking and capacity fading.

Question 70

How can tin anodes be protected to avoid damage from volume expansion (which leads to pulverisation and loss of electrical contact)?

Answer 70

Can coat Sn particles with various protective films, introduce additives into electrolyte e.g. FEC.

Particle size control - smaller particles can better accommodate volume expansions

Dispersion in carbon matrix -: for examples depositing tin on carbon nanofibers

Particle encapsulation - for example in a bigger carbon particle, so that if the Sn expands it still has space

Co-alloying to buffer volume changes - Other alloying materials: Sb, P, Ge, Bi, Si

Structured electrodes to accommodate volume change

Question 71

Pros and cons of sodium titanates, used for anodes in Na batteries:

Answer 71

Pros:
Abundant and low cost,
Non-toxic,
Stable,
Low operation voltage,
Low strain,
Decent cyclability depending on the specific compound,

Cons:
Low specific capacity,
Low electrical conductivity,
Poor ion diffusivity.

Question 72

How do Na ion battery cathodes store ions?

Answer 72

Most cathode materials store Na+ions by intercalation chemistry, which means that the number of storage sites is limited.

This suggests that it will be difficult to greatly increase the specific capacity of the cathode materials.

The use of cathode materials with high redox potentials is also limited because of electrolyte decomposition at high potentials.

Most commercialized electrolytes even for Li‐ion batteries are unstable and decompose at over 4.8 V versus Li/Li+.

Question 73

What are the 3 main Na ion battery cathode families?

Answer 73

Oxides

Polyanions

Prussian Blue Analogue (PBA)

Question 74

How do layered oxides compare to polyanions for Na ion battery cathodes?

Answer 74

Oxides were initially pursued because of their dominance in Li-ion batteries.

Na-based layered oxides show even richer crystal chemistry than Li-ion ones because Na ions can reside in more lattice sites.

However, they are more prone to structural phase transitions, reducing lifetime and limiting power density.

Meanwhile with polyanions, there is power switching from Li to Na for polyanion compounds due to open channels for ion diffusion.
However, it relies on the use of toxic vanadium.

Question 75

How does the solid electrolyte interphase (SEI) compare for LIBs and NIBs?

Answer 75

Cycling performance of NIBs still doesn’t compare to LIBs, implying continual consumption of electrolyte and an unstable SEI:

• Less compact layer,
• More resistive,
• Mechanically less robust,
• More soluble?

Factors affecting the SEI:
Type of solvent,
Type of salt,
Charging rate,
Temperature of formation.

Question 76

Comment on the use of K-ion batteries:

Answer 76

Similar redox potential to Li+ (-2.94 vs -3.04 V).
Larger ion size, lower energy density.
Fast ion diffusion – potentially faster charging batteries.
K+ can intercalate into graphite (unlike Na+) ~279 mAh g-1.
Combining graphite with soft carbon provides structural protection enhancing cycling behaviour.

Challenges:
Low capacity retention.
Need rigid cathode structures that can accommodate stable and repeatable K+ ion insertion.
High Lewis acidity of potassium results in fast electrolyte degradation.
Need electrolytes that can operate in the higher voltage window.

Question 77

What are the main steps in battery recycling?

Answer 77

Disassembly and discharging (onto direct recycling involving delamination and regeneration)
Shredding
Mechanical separation (onto indirect recycling involving hydrothermals, drying, and relithation)
Hydrometallurgy (leaching, extraction, Li precipitation)
Pyrometallurgy (heating and slag separation)

Question 78

Describe discharge in battery recycling and its pros and cons:

Answer 78

Discharge is the process of closing the battery circuit, allowing it to reach a stable open circuit voltage, to avoid short-circuiting later on.

Pros:
Takes electricity out of cells making them safe
Mature and safe
Electricity can be returned to the grid
Ni need for Li neutralization

Cons:
Only valid for large modules and packs
Li redeposition can occur if discharged too fast
Prone to thermal run away

Question 79

Describe shredding in battery recycling and its pros and cons:

Answer 79

In the indirect route, the cells are shredded and mechanically separated to obtain a black mass, composed of cathode and anode material

Pros:
Easy and fast to do

Cons:
Purity of recovered material is intrinsically lower

Question 80

Describe freezing in battery recycling and its pros and cons:

Answer 80

Another pre-treatment process is freezing, which consists of putting the cell in liquid nitrogen to freeze the electrolyte.

Pros:
Frozen in liquid N2 where the electrolyte becomes unreactive

Cons:
Only suitable for batch processes
Scale up difficult
Expensive reagents needed
Short processing time until the cells become active again

Question 81

Describe pyrometallurgy in battery recycling and its pros and cons:

Answer 81

In the pyrometallurgy route, the black mass (which contains all the active cathode and anode material) is exposed to high temperature, to remove carbon, binders, solvents etc.
What is left is a lithium-rich slag and an alloy containing Ni, Co etc.

Pros:
Allows charged cells to be processed
Burns off the electrolyte taking away the biggest hazard
Mature and scalable
No post process neutralization
Makes the hydrometallurgical process easier

Cons:
Inert atmosphere required-costly
Large quantity of toxic gases and extensive scrubbing is needed to treat them
vaporized electrolyte forms explosive mixture
Low recovery efficiency and you can’t recover electrolyte and graphite, as they combust
Large input of energy required
CO2 emission via combustion processes
Al current collectors will melt at these temperatures

Question 82

Describe hydrometallurgy in battery recycling and its pros and cons:

Answer 82

Hydrometallurgy is a leaching process, done with a variety of acid or alkaline solutions.
By adding other chemical precipitants or extractants, metals such as Co, Mn and NI can be separated via precipitation, extraction or adsorption.

Pros:
Higher recovery purity of active materials
Lower energy consumptions and lower emission of toxic gases compared to pyrolymetallurgy

Cons:
Highly corrosive solvents needed for the extraction
It requires longer pre-treatment compared to direct recycling

Question 83

Describe direct recycling in battery recycling and its pros and cons:

Answer 83

In the direct route, the battery cells are disassembled down to the individual electrodes. The separated electrodes can be then subject to delaminiation process (for example soaking in organic solvent) to separate the active material from the current collector.

Pros:
Higher Purity of the materials recovery
Higher collection efficiency
Lower energy consumption
Fewer chemicals required

Cons:
Longer pre-treatment process
It needs cells to be designed for easier disassembling

Question 84

What is the future of Li battery recycling:

Answer 84

Materials:
- Biodegradable binders
- No current collectors
- No critical materials
- Single crystal cathodes

Electrode/cell:
- Aqueous formulations
- Standardisation
- Digital materials labels
- Screw cap cells

Pack/module:
- No glues
- Design for disassembly
- Cell labels with components
- Easy lock release

Question 85

What is grey, blue, and green hydrogen?

Answer 85

Grey: Hydrogen from reforming and gasification technology using natural gas and coal, with emission significant amount of CO2 in the atmosphere

Blue: Similar to grey but with incorporation of CCS. Advanced reforming technology
producing blue hydrogen from
natural gas and biomass, with most CO2 emissions captured by carbon capture and storage (CCS).

Green: Electrochemical or photochemical technology using electricity or solar power to produce green hydrogen, which is highly flexible and stores hydrogen by capturing excess generation.

Question 86

Compare grey, blue, and green hydrogen:

Answer 86

Grey is from reforming and gasification of fossil fuels. Blue is similar but with CCS. Green is from electricity/solar power (e.g. electrolysis).

Grey hydrogen is the cheapest option (€1.50/kg), but CO2 emissions may make grey hydrogen more costly

Blue hydrogen can narrow the gap

Green hydrogen it the cleanest but it’s price depends on renewable electricity and electrolysis system

Question 87

What are the main drivers for switching to electrochemical hydrogen production?

Answer 87

• Use renewable energy (wind, solar, etc)
• Fossil fuel free
• No CO2 emission during the process
• Low temperature, low pressure
• Decentralised, on-demand on-site chemical synthesis
• Fast start-up and shut-down

Question 88

What are the issues with implementing a hydrogen economy at this stage?

Answer 88

High cost in H2 production

Challenges in Safe Storage

Transportation and distribution

Social awareness

Competition with other energy
Sources and storage systems

Question 89

What are the 2 main hydrogen storage methods?

Answer 89

1. REVERSIBLE ON-BOARD APPROACHES
   Compressed gas
   Liquefied hydrogen
   Metal Hydrides
   High surface area porous materials
2. CHEMICAL HYDROGEN STORAGE: REGENERABLE OFF-BOARD
   Hydrolysis reactions
   Hydrogenation/dehydrogenation reactions
   Ammonia
   Boron hydrides

Question 90

What are the limitations with using compressed hydrogen gas for H2 storage?

Answer 90

Volumetric capacity (0.039 kg/L at 700 bar)
Limits of high pressure
Costs
Refueling or filling time
Compression energy penalty (15-20% less than atmospheric pressure hydrogen)
Heat management requirements

Question 91

What are the advantages and limitations of storing H2 as a liquid?

Answer 91

Advantages:
Can store more H2 in a given volume than compressed tanks
The volumetric capacity of liquid hydrogen is 0.07 kg/L compared with 0.039 kg/L at 700 bar

Limitations:
Liquid H2 boil-off
Energy required for H2 liquefaction (-250C)
Tank cost

Question 92

What are the advantages and limitations of storing H2 as metal hydrides?

Answer 92

Advantages
High volumetric density
Low pressure
Safer

Limitations:
Very low gravimetric capacities 3-4 wt% (target for 2010 was 6 wt%)
Hydrogen release kinetics are too slow
Packing density of the powder is low
Need to lower cost and increase cycle life

Question 93

What are the characteristics of storing hydrogen using high surface area sorbents?

Answer 93

Fast hydrogen kinetics
Low hydrogen binding energies
Low thermal management issues during Hydrogen charging and discharging

Question 94

Why store hydrogen as ammonia?

Answer 94

Ammonia has a higher boiling point (-33.5C) than hydrogen, so it is easier to store as a liquid.

Has a high capacity for H2 storage (17.6 wt%)

The release of H2 from ammonia is endothermic

Disadvantages:
High fuel processing temperatures and large reactor mass and volume will be required

Is toxic

Traces of ammonia are incompatible with fuel cells

Given all these challenges, most likely ammonia reactors will never meet the requirements for commercially viable hydrogen-powered fuel cell vehicles but this could happen off-board

Question 95

What’s the Nerst equation?

Answer 95

E = E0 - RT/neF * lnQ

= E0 - RT/neF * ln(a.red / a.ox)

Where:
E - cell potential
E0 - standard cell potential
R - gas constant
T - temperature
F - Faraday constant
ne - number of moles of electrons transferred in the cell reaction
Q - reaction quotient
a - activity

We want an electrolyser potential to be smaller as it means less energy input is needed.

Question 96

What is PEMWE and AEMWE?

Answer 96

Proton Exchange Membrane Water Electrolyzer

Anion Exchange Membrane Water Electrolyzer

Question 97

List features of Proton Exchange Membrane (PEM) Water Electrolyzers and Anion Exchange Membrane (AEM) electrolyers:

Answer 97

PEM:
< 100 °C
The separator is a proton-conducting membrane
Polymeric Nafion membrane enables faster start up and shut down than traditional electrolysers
Perfect for coupling with intermittent renewables
Lower ohmic drop, higher current density
More compact thanks to solid membrane
Requires perfluorinated polymers and noble metal (Ir, Ru, Pt) catalysts

AEM:
< 100 °C
The separator is an OH-conductive anion exchange membrane
Least-mature technology
PGM-free anodes available
Catalyst and membrane stability limit the widespread
Combines operational ease of PEM with low-cost catalysts

Question 98

What’s Faraday’s law?

Answer 98

m = QM/Fn

Where:
m - mass of product produced during dt
Q - charge (= current * time)
F - Faraday constant
M - product molecular weight
n - number of e- transfer per unit product produced

Question 99

What’s the Butler-Volmer Equation?
What’s it for?

Answer 99

It calculates overall current density, j

j = j0(exp⁡((a.aveF)/RTη) − exp(−(a.cveF)/RT* η))

Where:
j - overall current density (A/m2)
j0 - exchange current density (A/m2)
T - absolute temp (K)
F - Faraday’s constant (C/mol)
R - gas constant (J/Kmol)
ac - cathodic charge transfer coefficient
aa - anodic charge transfer coefficient
η - overpotential (V)
ve- stoichiometric coefficient of the electrons

Question 100

What is U in electrochemistry?
How is it found?

Answer 100

U is the real potential, and can be determined with I-V curves.

U = E0.cell = E0.cathode - E0.anode

This is only true if there is no current through the cell and the resistance through the electrolyte solution is neglected.
Otherwise U ≠ E0.cell (the real potential is not equal to the thermodynamics potential)

For electrocatalytic cells U > E0.cell

Question 101

How does real potential, U, compare to the cell potential for electrolysers?

Answer 101

For electrocatalytic cells U > E0.cell

In particular, U = E0 + η(H2) + η(O2) + iR

Where:
η(H2) - overpotential for hydrogen evolution
η(O2) - overpotential for oxygen evolution
is - ohmic drop, due to electrolyte resistance and calculated as current*resistance

Question 102

Why is a plateau not observed in the current-potential curve of electrolysers even at high overpotential?

Answer 102

If the reactant has a low concentration in the electrolyte, at high overpotential the rate of reaction becomes limited by the transport of reactant to the catalyst surface.

As a result, at high overpotential, the current does not keep increasing with potential, but reaches a plateau which value is determined by the rate of reactant diffusion.

For the case of electrolysers, this plateau is not observed because the reactant (water and proton) is readily available.

Question 103

What parameters may be used to evaluate catalytic activity in electrolysers? (7)

Answer 103

1. The current density: Normalized to the geometric surface area of the electrode
2. The specific activity: Current normalized to the surface area of the catalyst
3. The mass-specific activity: Current normalized to the mass of the catalyst (for noble metal-based catalyst it is usually normalized to the mass of the noble metal)
4. Onset Potential: Highest potential at which a faradaic current is measured
5. Overpotential η: Difference between the onset potential and the theoretical thermodynamic potential
6. Diffusion-limited current: Measured current in the diffusion region
7. Half-wave potential: Potential at which the current is half of the diffusion-limited current

Question 104

Write the equations for the reactions taking place at the anode and cathode in a PEM and AEM electrolyser:

Answer 104

PEM:
Anode: 2H2O ⇌ O2 + 4H+ + 4ei
Cathode: 4H+ + 4e- ⇌ 2H2

AEM:
Anode: 4OH- ⇌ 2H2O + O2 + 4e-
Cathode: 4H2O + 4e- ⇌ 2H2 + 4OH-

Question 105

What are the main disadvantages of proton exchange membrane (PEM) electrolysers, alkaline electrolysers, and direct photocatalytic water splitting?

Answer 105

PEM:
- requires precious metals for the catalysts
- needs improvements in cost, stability, and conductivity of electrode materias

Alkaline:
- membrane stability and conductivity need improving
- poor catalyst activity

Photocatalysis:
- More efficient and stable photoelectrode materials needed

Question 106

Why are porous layers needed within membrane electrolysers?

Answer 106

Porous layers are used to remove the gaseous product from the surface of the catalyst and control bubble formation.

On the HER (H2 evolution reaction) side, carbon can be used as a transport later
On the OER (oxygen evolution reaction) side, the carbon would be quickly corroded so we need to use expensive and resistant materials, usually Ti
Cost of bipolar plate alone is approximately 30% of total cost
Since Ti passivates over time, currently coated with Pt and Au to prevent passivation

Question 107

Properties of PEM electrolyser membranes:

Answer 107

Per-fluoro sulfonic acid membranes (PFSA) are the state-of-the-art proton-exchange-membrane.
The most commoly used, commercially available is Nafion
It provides good proton conductivity (78 mS/dec)

But has several disadvantages, such as:
High production cost
High sensitivity to temperature and humidity
Long lasting in the environment

Question 108

Properties of AEM electrolyser membranes:

Answer 108

AEMs are solid polymer electrolyte membranes which contain positively charged cationic groups (e.g. quaternary ammonium functional groups) and mobile negatively charged anions (OH) as counter ions.

Requirements:
1. High ionic conductivity and selectivity
2. High mechanical stability
3. Low swelling ratio
4. Good water permeability
5. Chemical long-term stability
6. Cost-efficient
7. Environmentally-friendly membrane production process.

Question 109

What are some good electrolyser catalysts?

Answer 109

In acid conditions, Pt and Pd are the most active, but expensive and unsustainable.

In alkaline conditions, nickel alloys are active but the AEM electrolyzer has other limitations (i.e. the membrane).

To improve kinetics and durability of the electrode, extensive studies for highly active and stable oxygen electrocatalysts have been performed.
In acidic condition, noble metal compounds have been primarily utilized as electrocatalysts
Transition metals are not stable in acidic media, but some have shown good stability and activity in alkaline.

Question 110

Why is IrO2 a better catalyst in electrolysers than RuO2?

Answer 110

Whilst the potential for RuO2 is lower than IrO2 at first, over time, its function deteriorates and more V is needed, whilst IrO2 remains stable.

RuO2 can then dissolve into the electrolyte and deposit elsewhere.

Question 111

Compare HER and OER:

Answer 111

Hydrogen evolution reaction is facile:
- One intermediate: H*
- Ideal catalyst simply requires ∆G(H)=0
- The theoretical minimum overpotential is 0V

Oxygen evolution reaction is more challenging:
- At least three intermediates: OH, O, OOH*
- Suboptimal linear scaling relationships between ∆G(OH∗) and ∆G(OOH∗) impose at least 0.3 V
- Overpotential on even the most active catalyst.
- RuO2 is the most active OER catalyst in acid, but IrO2 is the most stable and hard to replace.

Question 112

Summary of water electrolysis:

Answer 112

H2 is a clean energy resource, and a suitable carrier to storage intermittent renewable energy
As a result, H2 production has been listed in the UK 10 points plan
H2 can be classified as grey, blue and green hydrogen
Water electrolysis constitute the cleanest option
There are four types of water electrolysers, with different working principle, operation conditions and efficiency.
The efficiency can be calculated by knowing the operation parameters.
Developing active and stable HER and OER electrocatalysts is the most effective way to improve the electrolyser efficiency. Benchmark catalysts are already available, but new materials need to be developed to overcome the major challenges.
Several factors can be used to evaluate the activity of catalysts: on-set potential, overpotential, faraday efficiency.
Nifion is the only suitable membrane for PEM water electrolyser, while AEM is still semi-commercial.
Besides catalysts and membranes, temperature and pressure can also affect efficiency.

Question 113

Compare PEM and AEM electrolysers:

Answer 113

Ion transport:
PEM transfers protons (H+) whilst hydroxide transfers hydroxide (OH-)

Temperature:
PEM often operates at lower temps

Catalyst:
PEM typically uses precious metal-based (e.g., Pt) whilst AEM uses non-precious metal catalysts (e.g., Ni-Fe)

Cost:
PEM has higher cost due to precious metal catalysts and membrane

Stabiltiy:
PEM susceptible to degradation under certain conditions. AEM generally more stable due to alkaline environment.

Question 114

Define Fermi level:

Answer 114

The Fermi level helps describe how electrons behave in materials, especially in metals or semiconductors.
It represents the highest energy level occupied by electrons at absolute zero temperature.

The Fermi level of a solid-state body is the thermodynamic work required to add one electron to the body. It is a thermodynamic quantity usually denoted by µ or EF for brevity.

When the LUMO level of the organic solvent or electrolyte lithium salt is lower than the Fermi level of the negative electrode, electrons will be injected into the LUMO orbit by that driving force, causing the electrolyte component to be reduced.
When the HOMO level is higher than Fermi energy level of the positive electrode, electrons are driven into the positive electrode, causing the solvent or lithium salt to be oxidized.

Question 115

Describe the Gravimetric storage density (mass %) vs Volumetric storage density (kg H2 /m3) of the following hydrogen storage techniques:

1. Transition metal hydrides
2. Light element hydrides
3. Hydrocarbons
4. Liquid hydrogen
5. Sorption
6. H2 hydrate
7. Pressurised tanks

Answer 115

1. Transition metal hydrides - Low GSD (1-8%), High VSD (100-160 kg H2/m3)
2. Light element hydrides - Medium GSD (5-20%), High VSD (50-160 kg H2/m3)
3. Hydrocarbons - Medium GSD (15-25%), Medium VSD (80-120 kg H2/m3)
4. Liquid hydrogen - High GSD (~100%), Medium VSD (~80 kg H2/m3)
5. Sorption - Low GSD (1-5%), Low VSD (1-40 kg H2/m3)
6. H2 hydrate - Low GSD, Low VSD
7. Pressurised tanks - Low-Med GSD, Low VSD

Question 116

Given the Nerst equation, how does the potential of a fuel cell change (increase, decrease, remains equal) if:

1. We increase the gas feed pressure
2. We increase the temperature
3. We change the pH

Answer 116

Cell potential increases with gas feed pressure and decreases with temperature.

Also, kinetics improve with temperature so that an increase in temperature usually leads to improved performance.

pH makes no difference.

For an electrolyser, we do not want a high cell potential.

Question 117

Which electrode forms hydrogen and oxygen?

Answer 117

Hydrogen: cathode (-ve potential)

Oxygen: anode (+ve potential)

Question 118

In radar diagrams comparing batteries, what 6 properties are compared?

Answer 118

1. Specific energy (capacity)
2. Specific power
3. Safety
4. Performance
5. Life span
6. Cost

Question 119

Why will the actual measured energy density of the battery be lower than the calculated value?

Answer 119

Irreversible capacity lost in first cycle due to SEI, imperfect charge balancing, mass of additional components in the cell not accounted for.

Question 120

Discuss the factors that can influence the solid electrolyte interphase (SEI) formation for such a battery.

Answer 120

Electrolyte (solvent/salt) - solvents with higher chemical stability and lower reactivity with electrode materials tend to form more stable SEI layers

C rate/formation cycle conditions - Higher C rates can accelerate SEI formation due to increased ionic flux

Electrode porosity - Higher electrode porosity can enhance electrolyte infiltration, promoting SEI formation

Type of Cathode - Different chemistries exhibit varying degrees of reactivity with electrolyte components, affecting SEI composition and stability

Full vs Half Cell configuration - The presence of both electrodes can influence SEI composition and morphology

Question 121

What properties of carbon electrodes effect the coulombic efficiency of batteries?

Answer 121

Pore structure
Surface chemistry
Graphitic structure
Particle size
Morphology
Impurity content

Question 122

What’s volume expansion in battery electrodes?

Answer 122

The physical increase in volume experienced by the electrode material during the charge and discharge processes of a battery.
This expansion is particularly common in materials used as anodes (negative electrodes).

This can be due to:
- Ion insertion
- Structural changes
- Phase transitions

And can lead to:
- Irreversible structural changes, degradation of the electrode material, and loss of active material, resulting in capacity fade and reduced battery performance over time.
- Loss of electrical contact
- Mechanical stress

Question 123

Tin-carbon composites are a promising anode material for sodium-ion batteries, as every tin atom can store almost 4 Na atoms, forming the compound Na15Sn4.

Commenting on this reaction, explain the advantages and drawbacks of using tin as an anode material, and mention at least one strategy that has been employed to address the challenges faced by tin anodes.

Answer 123

Sn reacts with Na to form Na15Sn4, resulting in extremely high theoretical capacity (847 mAh/g).

However, the alloying reaction accompanied by huge volume expansion.
This results in electrode pulverisation, continual consumption of electrolyte, loss of active material resulting in capacity loss.

To prevent this, we can coat Sn particles with various protective films, introduce additives into electrolyte e.g. FEC.

Question 124

In general, noble metal-free ORR catalysts are more active in acidic or alkaline conditions?

Answer 124

Alkaline

This is due to:
Lower overpotential
Favourable reaction mechanisms
Catalyst stability
Availability of active sites

Question 125

What’s Band theory in semiconductors?

Answer 125

Band theory in semiconductors helps explain the behaviour of electrons within materials.

In solids, electrons are not confined to individual atoms but instead form energy bands that extend throughout the material. These energy bands represent allowed energy levels that electrons can occupy.
In semiconductors, the energy bands are separated by energy gaps, where no electron states are permitted.

The two main energy bands in semiconductors are the valence band (lower energy) and the conduction band (higher energy).

The energy gap, also known as the band gap, is the energy difference between the top of the valence band and the bottom of the conduction band. It represents the minimum amount of energy required to move an electron from the valence band to the conduction band.

Question 126

What are the 2 types of doping in semiconductors?

Answer 126

N-type: donor impurity contributes free electrons

P-type: acceptor impurity creates a hole

These impact the Fermi level of the semiconductor

Question 127

Describe photocatalysis and photoelectrocatalysis:

Answer 127

Photocatalysis - particle-based semiconductor dispersed in a reaction media working under light irradiation without applied potential.

Photoelectrocatalysis - semiconductor films in a transparent electrode, immersed in an electrolyte and working under light irradiation and applied potential.

Question 128

What are the photocatalysis requirements for water splitting?

Answer 128

The semiconductor band gap must straddle the reduction and oxidation potentials for the proton reduction and water oxidation, respectively.
I.e. the conduction band should be more negative than the H+/H2 potential (CB < 0)
And the valence band should be more positive than the H20/O2 potential (VB > 1.23)

The conduction band edge must be more negative than 0 V vs RHE and the valence band edge more positive than 1.23 V vs RHE.

Question 129

What are the kinetic limitations of photocatalysis?

Answer 129

Fast charge recombination and poor charge separation hinders the performance of photocatalysts.

Question 130

Briefly describe the basis of photocatalysis:

Answer 130

Light absorption by a semiconductor (photocatalyst)

Photoexcited states generation (electron and hole)

Charge separation and migration

Transfer of excited states to water molecules

Charge recombination (unwanted)

Question 131

What improves charge separation in photocatalysis?

Answer 131

Using co-catalysts. Electrons/holes are transferred from the CB/VB from the SC to the metal. Improves electron-hole separation and inhibits recombination.

Question 132

What are heterojunctions?

Answer 132

Semiconductors in direct contact that allow the transfer of photoexcited states from one to another.

Question 133

What is the apparent quantum efficiency (AQE)?
How is it found?

Answer 133

The ratio of electrons transferred toward a certain product relative to incident photos at a given wavelength.

AQY =(Ne/Np)× 100% =(XM/Np)× 100%

Ne: Number ofreactingelectrons
Np: Number of incident photons
M: Number of product molecules
X: Number of reaction electrons

Question 134

What is the solar to fuel efficiency (SFE)?

Answer 134

The ratio of converted chemical energy relative to the incident solar energy.
AQE over the entire solar spectrum.

Question 135

What is AM1.5 G ?

Answer 135

AM1.5 G refers to the standard spectrum at the earth’s surface, derived from an standard test from the American Society of Testing Materials to facilitate the comparison of solar conversion efficiency measurements.

Question 136

List main parts of a photo-electrochemical cell:

Answer 136

1. Photoelectrode: The photoelectrode is the key component that absorbs light and initiates the photochemical reactions. It is typically made of a semiconductor material that can generate electron-hole pairs upon absorbing photons. Common semiconductor materials used for photoelectrodes include titanium dioxide (TiO2), silicon (Si), and metal oxides like hematite (Fe2O3) or bismuth vanadate (BiVO4).
2. Counter electrode (often Pt or IrO)
3. Electrolyte (often alkaline)

Question 137

How does the basis of an n-type and p-type semiconductor differ?

Answer 137

P-type: Electrons are the minority carriers (and holes are the majority)

N-type: Holes / electron vacancies are the minority carriers (and electrons are the majority)

Minority carrier (h+ in an n-type, e- in an p-type) transferred in the electrode surface.
Minority carriers drive the catalysis on the surface of the semiconductor.

The minority carrier is what brings about hydrogen evolution on the electrode surface.

P type will a photocathode.
N-type will be a photoanode.

Question 138

What are the minority carriers in n-type and p-type photoelectrocatalysis?

Answer 138

N type: holes/h+
P type: electrons/e-

Minority carriers drive the catalysis on the surface of the semiconductor.
The minority carrier is what brings about hydrogen evolution on the electrode surface.

P type will a photocathode.
N-type will be a photoanode.

Question 139

What is onset potential?

Answer 139

Working potential where the electrocatalytic current starts to take off from the background.

Question 140

What is the Incident photon-to-current efficiency (IPCE):

Answer 140

Ratio of produced photocurrent versus the incident photon flux at a given wavelength.

Question 141

What is the Faradaic efficiency?

Answer 141

Ratio of charges transferred to a given product relative to the total charges passed through the circuit, indicates the selectivity of an electrocatalysts towards a certain product. Desirable >80%

FE% = nFm/Q

Where:
n - # of reaction electrons
m - moles product
F - Faraday constant
Q - total charge passed in Coulombs

Question 142

What is doping (for semiconductors)?

Answer 142

The intentional introduction of impurities into an intrinsic (undoped) semiconductor for the purpose of modulating its electrical, optical and structural properties.
The doped material is referred to as an extrinsic semiconductor.

There can be N-type (e- charge carriers) or P-type (hole charge carriers).

Question 143

What does the band gap between valence and conduction bands suggest?

Answer 143

Narrower band gap = Better light harvesting, but also means that e- and h+ (charge) recombination is much faster

(Note - insulators have a large gap, conductors have no gap, semiconductors are in the middle)

Question 144

Pros and cons of photocatalysis:

Answer 144

Pros:
Cheap
Simple
Environmentally friendly
Versatile

Cons:
Not very efficient
Limited wavelength absorption
Catalyst recovery and recycling

Question 145

How does photocatalysis work?

Answer 145

1. Absorption of Light: Light is absorbed by a semiconductor material. This semiconductor is typically titanium dioxide (TiO2) or another similar material with a wide bandgap.
2. Generation of Electron-Hole Pairs: When photons are absorbed by the semiconductor, they excite electrons from the valence band to the conduction band, leaving behind positively charged holes in the valence band. This process generates electron-hole pairs.
3. Migration of Charge Carriers: The excited electrons and holes are free to move within the semiconductor material due to its crystalline structure. Electrons move towards the surface of the material, while holes migrate towards the interior.
4. Reduction and Oxidation Reactions: Once the electrons and holes reach the surface of the semiconductor, they can participate in reduction and oxidation reactions, respectively. The electrons can reduce certain compounds by transferring their energy, while the holes can oxidize other compounds by accepting electrons.
5. Degradation of Contaminants: In photocatalysis applications such as water purification or air treatment, the excited electrons and holes react with organic pollutants or other contaminants adsorbed on the surface of the semiconductor.
6. Regeneration of Catalyst: After the photocatalytic reactions occur, the semiconductor material remains unchanged and can be reused for further cycles of photocatalysis.

Question 146

What is the redox equation for H2 formation?

Answer 146

2H+ + 2e- → H2
E = 0V vs RHE
Reduction reaction

Question 147

What is the redox equation for H2O splitting?

Answer 147

2H2O → O2 + 4H+ + 4e-
E = 1.23 V vs RHE
Oxidation reaction

Question 148

List typical cocatalysts for photocatalysis

Answer 148

CrOx-Rh
SrTiO3:Al
CoOx

Improves electron-hole separation and inhibits recombination

Question 149

What is band bending?

Answer 149

The process in which the electronic band structure in a material curves up or down near a junction or interface.

Imagine you have two different materials joined together, like a semiconductor and a metal. Each material has its own “energy levels” for its electrons, which are like steps on a staircase.

Now, when you bring these materials close together to form a junction or interface, something interesting happens to the electron steps. They start to bend near the junction, like a ramp going up or down.

This bending happens because the two materials have different properties. One might have more electrons available than the other, or they might have different abilities to conduct electricity.

Because of this bending, the electrons in one material want to flow into the other material to balance things out, like water flowing downhill. This movement of electrons continues until a balance is reached, and the bending stops.

When this balance is achieved, there’s no more net movement of electrons, and we say that the potential difference, or the energy difference, between the materials has vanished. This state is called equilibrium.

Schottky contacts have a potential barrier at the metal-semiconductor junction, allowing current flow predominantly in one direction, while Ohmic contacts have no barrier, enabling easy flow of current in both directions.

Question 150

What is a space charge layer?

Answer 150

A region that forms at the interface between two materials, typically a semiconductor and another medium such as an electrolyte or another semiconductor with different doping characteristics.
This region is characterized by an excess or deficit of charge carriers, which creates an electric field.

N-type semiconductors have a Fermi level energy higher than the water chemical potential.
Equilibration is achieved by electron flow away from the semiconductor-electrolyte interface. This loss of majority carriers in the SC leads to the formation of a space charge layer.

A region at the surface of the photoanode that is depleted of electrons, which is balanced by an equal negative charge in the electrolyte at the electrode surface, termed the Helmholtz layer.
The SCL bends the bands and gives directionality of excited charge-carrier movement.
Electrons are repelled by the field and drawn to the bulk where holes are driven to the surface.

Question 151

What are scavengers?

Answer 151

A material added to a semiconductor that helps oxidation.
The scavenger is easier to oxidise, which helps collect holes faster than water formation, preventing the holes from recombining with the electrons.

They’re usually small organic molecules e.g. methanol and ethanol. Instead of oxidising water, the molecules are oxidised which are easier, faster, and require less oxidation potential than water.

Question 152

How is electron-hole separation increased?

Answer 152

Cocatalysts
Scavengers
Heterojunctions

Question 153

What are the 3 types of heterojunctions?

Answer 153

I) Semiconductor with much wider band gap that straddles the CB and VB of the other (not very useful)

II) Similar band widths (ok) Photooxidation in both

Z-scheme) The CB of one semiconductor is more negative than the VB of the other. Following photooxidation in both SCs, the electron from one recombines with the hole of the other, so there remains an e- on one SC with a huge reducing power, and a hole on the other SC with a huge oxidation power, and the recombination change is hugely decreased as the e- and h+ are greatly separated.

Question 154

What materials may be used as photocathodes?
Comment on their characteristics:

Answer 154

Chalcogenides: Earth abundant, high absorption, stable, easy to make

Cu2O: Easy to make, abundant, low stability.

Question 155

What reaction takes place at the photocathode and photoanode?

Answer 155

Photocathode: H2 evolution reaction (HER)

Photoanode: O2 evolution reaction (OER)

Question 156

List typical materials used for photoanodes:

Answer 156

TiO2. Titanium dioxide (3.2 eV)

Fe2O3. Hematite (2.1 eV).

BiVO4. Bismuth Vanadate (2.4 eV)

Question 157

What is the timescale of the charge carrier processes?

Answer 157

Carrier lifetimes are typically in the range of pico to nanoseconds, whereas the timescale for water oxidation is in the range to milliseconds to seconds.

Question 158

What are overlayers?

Answer 158

Additional thin layers of material deposited onto the surface of a semiconductor substrate.

They can induce band bending at the junction, which, depending on the relative Fermi levels of the materials involved may aid or hinder the separation of photogenerated charges

Question 159

What are tandem PECs?

Answer 159

Tandem photoelectrochemical cells (PECs) are a type of solar cell configuration designed to enhance the efficiency of light-to-electricity conversion by combining multiple photoelectrodes with complementary light absorption properties.

In a tandem PEC, two or more photoelectrodes with different bandgap energies are stacked or arranged in series to capture a broader spectrum of sunlight and efficiently utilize photons with different energies. Each photoelectrode is typically made of a semiconductor material.

Suitable photocathode material for water reduction (Cu2O)
Suitable photoanode material for water oxidation (TiO2, Fe3O4, BiVO4)
Evolution of H2 and O2 between applied vias
Operating current limited by the photoelectrode producing the least photocurrent

Question 160

What are the challenges with electrochemical conversion of CO2 into other products (to reduce CO2 emissions)?

Answer 160

CO2 has a high activation potential, requiring larger amounts of energy. (Electrochemical, Photochemical, or Thermochemical systems can be used to provide such energy.)

Selectivity is also an issue - lower selectivity for larger products

Hydrogen formation is easy and quick and competes with CO2 reaction processes.

Overpotentials - Both OER and CO2R have very large overpotentials, meaning CO2R will run at a larger cell potential than this thermodynamic potential

Question 161

What happens in electrochemical CO2 reduction?

Answer 161

Electrodes, catalyst, electricity needed.

Reduction occurs at cathode - CO2 is bubbled into the system so, at the cathode, CO2 forms other products such as CH4, CH3OH etc.
At the anode, water is oxidised to form oxygen.

Question 162

Comment on the selectivity of electrochemical CO2 reactions:

Answer 162

Low overall selectivity

Lower selectivity for larger products

Hydrogen formation is competing reaction

Question 163

Describe a H type cell:

Answer 163

Catholyte and anolyte separated by a membrane

Usually neutral electrolyte (KHCO3 0.5 M)

Low current densities (20 mA cm-2) owing to low CO2 solubility (34 mM at ambient conditions) (at high current densities, CO2 runs out and the process reverts to hydrogen evolution)

Ag/AgCl reference electrode, water oxidation catalyst in the anode (IrO, Pt/C)

Looks like a capital H shape

Question 164

Describe Linear Sweep Voltammetry and what it is used for:

Answer 164

Linear sweep voltammetry can be used to calculate the peak current, calculate the peak current potential, and calculate the half-peak current potential.

The current at a working electrode is measured while the potential between the working electrode and a reference electrode is swept linearly in time.
In the context of linear sweep voltammetry, “swept” means that the potential between the working electrode and the reference electrode is changed continuously and linearly over time.

Screening potential

Recording current

Catalyst comparison

Question 165

Describe Chronoamperometry

Answer 165

The electric potential of the working electrode is stepped and the resulting current from faradaic processes occurring at the electrode (caused by the potential step) is monitored as a function of time.

Constant potential

Recording current

Stability

Product quantification

Question 166

What is the partial current density?

Answer 166

The current density used to produce one single product.

Proportional to product formation (>200 mA cm-2 for industrially relevant application). Calculated multiplying the total geometric current density by the faradaic efficiency.

Question 167

What is the Turnover frequency?

Answer 167

Defined as the rate of electrochemical conversion per electrocatalytic site at a certain potential.
Highly challenging to calculate due to the structural ambiguity of the active sites and the difficulty of accurately quantifying them.

TOF = J / (nFASN)

Where:
J is the partial current density of the product
n is the number of electrons
F is the Faraday constant
ASN is the active site number

Question 168

Describe cell voltage:

Answer 168

Difference in electrical potential between anode and cathode, controlled by kinetic overpotentials of each half reaction and ohmic losses due to ion transport through the device. (<3 V)

Question 169

What does the CO2 utilisation degree show?

Answer 169

It determines the extent to which CO2 is effectively converted to the product of interest.
Defined as the molar ratio of the CO2 converted to the product of interest and the total CO2 inlet to the cathode.

Hindered by the formation of carbonates (HCO3- and CO32-) by reaction of CO2 with OH-.

CO2 UD = V(CO2 to product) / (V((CO2 to product) + V(CO2 lost)) *100

Where V are volumes

Question 170

Describe a flow electrolyser:

Answer 170

It possesses a porous electrode where the catalyst is coated and, instead of dissolving CO2 in the electrolyte, the CO2 goes through the porous layer, through the catalyst, into the electrolyte. GDL (gas diffusion layer) is hydrophobic.

This means there is a triple phase boundary between catalyst, electrolyte, and product.

No longer limited by solubility.

Gaseous CO2 is reduced at a gas/electrode/electrolyte interface

Deposition of a catalyst onto a GDL (typically a commercially available carbon-based substrate with hydrophobic treatment).

GDL is more resistive compared to pure metals

Conductive tape is applied around the electrode to minimize distance electrons need to travel

Sufficient current collection is extremely important for high current density

Flooding still an issue. Attributed to salt penetration into the GDL that compromises hydrophobicity

Question 171

Describe gas diffusion electrodes:

Answer 171

Gas diffusion electrodes (GDE) are electrodes with a conjunction of a solid, liquid and gaseous interface, and an electrical conducting catalyst supporting an electrochemical reaction between the liquid and the gaseous phase

Enhanced mass transport. CO2 in contact with catalyst and electrolyte simultaneously.

Large electrochemical surface area - Industrial current densities (200 mA cm-2)

With or without electrolyte (flow cell vs zero gap electrolyser)

Membrane and flowing aqueous electrolyte to separate the electrodes (hybrid electrolysers, flow electrolyser).

Control cathode and anode pH, overcomes product crossover

Deposition of a catalyst onto a GDL (typically a commercially available carbon-based substrate with hydrophobic treatment).

Flooding still an issue. Attributed to salt penetration into the GDL that compromises hydrophobicity

Question 172

Key aspects of gas diffusion electrodes without electrolyte:

Answer 172

Membrane electrode assembly (MEA)

Humidified CO2, is fed to the cathode.

Cathode and anode sandwich an ion-exchange membrane

Polymer electrolyte membrane reduces ohmic losses by decreasing the interelectrode distance.

Ionic species transport to and from the anode through a membrane

Question 173

How does CO2 conc and pH vary in electrochemical CO2 reduction devices with current density.

Answer 173

As current density, increases, CO2 conc decreases to 0 at ~ 20mA/cm2 in H cells, whilst is decreases more gradually in GDLs (above 200 mA/cm2).

pH increases as more OH- ions are formed.

GDL allows high CO2 concentration near the catalyst at high current densities, but the pH increases quicker as OH- is generated at a faster rate.

Question 174

What is carbonate crossover?

Answer 174

When CO2 reacts with OH- ions to form HCO3- and CO3 2- ions

Carbonate crossover in CO2 electrochemical devices refers to the unintended transport of carbonate ions across the membrane or separator.
During the electrochemical reduction of CO2, carbonate ions (CO3^2-) can be formed as intermediates.
- Carbonate ions reaching the cathode can react with hydrogen ions or undergo reduction reactions, affecting the purity of the desired product.
- Carbonate ions may interfere with the electrochemical reactions occurring at the cathode, reducing the efficiency of CO2 reduction or altering the selectivity of the process.
- Prolonged exposure to carbonate ions can lead to degradation or fouling of the membrane, affecting the overall durability and performance of the CO2 electrochemical device.

This is dealt with with bipolar membranes.

Question 175

What is used to deal with carbonate crossover?

Answer 175

Bipolar membranes

[Without bipolar membranes: The carbonate ions pass an AEM (anion exchange membrane) where the carbonates react with H+ ions to reform CO2. This CO2 formed is known as lost CO2.]

WITH BIPOLAR MEMBRANES:
An AEM (anion exchange membrane) and CEM (cation exchange membrane) are present, which separate the H+ and OH- ions from water reactions.
Carbonate is formed in the cathode side.
(Before, carbonate would travel to the anode and reform CO2).
Now, CO2 is regenerated at the cathode, preventing CO2 loss, carbonate crossover, and maintaining alkaline conditions at the anode.

However, there will be a lot of protons at the cathode, encouraging hydrogen formation.
This reduces the Faradaic efficiency for CO2 formation.

Question 176

Outcomes of bipolar membranes:

Answer 176

Lower faradaic efficiency and higher cell voltage H+ flux benefits HER over CO2

Higher cell voltage arising from membrane resistance

Higher CO2 utilization owing to lack of carbonate and product crossover

Question 177

How does catalyst strength impact product formation with CO2 reduction?

Answer 177

If C binds weakly to the catalyst, you have a catalyst that makes CO i.e. CO is released easily.

If it binds strongly enough for another CO to come, the 2 CO molecules can be coupled and hydrogenated, forming hydrocarbons.

If CO binding is too strong: Catalyst poisoning, leading to HER

If CO* binding is not strong enough: CO* desorbs before C-C or C-H coupling. CO production

Cu is the only catalyst that binds CO* strongly and H* mildly, therefore is the only material capable to make C2+/coupled products

Question 178

List potential electrode materials to form H2, CO, Formates, and C1-C3:

Answer 178

H2 = Ni, Fe, Pt, Ti
CO = Au, Ag, Zn, Pd, Ga
Formate = Pd, Hg, Tl, In, Sn, Cd
C1-C3 = Cu

Question 179

How does pH impact CO2 electrolyser performance?

Answer 179

Low pH (high H+ concentration) decreases the selectivity towards CO2R versus HER

High pH (high OH- concentration) increases carbonate crossover

Local pH changes during operation. Depletion of protons/production of hydroxide ions from HER and CO2RR.

pH strongly affects CO2R selectivity

CH4 production is favoured in a more acidic pH

Alkaline conditions supress methane and hydrogen formation, promote C2+

Question 180

What are the 3 main families of abode material used in Na ion batteries?

Answer 180

Hard carbon
Tin-based
Sodium titanates

Question 181

What contributes to voltage losses in electrolysers?

Answer 181

Inefficiencies (overpotentials η):
1. ion/electron transport (ηcond)
2. electrode kinetics (ηkin)
3. mass-transport induced thermodynamic effects (ηmt)
4. Cathode/membrane interface
5. Ohmic losses and resistance

Question 182

Describe electrolyser electrode materials (Bi, Sn) to make formates:

Answer 182

• Adsorption of bent radial anion through oxygen (oxyphilic surfaces) and hydride transfer.
• TRL 3-5, Sn NPs catalyst in GDE, AEM
• Acidic centre compartment for direct production of FA and not the deprotonated form (HCOO-)
• OER at anode, protons travel towards centre compartment
• 2000 A m-2 (200 mA cm-2), 3.75 V, 73-91% FE
• 10 wt% FA solution that is distilled-extracted to achieve 85 wt%
• CO2UD 37-46%

Question 183

Describe electrolyser electrode materials (Au, Ag) to make CO:

Answer 183

TRL 5-6, GDEs in a membrane electrode assembly

Humidified CO2 stream is fed at the cathode producing CO and (undesired) H2.

Carbonate crossover in the cathode, carbonates cross membrane to anodic compartment.

2000 A m-2 (200 mA cm-2), 3 V, 98% FE

CO2UD 49%

Question 184

Describe electrolyser electrode materials (Cu) to make C1-C3 compounds:

Answer 184

Copper is the only heterogeneous catalyst (to date) that has shown a propensity to produce valuable hydrocarbons and alcohols, such as ethylene and ethanol from electrochemical CO2 reduction.

Activity and selectivity is highly sensitive to catalyst surface structure, morphology, composition, electrolyte and pH.

TRL 3-4, GDEs in a membrane electrode assembly (analogous to CO electrolyser)

Humidified CO2 stream is fed at the cathode which is separated with an AEM from the anode.

Alkaline aqueous stream is fed in the anode for the OER

The cathode outlet contains CO2, H2O, C2H4, other gas and liquid products

1200 A m-2 (120 mA cm-2), 3.7 V, 64% FE

CO2UD 20-33%

Question 185

List key factors of a fuel cell (compared to a battery):

Answer 185

Fuel cells:
Require a fuel
Faster recharging time for fuel cells vehicles, compared to battery vehicles
Higher specific energy than batteries
Lower specific power than Li-ion batteries
No self-discharge concerns

Question 186

Pros and cons of PEM fuel cells:

Answer 186

Pros:
Fast start-up
Low temperature
Non-corrosive electrolyte
High power density
Operates at low pressure  safer
Higher Performance
Stable and active ORR catalyst available (Pt)
Low Pt loading for HOR
Durable and highly conductive membrane available (but perfluorosulfonate based)

Cons:
Relatively low tolerance to carbon dioxide (50ppm) and sulfur (few ppm)
Need reactant gas humidification
Need expensive and unsustainable platinum as ORR catalyst
Need Nafion membrane, expensive, complex production
Noble metals needed for OER (transition metals are stable but don’t perform well in acid)

Question 187

Pros and cons of AEM fuel cells:

Answer 187

Pros:
Low temperature
Fast start-up
High efficiency
Minimal corrosion
Noble-metal free alternative for the ORR catalyst
Noble metal-free ORR catalysts exists

Cons:
Intolerant to CO and CO2 (350 ppm)
Liquid electrolyte
Complex water management
Short life-time
Lower performace
Slightly higher Pt loading needed for HOR
Lower conductivity and durability of the membrane (but fluorine-free options available)

Question 188

How is water production at the cathode of PEM fuel cells managed?

Answer 188

In PEMFCs, water is only generated (at the cathode as a product of the oxygen reduction reaction, ORR) and not electrochemically consumed.
Water is also moved to the cathode from the anode by electro-osmotic drag as H2O produced by the hydrogen oxidation reaction (HOR) moves through the PEM. Thus, removing cathode water is the prime concern in the PEMFC to avoid flooding.

To avoid cathode flooding and anode dry out, water need to back diffuse to the anode through the membrane.
The role of water is also to hydrate the membrane and to facilitate H+conduction.

Question 189

Describe properties of anion conductive membranes:

Answer 189

Hydroxide Conductivity
Alkaline Stability
Water Management

Despite recent advances, anion exchange membranes still show lower conductivity and stability compared to Nafion, hindering the development of AEMFCs.
In addition, water management in AEMFCs is more complex than in PEMFCs

Question 190

How is water production at the cathode of AEM fuel cells managed?

Answer 190

Water is both generated at the anode and consumed at the cathode.

Water is moved from the cathode to the anode by electro-osmotic drag.

Water management includes providing enough water to hydrate the membrane, while avoiding flooding and dry out at both electrodes.

The balance between membrane hydration and flooding is thin and more complex than in PEMFCs

Question 191

What is the best catalyst for hydrogen oxidation (HOR) at the anode of fuel cells?

Answer 191

The state-of-the-art catalyst for HOR is platinum

But with lower noble metal loading (0.05 𝑚𝑔_𝑃𝑡 𝑐𝑚^(−2)) compared to the oxygen reduction catalyst (0.13 𝑚𝑔_𝑃𝑡 𝑐𝑚^(−2))

The kinetics of hydrogen oxidation is slower in alkaline than acidic media, resulting in higher platinum loadings at the anode for AEMFCs compared to PEMFCs.

Question 192

What are the difficulties with Oxygen Reduction Reaction (ORR) at the Cathode?

Answer 192

Several reactions possible, producing multiple intermediates.

Peroxides may form which are reactive and may destroy the catalyst.

Question 193

What approaches have been made to improve the sustainability of fuel cells?

Answer 193

One approach to improving the sustainability of fuel cells is decreasing platinum loading, this can be achieved by:

1. Increasing the activity of platinum-based catalysts
2. Increasing platinum utilization

Question 194

How can the activity of platinum-based catalysts be increased?

Answer 194

By alloying Pt with other metals one can modifying the binding energy of the intermediates and get closer to the top of the volcano.

Pt3Ni(111) is the Pt alloy with the highest reported activity so far.

Question 195

How can platinum utilisation be increased?

Answer 195

Improving the catalyst support

Avoiding platinum aggregation

Question 196

What Platinum-free Oxygen Reduction Catalysts are being made?

Answer 196

M-N-C catalysts consist of nitrogen and transition metals on a carbon support.

They are generally synthesised by:
1. Carbon support, nitrogen-containing molecule or polymer and Transition metal precursor
2. Pyrolysis
3. Washing

Question 197

What pH do Pt-free catalysts work best?

Answer 197

Alkaline

Question 198

Describe fuel cell stability:

Answer 198

To produce a fuel cell that can run for hundreds of hours, the stability of the oxygen reduction catalysts is of paramount importance.

Unfortunately, all ORR catalysts lose activity with time to a certain degree.

The loss in activity results from
- The dissolution of active sites (the metal)
- A rearranging of the structure of the active site to a less performing form
- The degradation of the support

In general, stability in fuel cells is less of an issue compared to electrolysers. Carbon can always be used as a support and porous layer, it does not corrode as for the case of oxygen evolution. In addition, the bipolar plates are less subject to corrosion, so it does not require Ti (as for the case of OER in electrolysers).

Question 199

How does the potential we need to apply to start reactions relate to the thermodynamic theoretical cell potential in fuel cells?

Answer 199

Opposite to electrolysers, the potential out of a fuel cell is always lower than what is thermodynamically predicted.

Question 200

Describe a rotating disk electrode (RDE):

Answer 200

RDE: rotating disc electrode
RRDE: rotating ring-disc electrode

3 electrode set-up:
- Working electrode – where the catalyst is deposited and depending on the reaction we are looking at, we can observe oxygen reduction to water/peroxide
- Counter electrode – to complete the cell circuit
- Reference electrode – used to measure the potential at the working electrode

Oxygen is bubbled through the electrolyte and is reduced at the catalyst, deposited on the working electrode.
This set-up studies only the half reaction at the cathode.

This electrodes have a conductive glassy carbon substrate where the catalyst is deposited .
They can rotate, to provide defined mass transfer.

The RRDE has in addition a platinum ring around the glassy carbon disc

Question 201

Limitations of rotating disk electrodes:

Answer 201

Only looks at the half reaction at the cathode

Has a different geometry from fuel cells, as oxygen diffuses through the electrolyte, instead of coming in gaseous form from the other side of the electrolyte

Limited mass transport and current density

Question 202

What’s a reference electrode?

Answer 202

A reference electrode is an electrode that has a stable and well-known electrode potential.

The overall chemical reaction taking place in a cell is made up of two independent half-reactions, which describe chemical changes at the two electrodes.

To focus on the reaction at the working electrode, the reference electrode is standardized with constant (buffered or saturated) concentrations of each participant of the redox reaction.

Question 203

Describe membrane electrode assembly (MEA):

Answer 203

This is the closest geometry to real fuel cells, it contains an anion/proton exchange membrane and a gas diffusion layer.

Both the cathodic oxygen reduction and the anodic hydrogen oxidation are investigated and the gas transport is similar to that in fuel cells.

However, they are very time consuming and complicated to assemble and test.

MEA consists of three key elements: a proton-conducting membrane (typically made of perfluorosulfonic acid), an anode catalyst layer, and a cathode catalyst layer.
The membrane allows ions to pass through while blocking electrons, facilitating the electrochemical reactions of hydrogen oxidation at the anode and oxygen reduction at the cathode.
Catalyst layers on either side of the membrane enhance reaction rates by providing active sites for electrochemical reactions.

Question 204

Describe gas diffusion electrodes (GDE):

Answer 204

In between RDE and MEA. Has a more similar geometry to fuel cells, compared to RDE. Offers faster mass transport and higher current density, compared to RDE.
Is easier and faster to assemble and used compared to MEA.

Gas diffusion electrodes (GDE) are electrodes with a conjunction of a solid, liquid and gaseous interface, and an electrical conducting catalyst supporting an electrochemical reaction between the liquid and the gaseous phase.

Question 205

Describe the difference in linear sweep voltammetry curves for rotating disc electrodes (RDE) and gas diffusion electrodes (GDE):

Answer 205

GDEs have a continuous flow of gas, so do not experience mass transfer / diffusion limitations.
Thus, as potential is decreased, current density continues to increase exponentially.

RDEs experience a kinetic region as well as a mixed and diffusion region (regions explain the limiting steps).

Question 206

Explain linear sweep voltammetry:

Answer 206

In linear sweep voltammetry the potential is varied linearly.

At the initial potential no reaction happens and the measured current is zero.

As the potential is decreased, oxygen is reduced and a negative current is measured. The kinetic of reaction increases with overpotential and so does the current -> Kinetic Region.

When the reaction kinetics is so high that all oxygen at the surface is immediately consumed, the reaction becomes controlled by the oxygen diffusion and the current is independent on the potential -> Diffusion Region

Question 207

What ways is current normalised to give different parameters?

Answer 207

The current can be normalized with respect to different parameters, giving:

The current density: Normalized to the geometric surface area of the electrode

The specific activity: Current normalized to the surface area of the catalyst

The mass-specific activity: Current normalized to the mass of the catalyst (for Pt-based catalyst it is usually normalized to the mass of platinum)

Question 208

Define:
Onset Potential
Overpotential η
Diffusion-limited current
Half-wave potential

Answer 208

Onset Potential: Highest potential at which a faradaic current is measured

Overpotential η: Difference between the onset potential and the theoretical thermodynamic potential

Diffusion-limited current: Measured current in the diffusion region

Half-wave potential: Potential at which the current is half of the diffusion-limited current

Question 209

How can the amount of hydrogen peroxide (H2O2) formed on a rotating ring-disc electrode be determined?

Answer 209

In a rotating ring-disc electrode, oxygen that has only partially reacted to hydrogen peroxide on the disc catalyst is further reduced to water on the platinum ring, with a collection efficiency ε.

The current generated at the ring (𝐼_𝑅) and at the disk (𝐼_𝐷) are collected separately, allowing the determination of the hydrogen peroxide production and the average number of electrons (n).

Equations:
n = 4*I_D/(I_D + I_R/ε)

Y_peroxide = 100(2I_R/ε)/(I_D + I_R/ε)

Question 210

Why is biomass sustainable?

Answer 210

Its inherent energy comes from the sun

It absorbs carbon dioxide as it grows

Can be used directly to displace fossil fuels such as oil, coal, and natural gas for the production of energy or other products

Can regrow in a relatively short time (unlike fossil fuels)

The carbon that is released from the organic material was sequestered recently, compared to fossil fuels where the carbon was sequestered hundred of millions of years ago.

Biomass residues are wastes that would be combusted or left to decompose.

Question 211

What occurs in biomass electrolysis?

Answer 211

Biomaterial (CxHzOz) oxidised at the anode to form H+, e-, and CO2, and the protons are then reduced at the cathode to form H2.

Oxidising an organic material is much more favourable thermodynamically (more easily oxidised) than water.

For biomass, thermodynamically < 1.23V, in practise < 1 V

Question 212

How is biomass electrolysis cell voltage found?

Answer 212

V = ∆E.eq + Ƞa + Ƞc + IR

Where:
∆E.eq is the difference between equilibrium electrode potentials
Ƞa is the overpotential at the anode
Ƞc is the overpotential at the cathode
IR is the ohmic drop

Oxidising biomass is much easier than water.
∆E.eq differs owing to the thermodynamics of an organic molecule oxidation.

Biomass efficient transformation is governed by kinetics, which can be achieved with a suitable electrocatalyst to compensate for the involved overpotentials

Thermodynamics of oxidation determines ∆E.eq.

Question 213

What are the anode and cathode reactions in biomass electrolysis?

Answer 213

Anode:
CxHyOz + H2O → oxidation products + H+ + e-

Cathode:
H+ + e- → H2

Alkaline (NaOH, KOH) often used, though acid can also be used.

Question 214

What by product does biomass electrolysis produce?

Answer 214

Complete biomass oxidation produces CO2 (under acidic conditions) or CO2-3 (under alkaline conditions).

Partial oxidation generates value-added chemicals.

OER involves multi-proton-coupled electron-transfer steps, suffering from the limitation of suboptimal scaling relation between OOH and OH adsorption energies, which is the major bottleneck and requires a cell potential above 1.23 V.

Question 215

If the activity is instead plotted as a function of the binding energy of OH would one obtain the same volcano shape?

Answer 215

The volcano plot would have the same shape if plotted as a function of the binding energy of another intermediate, as the binding energies of key intermediates are linearly dependent.

Question 216

Is the top of the volcano equal to the thermodynamic potential for oxygen reduction (zero overpotential)?

Answer 216

No, the top of the volcano is not equal to zero overpotential.

This is because the binding energy of O* and HO* cannot be tuned separately and even the optimal O* binding energy results in the presence of two ORR steps which are uphill in free energy.

This results in the requirement of a finite overpotential for the reaction to occur.

Question 217

In PEMFC, the cell voltage increases with temperature up to around 80°C and decreases rapidly for temperatures above this value.

Explain this behaviour and discuss why solid oxide fuel cells are operated at higher temperature.

Answer 217

As the temperature increases the electronic conduction decreases, while the ionic conduction in the electrolyte increases, resulting in an overall increase in the rate of reaction up to 80°C.

Above this temperature, water starts to boil causing oxygen starvation at the catalyst and the conductivity in the Nafion membrane decreases significantly.
As a result, the optimal operating temperature for PEMFC is 70-90°C.

For the case of solid oxide fuel cell, the presence of a solid electrolyte removes the issues associated with water management and allows operating at higher temperature and reaction rates.

Question 218

State three limitations/remining challenges for alkaline exchange-membrane fuel cells to became widely commercial:

Answer 218

• The higher requirement of platinum for the HOR catalyst
• The lower activity and stability of oxygen reduction catalysts compared to platinum in acidic conditions
• The limited conductivity of anion exchange membranes.

Question 219

How can you determine the activity of a new catalyst?

Answer 219

Using linear sweep voltammetry, a polarisation curve can be made in oxygen saturated atmospheres.

Impedance spectroscopy can be also used to characterize the main resistances in the system.
Finally, a durability test should be performed to assess the stability of the catalyst.

Question 220

Provide three different approaches for improving the performance of a semiconductor photocatalyst in CO2 reduction:

Answer 220

Adding a co catalyst

Enhancing the specific surface area of the semi conductor

Creating a Z scheme heterojunction

Adding a hole scavenger to the reaction media

Doping the semiconductor

Question 221

Write the reaction equations for the four-electron pathway for OER, starting from H2O:

Answer 221

H2O ↔ HO* + H+ + e-

HO* ↔ O* + H+ + e-

O* + H2O ↔ HOO* + H+ + e-

HOO* ↔ * + O2 + H+ + e-

Question 222

IrO2 is a better / more stable option as an OER catalyst over RuO2.
Why?

Answer 222

At lower applied potentials, there’s a process where a layer forms on the surface of the Ru catalyst.
This layer, called a hydrous amorphous layer, makes it easier for the ruthenium ions to escape into the solution. This leads to catalyst degradation and loss of efficiency.

The same happens for IrO2, but at a much higher applied potential.

M²ⁿ⁺Oₙ²⁻ + nH2O <-> M²ⁿ⁺(aq) + 2n OH⁻(aq) O + 2n e-

Question 223

Design an electrochemical cell with all its components which displays high faradaic efficiency towards CO and another one towards C2H4 and explain briefly how you would assess their performance:

Answer 223

Cell: H-type with anion exchange membrane, gas diffusion electrode
Electrolyte: 0.1 M KHCO3 with CO2 inlet and outlet
Working electrode: Au, Ag, or FeNC for CO and Cu for ethylene
Counter electrode: Pt foil or IrO
Reference electrode Ag/AgC

The CO2 reduction performance would be assessed by performing chronoamperometry measurements at different negative working potentials (between -0.2 and -1 V approximately) for a significant time interval (15-60 min) and constant CO2 purge into the electrolyte.

The head space would be analysed by gas chromatography to quantify the amount of product and the faradaic efficiency would be calculated by FE% = nFm/Q
Where:
* n is the number of reaction e- (2 for CO, 12 for C2H4)
* m is the moles of product
* Q is the total charge in coulombs (Q = It)

Question 224

What happens during the first electron transfer in CO2 reduction?

Answer 224

CO2 + e- → CO2*-

The first electron transfer is employed to create a CO2- radical anion.
When this happens the C-O bond gets weakened and the molecule reorganizes from the original linearity to a bent radical owing to the unpaired electron.

Carbon dioxide is a highly thermodynamically stable molecule owing to the strong C=O bond, and therefore its reduction is kinetically sluggish.

This radical intermediate can then undergo further reduction steps to produce various products depending on the reaction conditions and the specific catalyst involved in the CO2 reduction process.

Question 225

How do you derive the water stability diagram?

Answer 225

Start with the Nerst equation:
E = Eo − (RT/nF)*lnQ

Simplifying and considering that ln x = 2.303 log x,

E = Eo − (0.059/n)*log10(Q)

And Q is the activity coefficients, a, of the products/reactants, = partial pressures:
Q = p.products / p.reactants

Considering Water at 1 bar and 1M solution:
O2 + 4H+ + 4e- -> 2H20 (so 4e- used)

E = 1.23 - (0.059/4)log10([H2O]^2 / [O2][H+]^4)
E = 1.23 - (0.059/4)log10(1/[H+]^4)

pH = -log10 (H+)

E = 1.23 - 0.059pH

Question 226

Explain the working principle of rotating ring-disc electrode and how they differ from rotating disc electrode:

Answer 226

RDE consists of a glassy carbon disk where the catalyst is deposited. They allow study of a single reaction in a half-cell configuration and can rotate to provide defined mass transport.

RRDE has a Pt ring where peroxide produced at the disc is transported.

Oxygen travels to the first disc where it is reduced to peroxide (2e-) or water (4e-)

If peroxide is produced, it goes to the disk where it is oxidised back to O2

Monitoring the current at the disc and ring, we can determine how much peroxide was produced at the disc

Question 227

Explain the meaning of the average number of electron transfer. What is the desired value, and what are the potential issues associated with high peroxide production?

Answer 227

The average number of electron is a measure of how much of the reacted oxygen is converted to water (4 electron) or peroxide (2 electrons).

The ideal ORR catalyst should have n=4, i.e. only produce water.

This is because peroxide production accelerate the degradation of the catalyst.