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1

What is the definition for lattice enthalpy? 

ΔLEHø is the enthalpy change that accompanies the formation of one mole of an ionic lattice from its gaseous ions under standard conditions.

2

What is the definition for standard enthalpy change of formation? 

ΔfHø, is the enthalpy change that accompanies the formation of 1 mole of a compound from its elements. 

3

What is the definition of the first ionisation energy? 

is the energy change that accompanies the removal of 1 mole of electrons from 1 mole of gaseous atoms. 

4

Why do stable ions go on to form ions when a huge amount of energy is required? 

When oppositely charged ions attract one another, forming a giant ionic lattice, there is a huge lowering of the energy through very strong attraction. So, aulthough the amount of intial energy required to fomr ions is large, the lowering of the enrgy on forming the lattice more than compensates for it. 

This is the reason ionic substances have strong ionic bonds and high melting and boiling points. 

5

Give an example of an equation of lattice enthalpym and explain the enthalpy change.  

K(g)  + Cl- (g) → KCl (s) 

  • The ions are both gaseous.
  • One mole of the substance is formed.
  • The enthalpy change is negative - energy is released to the surroundings.
  • Ionic lattice formation is exothermic. 

6

What do exothermic lattice enthalpy values indicate?

  • More exothermic lattice enthalpy values mean stronger ionic bonds (stronger electrostatic interactions)
  • More exothermic lattice enthalpy values mean higher melthing and boiling points as more energy is required to overcome the interactions present. 
  • The most exothermic lattice enthalpies arise when ions are small and have large charges - as the charges cause large electrostatic forces and smaller ions can get closer togther. 

7

Why is it not possible to measure lattic enethalpy directly? 

Because it is impossible to form one mole of an ionic solid from its gaseous ions. 

8

How can you calculate lattice enthalpy? 

By constructing a Hess' cycle called Born - Haber cycle - showing intermediate steps between the elements that form the ionic substance. 

9

What are the key features of a Born - Haber cycle? 

  • A continuous cycle is forme that can start at the elements and end at the elements.  
  • Includes a step that shows the formation of one mole of the solid ionic lattice from the gaseous ions - this corrosponds to the lattice enthalpy. 
  • The remaining steps show intermediate changes that corrospond to key enthalpy changes that can be measured. 
  • The lattice enthalpy is calculated by applying Hess' law. (the enthapy change is independant of the rout of reaction) 

10

Whta are key enthalpy changes? 

Elements have have to go through a series of steps before they are ready to form ionic lattices. These changes have enthalpy chnages associated with them. The key enthalpy changes are given below. 

11

What is the symbol and defintion of the standard enthalpy change of formation?

ΔfHø The enthapy change that occurs when one mole of a comound is formed from its constituent elements in their standard states under standard conditions. 

12

Give an example of a symbol equation for standard enthalpy change of formation, and is it exo or endothermic? 

K(s) + 1/2 Cl2(g) → KCl(s)

It is an exothermic process.

13

What is the symbol and definition for the standard enthalpy change of atomisation?

ΔaHø The enthalpy change that takes place when one mole of gaseous atoms forms from the element in it's standard state. 

14

Give an example of a symbol equation for standard enthalpy change of atomisation, and is it exo or endothermic? 

For potassium metallic bonds are broken; 

K(s) → K(g) 

For chlorine, covalent bonds are broken:

1/2Cl2 (g) → Cl(g)

The process is always endothermic.

 

15

What is the symbol and definition for the first  ionisation energy?

ΔI1Hø The enthalpy change accompanying the removal of one electron from each atom in one mole of gaseous atoms to form 1 mole of gaseous 1+ ions.

16

Give an example of a symbol equation for the first ionisation energy and is it exo or endothermic? 

K+(g) → K+(g) + e-

The process is endothermic. 

17

What is the symbol and definition for the second ionisation energy? 

The enthalpy change accompanying the removal of one electron from each ion in one mole of gaseous 1+ ion to form one mole of gaseous 2+ ions. 

18

Give an example of a symbol equation for the second ionisation energy, and is it exo or endothermic? 

Ca+(g) → Ca2+(aq) + e-

The process is endothermic. 

19

What is the symbol and defintiion for the first electron affinity?

ΔEA1Hø The enthalpy change accompanying the addition of one electron to each atom in one mole of gaseous atoms to form one mole of gaseous 1- ions. 

20

Give an example of a symbol equation for the first electron affinity, and is it exo or endothermic? 

Cl(g) + e→ Cl(g)

The process is exothermic.

21

What is the symbol and definition for the second electron affinity?

ΔEA2Hø 

One mole of gaseous 2- ions is formed from gaseous 1- ions.

22

What is the overall aim when constructing a Born-Haber cycle?

  • Make sure each original element is gaseous and on its own as separated atoms.
  • Ionise relevant elements to give the appropriate positive charge needed (first and second ionisation energies).
  • Ionise relevant elements to give the appropriate negative charge needed ( i.e. first and second electron affinity).

23

What can Born-Haber cycles be used to calculate? 

Any unknown enthalpy change. We can apply Hess' law to the cycle:

  • ΔfHø = the sum of all other enthalpy changes or
  • the sum of anticlockwise enthalpy changes = sum of clockwise enthalpy changes.

(on pages 56 - 61 there is more on Born-Haber cycles)

24

What is the definition of the standard enthalpy change of solution? 

ΔsolHø , is the enthalpy change that takes place when one mole of a solute is completely dissolved in water under standard conditions.

 

25

What is the definition of the standard enthalpy change of hydration?

ΔhydHø, is the enthalpy change that takes place when dissolving one mole of gaseous ions in water. 

26

What happens when a solid dissolves? 

When a solid dissolves, two processes take place.  

  • The ionic lattice breaks down.
  • The free ions become part of the solution (hydration).

A change in enthalpy occurs when this overall process happens (ΔsolHø).

27

Is the dissolving of a substance exo or endothermic?

The dissolving of a substance can be endothermic (for example ammonium nitrate dissolves in water) or exothermic (for example, when calcium chloride dissolves in water). 

28

Why is the enthalpy change of ionic lattice breakdown equal to -ΔLEHø?

The lattice enthalpy is the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions. The process that occurs when the lattice breaks down during dissolving is the reverse of this process. We imagine the lattice becoming free gaseous ions. 

29

30

The breakdown of an ionic lattice;

How do the processes of lattice formation and break down compare?

  • the processes are identical but the reverse of one another
  • the enthalpy change has the same value but different signs 
  • lattice enthalpy has a negative sign and is exothermic
  • the breakdown of the ionic lattice has a positive sign and is endothermic. 

31

What dictates the size of lattice enthalpy? 

The magnitude of any lattice enthalpy is dependent on:

  • the size of the ions involved
  • the charges on the ions
  • ionic bond strength (which is dependent on ionic size and charge)

32

Why does the size of ions affect the size of the lattice enthalpy?

Smaller ions, i.e. ions with a smaller ionic radius, can get closer together. They will attract one another more strongly and give rise to more exothermic lattice enthalpy values, i.e. more negative values. 

Lattice enthalpy becomes less exothermic and less negative ad the size of negative ions increase.

33

How do the charges on the ions affect the lattice enthalpy? 

Ions with higher charges cause greater electrostatic attraction and in turn more exothermic lattice enthalpy values. The most exothermic lattice enthalpy values arise from small, highly charged ions.

The smallest, most highly charged ions will give rise to the largest lattice enthalpies, as they can pack closer to oppositely charged ions, with higher attractions. Lattice enthalpies will become more exothermic - more negative. 

34

State and explain which compound has the most exothermic lattice enthalpy, MgCl2 , MgBr2 or MgI2 ?

Magnesium chloride has the most exothermic lattice enthalpy because the chloride ion is smaller than both bromide ions and iodide ions. This means the Mg2+ and Clions in the MgCl2 lattice can pack closer together and exert a greater attraction on each other than the ions in MgBr2 or MgI2.

(when discussing situations such as the one shown in the worked example, it is essential that you refer to the actual ions involved. For example, when you compare the lattice enthalpy of MgCl2 and MgBr2 you should say 'the magnesium 2+ ions can pack closer in MgCl2 ...' rather than just magnesium can get closer to chlorine'  - the latter would be incorrect, as it is the ion not the element forming the lattice.)

35

Explain charge density.

Charge density describes how 'spread out' a charge is across an ion. If we consider two ions with the same charge, the smaller of the two ions would have a greater charge density as the charge is found in a smaller area. The larger of the two would have a lower charge density and be less attractive, as the charge is more dispersed. 

36

What is hydration? 

  • Once the ionic lattice has broken down into its constituent ions, these have to become part of the solution. The ions are able to do this if the solvent (water in the case of hydration) can interact with them in similar ways to the bonding in the lattice.
  • 'like dissolves like' 
  • Ionic solids are therefore able to dissolve in polar solvents, such as water.

37

What happens during hydration? 

  • the positive ions will be attracted to the slightly negative oxygen in water molecules.
  • the negative ions will be attracted to the slightly positive hydrogen in water molecules. 
  • the water molecules will completely surround the ions. 

38

39

What enthalpy change occurs when ions become hydrated?

Energy is released when new bonds are between ions and water molecules. This is the standard enthalpy change of hydration ΔhydHø, and is the enthalpy change when one mole of aqueous ions are formed from their gaseous ions, under standard conditions. 

40

Is the enthalpy change of hydration an endo or exothermic process?

Hydration is an exothermic process. 

41

What is the magnitude of the enthalpy of hydration dependent on?

  • the size of the ions involved 
  • the charges on the ions

42

How does the size of ions involved affect the magnitude of the enthalpy of hydration? 

Ions with smaller ionic radii can get closer to the water molecules and are able to attract them more strongly. This means that on hydration, more energy is released and ΔhydHø becomes more exothermic.

43

How do the charges on the ions affect the magnitude of the enthalpy of hydration?

The higher the charge on an ion, the greater the attraction it will have with the water molecule. This will give a more negative and hence more exothermic value for the enthalpy of hydration.  

The size and charge of ions affecting the magnitude if the enthalpy of hydration is represented by this example.  

44

When creating a born-Haber cycle for lattice enthalpy where do  

ionic solids go and gaseous ions go?

  • The ionic solid is at the bottom of the cycle
  • The gaseous ions are at the top of the cycle

45

When creating a born-Haber cycle for lattice enthalpy where do the routes via lattice enthalpy and the route via enthalpies of solution and hydration go?

  • the route via lattice enthalpy is shown on the left
  • the route via enthalpies of solution and hydration is shown on the right
  • (example on pg 65)

46

What is the definition of entropy?

Entropy, S, is the quantitive measure of the degree of disorder in a system. 

47

What is the definition of the standard entropy? 

The standard entropy, Sø, of a substance  is the entropy content of one mole of the substance under standard conditions.

48

What is the definition of standard entropy change of reaction?

Standard entropy change of reaction, ΔSø, is the entropy change that accompanies a reaction in the molar quantities expressed in a chemical equation under standard conditions, all reactants and products being in their standard states. 

49

When talking about entropy what do the terms surroundings and system describe? 

System is used to describe the actual particles involved in a reaction or process, surroundings describes everything outside the system. 

50

What units do standard entropies have?

J K-1 mol-1

51

As the particles in a system become more disordered how does this affect entropy?

It increases the entropy of a system. If you think of solids compared to gases, we would say that the solid has a lower disorder compared to the gas - the gas id more disordered. 

52

Entropy is a key _______ a factor that can be used to describe chemical processes, usually alongside enthalpy.

thermodynamic 

53

All substances above 0 K possess a certain degree of disorder because they are in constant motion. What does this mean?

  • entropy, S, is always a positive number above 0.
  • at 0 K entropy is zero for perfect crystals.

54

Does entropy tend to increase or decrease?

Most substances are thermodynamically stable at the lowest energy state, which would correspond to a low entropy. However, entropy always tends to increase; liquid water naturally evaporates into gaseous water, increasing entropy; heat energy spreads out from hot objects, increasing entropy; salt particles dissolve in water, increasing entropy. It is always more probable that a more disordered system will be found instead of a more ordered system. 

55

Can entropy decrease during a reaction? 

Entropy can change during the course of a chemical process or reaction and these changes would be described in terms of the change in entropy ΔS .

There is always a tendency towards higher entropy (the second law of thermodynamics)  although it can also decrease giving a negative value of Δ S. For example, water freezing would represent a decrease in entropy as a liquid has become a more ordered solid, with lower levels of energy disposal. 

56

How is entropy affected by temperature? 

The entropy of pure substances increases with increases temperature.

  • Particles at higher temperatures have higher energy and move more
  • The arrangement of particles at higher temperatures becomes more random.
  • Entropy of solids < entropy of liquids < entropy of gasses

57

Describe the entropy change when dissolving solids? 

If a solid ionic lattice dissolves, ions can spread out and the positions of the ion are far more disordered than within the lattice. This means entropy increases. 

58

How do the number of gas molecules affect entropy?

  • An increase in the number of gas molecules causes an increase in entropy.
  • A decrease in the number of gas molecules causes a decrease in entropy. 

59

How is the standard entropy change of a reaction calculated?

Using the entropies of the reactants and products.

60

This equation shows the formation of ammonia from nitrogen and hydrogen. 

N2 + 3H2 → 2NH3

Use the data in the table to calculate ΔS for this reaction.

61

What is the definition of the free energy change? 

  • ΔG, is the balance between enthalpy, entropy and temperature for a process:
  • ΔG = ΔH - TΔ
  • A process can take place spontaneously when ΔG < 0.

 

62

How do we calculate total change in entropy? 

63

What must happen for spontaneous changes to occur? 

For spontaneous changes to occur, the total change in entropy must be positive.

Reactions that have a decrease in entropy can occur spontaneously if the change in entropy of the surroundings is positive enough to make the total change in entropy positive. 

64

What is the energy that becomes 'free' during a reaction is known as?

The energy that becomes 'free' during a reaction is known as Gibbs free energy after the theoretical physicist and mathematician, Josiah Willard Gibbs, and is given the symbol G. 

65

Using the Gibbs equation how can you predict the feasibility of reactions? 

  1.  Large increases in entropy will cause decreases in ΔG, because the term - TΔS will become larger. 
  2. Large negative values for ΔH (i.e. exothermic reactions) will result in more negative values for ΔG.

There is a balance, for example, a highly exothermic reaction causes a large decrease in entropy, ΔG may be positive, in which case the change would not be spontaneous. 

66

How do the different possible combinations of ΔH and ΔS affect the feasibility of spontaneous change in the Gibbs equation? 

67

Explain why exothermic reactions are generally spontaneous and endothermic reactions are generally only spontaneous in certain situations? 

  • exothermic reactions ate generally spontaneous - the negative value of ΔH is usually still able to make ΔG negative, even  if the entropy change is positive
  • endothermic changes are only spontaneous if the entropy is high enough to make TΔS large and positive, i.e. greater the ΔH.

68

Describe how to work out the correct units in a free energy related calculation? 

In free energy related calculations, you need to get the units of joules for entropy and enthalpy the same. 

  • ΔH is usually given in kJmol-1
  • ΔS is usually given in JK-1mol-1​​

First get ΔS into kJ K-1 mol-1:

  • to convert l to KJ, divide by 1000

Also remember that entropy is worked out using temperature min K:

  • to convert oC  to K, add 273
  • to convert K to  oC, subtract 273

 

69

What are the limitations of using ΔG to predict the feasibility of reactions?

 

Calculating the value of free energy, ΔG, gives a theoretical answer for whether a reaction or process will react spontaneously, i.e. whether it is thermodynamically possible.

Whether or not a reaction proceeds spontaneously also depends on kinetic factors;

  • The reaction may have a high activation energy - energy needs to be initially supplied to overcome this, e.g. igniting fuel. 
  • The rate of the reaction may be extremely slow. 

Equally, reactions that have positive values of ΔG are considered not feasible. They can be made to takes place, however, usually by changing the temperature of the reaction. 

70

Define oxidation 

Oxidation is the loss of electrons or increase in oxidation number. 

71

Define reduction 

Reduction is gain of electrons, or a decrease in oxidation number. 

72

Define an oxidising agent.

An oxidising agent is the species that is reduced in a reaction and causes another species to be oxidised. 

73

Define a reducing agent. 

A reducing agent is the species that is oxidised in a reaction and causes another species to be reduced. 

74

Electrons must be transferred during a redox reaction - they cannot just disappear, they must be transferred between species.

Describe oxidising agents.  

When a species is oxidised, it loses electrons. These electrons are gained by the species being reduced. If a chemical is readily reduced it will gain the electrons in the reaction and is known as the oxidising agent

75

Electrons must be transferred during a redox reaction - they cannot just disappear, they must be transferred between species.

Describe reducing agents.  

When a species is reduced, it gains electrons. These electrons have been removed from the species being oxidised. If a chemical is readily oxidised it will provide the electrons in the reaction and is known as the reducing agent.  

76

Break down this redox reaction into the two half equations showing oxidation nd reduction. 

Mg + 2HCl → MgCl2 + H2

  1. Mg → Mg2+ + 2e-   This is the oxidation half-equation. 
  2. 2H+ + 2e- → H2   This is the reduction half-equation. 
  • The number of electrons lost in the oxidation half-equation is the same as the number of electrons gained in the reduction half-equation. This must always be the case - the electrons must balance. 
  • Cl is not included in either of the half equations. This is because chlorine ions are not involved in the redox reaction - its oxidation number remains  -1 throughout the reaction. It is known as a spectator ion and does not need to be included in the half equation. 
  • (electrons can only be added)

77

We can construct an overall equation for a redox reaction if we know the oxidation and reduction half-equations. 

What series of steps must be followed to construct an overall equation? 

  1. Identify the oxidation and reduction half equations. - Equations for the separate oxidation and reduction reactions are shown lined up. 
  2. Balance the electrons. - The half-equations are scaled up or down (by multiplying the entire half-equation) so that the number of electrons is the same in each.
  3. Add the two half-equations together and cancel the electrons. 

(worked example on pg 72)

78

Assigning oxidation numbers to each substance involved in a redox reaction allows you to track the movement of electrons.

What does the increase and decrease in oxidation numbers show you? 

 

  • An increase in oxidation number shows a substance has been oxidised.
  • A decrease in oxidation number shows a substance has been reduced.   
  • The overall increase in oxidation number will equal the overall decrease in oxidation number. 

79

It is possible to predict the whole reaction that will occur and write a full redox equation by knowing each species involved and how they will behave.

Give some examples of how different species behave in a reaction. (tendency towards losing or gaining electrons etc)  

  • Metals generally will be the species that lose their electrons 
  • non-metals usually gain electrons 
  • the group number (number of outer electrons) can be used to predict the ion change for most elements - along with the other rule for assigning oxidation numbers to species involved in redox reactions.
  • Worked example on page 72. 

80

What are the common examples of species you that need to add to balance a redox equation?

  • H+ usually when reactions are carried out under acidic conditions.
  • OH- usually when reactions are carried out in alkaline conditions
  • H2 O usually when the equation needs extra O and H to be added. Simply add the number of ions needed to balance the equation. 

81

Why are titrations useful? 

Titrations are a way of determining amounts of substances, for example, the concentration of an unknown acid. They can also be used to determine the amounts of species being oxidised or reduced - this is known as a redox titration. 

82

How do you carry out a redox titration? 

A known concentration of either a reducing agent or oxidising agent is placed in a burette and titrated against an unknown concentration of the chemical that is being oxidised or reduced respectively.

Whilst for acid-base reactions we use an indicator to identify the end point, often for redox titrations this is not necessary. Many redox titrations involve species that self-indicate - they change colour between different oxidation states. 

MnO4is commonly used to oxidise solutions containing iron(II), it can be used as an oxidising agent in many other chemical reactions.   

83

Describe the use of manganate in the redox titration between Fe2+ and MnO4-

Manganate(VII) MnO4- is a common oxidising agent, usually obtained from potassium permanganate(VII), KMnO4 . It has a deep purple colour but becomes colourless when it is reduced from +7 to +2 oxidation states.

MnO4- → Mn2+  Purple to colourless

This usually occurs in the presence of H+ ions, i.e. in acidic solutions (typically sulphuric acid is used, as hydrochloric acid reacts with MnO4-.)

84

What is the balanced symbol equation for the redox titration between Fe2+ and MnO4- .

MnO4- + 8H+ + 5Fe2+ → Mn2+ +5Fe3+ + 4H2​O

85

How is the end point of the titration between Fe2+ and MnO4 ?

The end point is seen when excess MnO4ions are present - indicated by a faint pink colour appearing. This occurs because all the Fe2+ ions have reacted and the MnO4can no longer be reduced to the colourless Mn2+ .

86

What can the redox titration between Fe2+ ions and MnO4- be used to determine?

We can use this reaction to determine the concentration of iron in an unknown solution, or indeed other ions that can be oxidised by MnO4-, or to calculate the percentage composition of a metal in a solid sample if a compound or alloy.  

87

How do you calculate the mass of iron in the redox titration between Fe2+ ions and MnO4 ?

When calculating the mass of iron in a substance, the molar mass of a Fe2+ ion is taken to be the same as the molar mass of Fe- therefore the titration results can be used directly to convert the number of moles into the mass present.  

(worked example on page 74) 

88

Redox titrations between I2 and S2O3 ;

Why is iodine a useful substance for a redox titration? 

Iodine is a useful substance to use in redox titrations as it has a dark blue-black colour in the presence of starch but when it is reduced to iodide ions this colour disappears.0

I2 → 2I- + 2e- 

blue-black colour with starch → blue-black colour disappears  

89

The redox titration between I2 and S2O32- ;

What will the reduction of iodine to iodide ions occur in the presence of?

  • This reduction of iodine to iodide ions will occur in the presence of thiosulphate ions, S2O32-.
  • 2S2O32- + I2 →  S2O32- + 2I-    (thiosulfate to tetrathionate)
  • We can use aqueous iodide ions and aqueous thiosulfate to determine the concentration of unknown reducible species.

90

Redox titrations;

We can use aqueous iodide ions and aqueous thiosulfate to determine the concentration of unknown reducible species.

Often this involves an initial reaction between the unknown oxidising agent and iodide ions, which has iodine as a product. 

Describe this reaction. 

Often this involves an initial reaction between the unknown oxidising agent and iodide ions (for example by mixing it with potassium iodide), which has iodine as a product. 

For example; Cl2 + 2I- → 2Cl- + I2

This liberated iodine then goes on to react with thiosulphate ions, from a solution with a known concentration being added from a burette. 

2S2O32- + I2 → S4O62- + 2I-

In the presence of starch, a blue-black colour will remain present as long as there is any iodine. Once this has all reacted with the thiosulphate ion, this will disappear, marking the end point of the reaction. 

(worked example on page 74)

91

Sometimes you will be required to interpret the results from redox systems you are not familiar with. You will be given any relevant equations and experimental results that you need. The procedures will be the same.

Give some general steps you may need to use. 

  • Determining numbers of moles used for any substances you have concentrations and volumes for, using n=cV and number of moles = mass/molar mass, where necessary
  • identifying the reaction stoichiometry using balanced equations
  • deciding on the amounts that have reacted for known substances and deducing the amounts of unknows 
  • using the equations discussed above to calculate unknown quantities or concentrations. 
  • using the equations discussed above to calculate unknown quantities or concentrations. 

(Worked example pg 75)

92

Redox titrations can be used to estimate the concentration of a solution containing Cu2+ ions. The copper(II) solution is mixed with aqueous iodide ions, I-, and a redox reaction occurs. What two half equations show this?

  1. iodide ions are oxidised to iodine, 2I- → I2 + 2e-
  2. copper(II) ions are reduced to copper(I) ions, Cu2+ + e- → Cu+

So the overall ionic equation is: 2Cu2+ + 4I- → 2CuI + I2

93

Describe the reaction of the redox titration to determine the copper content of solutions and alloys. 

The reaction produces a light brown/yellow solution and a white precipitate of copper(I0 iodide, but the precipitate appears light brown due to the iodine in the solution. 

The mixture of copper iodide and iodine can be titrated against sodium thiosulfate of known concentration. As the iodine reacts, the iodine colour gets paler during the titration. When the colour is a pale straw, a small amount of starch is added to help with the identification of the end point. A blue-black colour forms. The blue-black colour disappears sharply at the end point because all the iodine has reacted.   

94

How do you calculate the copper content of solutions and alloys from the visual results of a redox titration? (copper iodide and iodine being titrated against sodium thiosulfate)

You can use the amount of thiosulfate to determine the concentration of iodine in the titration mixture, and hence the concentration of copper(II) ions in the original solution.

One application of this method is to calculate (within experimental error) the concentration of copper(II) ions in an unknown solution, or the percentage of copper in alloys such as brass and bronze.

To obtain the copper(II) in solution, the alloy is first reacted with nitric acid. Brass and bronze can both be reacted to produce a solution containing copper(II) ions, which can then be reacted with potassium iodide solution. The iodine that forms is then titrated against sodium thiosulfate. 

(worked example page 76)

95

Define a standard electrode potential of a half-cell, Eø .

The standard electrode potential of a half-cell, Eø , is the e.m.f. of a half cell compared with the standard hydrogen half-cell, measured at 298 K with solution concentrations of 1 mol dm-3 and a gas pressure of 100 kPa.

96

Explain the electricity from redox reactions. 

During a redox reaction, electrons will be transferred. The flow of electrons involves electrical energy, so redox reactions are of huge importance in the utilisation of electrical energy - in fact, batteries and cells rely on redox reactions.

When we consider redox reactions as part of a cell, each half-equation involved is known as a half cell. Different half cells can be connected together to make a cell. This is because, together, the half cells cause electrons to flow between each other, releasing electrical energy.

97

Describe the half cells used to create standard electrode potentials. 

A half cell comprises an element in two oxidation states. The simplest half cell has a metal placed in an aqueous solution of its own ions.

For example, a copper half cell comprises a solution containing Cu2+ ions (oxidation state 2+) into which a strip or rod of copper metal (oxidation state 0) is placed.

An equilibrium exists at the surface of the copper between these oxidation states of copper.

Cu2+ + 2e- ⇔ Cu

  • The forward reaction involves electron gain (reduction)
  • The reverse reaction involves electron loss (oxidation).

By convention, the equilibrium is written with the electrons on the left-hand side. The electrode potential of the half-cell indicates its tendency to lose or gain electrons in this equilibrium. 

98

In a half-cell, how are the metals put in equilibrium with their ions? 

This is achieved, in practice, by placing the metal into a solution of the metal ions (i.e. a compound of the metal in aqueous solution). The piece of solid metal acts as an electrode when the half cell is connected  to another half cell to form a cell. 

99

What is the hydrogen half cell? 

Half cells can also be made from non-metals in equilibrium with non-metal ions. 

The moist common example of a half-cell comprising a non-metal and non-metal ions is the hydrogen half cell, comprising hydrogen gas, H2, in contact with hydrogen ions, H+

2H+ + 2e- ⇔ H2

If a hydrogen half cell were to be connected to another half-cell, to form a cell, there is no solid piece of metal that can act as the electrode. This is overcome by the use of a platinum electrode. 

100

Apart from hydrogen give another example of half cells containing non-metals in equilibrium with non-metals. 

A half cell of bromine and its ions;

Br2 + 2e- ⇔ 2Br-

101

What is the purpose of a platinum electrode in a hydrogen half cell? 

The platinum is inert and does not react at all - its sole purpose is to be in contact with both the H2 and the H+ ions and to allow the transfer of electrons into, and out of, the half cell via a coating in which electrons can be transferred between the non-metal and its ions. 

102

What does a standard hydrogen half-cell comprise of? 

  • Hydrochloric acid, HCl, of concentration 1 mol dm-3, as the source of H+
  • hydrogen gas, H2, at 100kPa pressure
  • an inert platinum electrode to allow electrons to pass into or out of the half cell via a connecting wire. 

103

Describe a metal ion/metal ion half cell. 

  • This type of half cell contains ions of the same element in different oxidation states. For example, a half cell can contain Fe3+ + e- ⇔ Fe2+

Fe3+ + e- ⇔ Fe2+

  • This type of half cell would again need to involve a platinum electrode as there is no solid piece of metal that could act as an electrode.

A standard Fe3+/Fe2+ half cell is made up of:

  • a solution containing Fe2+ ad Fe3+ ions with the same concentrations ('equimolar')
  • An inert platinum electrode to allow electrons to pass into or out of the half cell via a connecting wire. 

104

What is the standard electrode potential? 

Different half cells have different electrode potentials. When two half cells are connected together to form a cell, they will have an overall cell potential. This is a measure of how well electrons can be 'pushed around' the cell. The larger the overall cell potential, the more the electrons are 'pushed around'. The actual value of the overall cell potential will depend on the electrode potential of the half cells involved. 

105

How can you determine the standard electrode potential of a half cell? 

We can determine the standard electrode potential of a half cell by connecting it to a hydrogen half cell. The tendency for different half cells to accept or release electrons is measured as an electromotive force (e.m.f.), or voltage, measured in volts, V. The hydrogen half cell has an e.m.f. a value of 0 V so it can be used as a reference to measure other half cells against. 

106

How are two cells joined and why? 

  • a wire - this allows the electrons carrying charge to flow through it. 
  • a salt bridge - this connects the two solutions and allows ions carrying charge to be transferred between the half cells. Salt bridges are usually made from a piece of filter paper soaked in an aqueous solution of an ionic substance, usually KNO3 or NH4NO3 .

107

What do these half cells show? 

A hydrogen half cell being used to measure the potential of a Zn2+/Zn half cell. The reading on the voltmeter gives the standard electrode potential of the zinc half cell. 

108

What is the electrochemical series? 

We can list half cells in order of their standard electrode potentials, this is known as the electrochemical series. 

109

How can you tell if the forward or backwards reaction will happen according to the electrochemical series? 

The horizontal arrows show each reaction as a reduction reaction as electrons are gained. If the electrode potential is a negative value, the backwards reaction will occur when compared with the standard hydrogen electrode. If the electrode potential is a positive value the forward reaction will occur when compared to the standard hydrogen electrode. 

110

What is the correlation between the Eø value and the tendency to undergo oxidation in the electrochemical series of half cells.  

The more negative the Eø value, the greater the tendency towards the half cell undergoing oxidation; the more positive the Eø value, the greater the tendency towards the half cell undergoing reduction, when connected in a cell. 

111

Out of this small electrochemical series which half cell has the greatest tendency to release electrons? 

From this small series, we could say that the half cell

Fe 2+ + 2e- ⇔ Fe

  • has the most negative Eø value
  • therefore has the greatest tendency to release electrons and shift the equilibrium to the left.

Eø values for standard electrode potentials can be used to predict what will happen when half cells are connected together to form a cell.  

112

How can a simple electrochemical cell be made? 

A simple electrochemical cell is made by connecting together two half cells with different electrode potentials:

  • one cell releases electrons
  • the other cell half gains electrons
  • the difference in electrode potential is measured by a voltmeter
  • The two half-cell are joined using a wire and a salt bridge to allow to charge to be carried between each cell via the electrons and the ions.  

113

What can you infer from the results of this half cell; 

  1. Zn2+ + 2e- ⇔ Zn        Eø = -0.76V
  2. Cu2+ + 2e- ⇔ Cu        Eø= +0.34V

  1. The Zn2+/Zn equilibrium releases electrons more readily than the Cu2+ /Cu equilibrium. We know this because the half cell has the more negative value, so has a greater tendency towards the equilibrium shifting left.
  2. The Zn2+/Zn equilibrium releases electrons into the wire, making zinc the negative electrode. 
  3. Electrons flow along the wire to the Cu electrode fo the Cu2+/Cu half cell. 
  • The Zn2+/Zn equilibrium loses electrons and moves to the left 

Zn2+ + 2e- ⇔ Zn

  • The Cu2+/Cu equilibrium gains electrons and moves to the right:

Cu2+ + 2e- ⇔ Cu ⇒

114

What changes the reading on the voltmeter of an electrochemical cell? 

The reading on a voltmeter measures the potential difference of the cell - the difference between the electrodes potentials of the half cells. The bigger the value, the further away from the equilibrium position the reaction moves.

115

When can the reading on a voltmeter of an electrochemical cell be taken as the cell potential? 

The reading on the voltmeter can be taken as the cell potential as long as any ions of the same element have a concentration of 1 mol dm-3 or are equimolar. 

116

Standard cell potentials;

How can the standard cell potential be calculated? 

Eøcell = Eø (positive terminal) - Eø(negative terminal)

117

Standard cell potentials;

What is the cell reaction? 

The cell reaction is the overall chemical reaction taking place in the cell - the sum of the reduction and oxidation half-reactions taking place in each half cell. 

Follow worked examples on page 80-81, examples of using electrode potentials to predict a reaction and then writing the cell reaction.

118

Standard cell potentials;

By calculating the cell potential for a reaction, using the standard electrode potentials for each half cell what can you determine?

By calculating the cell potential for a reaction, using the standard electrode potentials for each half cell we can determine whether electrons are likely to flow and hence the feasibility of any reaction. 

The electrochemical series can give a quick indication of how species will react. 

119

How can the electrochemical series give a quick indication of how species will react?  

Half cells towards the top ⇒  more negative values ⇒ higher tendency towards being oxidised. 

When half cells are in combination, the one with more negative values - or less positive - will be the species to follow the oxidation reaction, the other half following the reduction reaction and gain electrons. 

A species will react with the species below it and to the left of the equation. 

120

From the electrochemical series,  which species can you predict to react with each other species?

Use the examples of iron and copper. 

  • Iron would react with H+ to form H2 - iron would react with acids. 
  • Copper would not react with the H+ - i.e. copper would not react with acids
  • Copper would react with Cl2

Worked example on page 81. 

121

Limitations of predictions of feasibility from cell potentials:

How could electrode potentials and concentration produce a limitation of the feasibility from cell potentials? 

Non-standard conditions alter the value of an electrode potential. The half-equation for the copper half cell is:

Cu2+ + 2e- ⇔ Cu

From Le Chateliers principle, on increasing the concentration of Cu2+.

  • the equilibrium opposes the change by moving to the right
  • electrons are removed from the equilibrium
  • the electrode potential becomes less negative, or more positive. 

A change in electrode potential resulting from concentration changes means that predictions made on the basis of the standard value may not be valid.  

 

 

122

Limitations of predictions of feasibility from cell potentials:

Why may reactions not actually take place and form a limitation of the feasibility from cell potentials? 

 

  • Predictions made about the equilibrium position but not reaction rate, which may be extremely slow because of high activation energy.
  • The actual conditions used for a reaction may be different from the standard conditions used to measure Eø values. This will affect the value of the electrode potential.
  • Standard electrode potentials apply to aqueous equilibria - many reactions take place under very different conditions.

123

Limitations of predictions of feasibility from cell potentials:

What is a general working rule for working out if a reaction will actually take place? 

  • The larger the difference between Eø values, the more likely it is that a reaction will take place. 
  • If the difference between Eø values, the more likely it is that a reaction will take place.

124

Electrochemical cells are used as our modern-day cells and batteries. Cells can be divided into which three main types? 

  • Non-rechargeable cells - provide electrical energy until the chemicals have reacted to such an extent that the voltage falls. The cell is then flat and discarded. 
  • Rechargeable cells - the chemicals in the cell react, providing electrical energy. The cell reaction can be reversed during recharging. The chemicals in the cell are regenerated and the cell can be used again. Common examples include: 
  1. nickel and cadmium (NI-Cad) batteries, used in rechargeable batteries
  2. Lithium-ion and lithium-polymer batteries, used in laptops.
  • Fuel cells - the cell reaction uses supplies of a fuel as an oxidant, which are consumed and needs to be continuously supplied. The cell will continue to provide electrical energy so long as there is a supply of fuel and oxidant. 

125

What are the risks of using electrochemical cells? 

The risks and benefits of using electrochemical cells must be weighed up when cells are designed. 

For example, lithium cells, which are often found in the home, can provide high levels of battery life but have been found to have several drawbacks. These include toxicity on being ingested and rapid discharge of current, which can cause fires and even explosions.

This has led to some controls on such batteries, including restrictions on the transport of lithium-based batteries and limited sales of batteries to individual consumers in some countries.  

126

What are modern cells based on? 

Hydrogen, or hydrogen-rich fuel cells such as methanol, CH3OH. 

127

How does a fuel cell create a voltage? 

A fuel cell uses energy from the reaction of a fuel with oxygen to create a voltage. 

  • The reactants flow in and products flow out while the electrolyte remains in the cell. 
  • Fuel cells can operate virtually continuously so long as the fuel and oxygen continue to flow into the cell. Fuel cells do not have to be recharged. 

128

This is a simple hydrogen-oxygen fuel cell with an alkaline electrolyte. What is the negative terminal of the cell? The redox equilibria are shown below. 

2H2O + 2e- ⇔ H2 + 2OH-    Eø = -0.83 V

1/2O2 + H2O +2e- ⇔ 2OH-  Eø = + 0.40 V

The more negative of the two systems is the negative terminal of the cell. 

2H2O + 2e- ⇔ H2 + 2OH-    Eø = -0.83 V   Negative terminal 

1/2O2 + H2O +2e- ⇔ 2OH-  Eø = + 0.40 V  Positive terminal 

The more negative hydrogen system provides the electrons. 

129

Work out the overall reaction and the voltage for a simple hydrogen - oxygen fuel cell. 

2H2O + 2e- ⇔ H2 + 2OH-    Eø = -0.83 V   Negative terminal 

1/2O2 + H2O +2e- ⇔ 2OH-  Eø = + 0.40 V  Positive terminal 

This equilibrium is reversed when writing the half equations at each electrode. The half-equations are added together and electrons cancelled to give the equation for the overall cell reaction:

H+ 2OH- ⇒ 2H2O + 2e- 

1/2O2 + H2O +2e- ⇔ 2OH- 

Overall; 

H2 + 1/2O2 → H2O

Eøcell Eø (positive terminal) - Eø (negative terminal)

   = 0.4 - (-o.83)

   =1.23 V

 

130

You're done!

Take a break - you deserve it, have a snack, cup of tea or go for a quick walk.