5.2

Question 1

What is the definition for lattice enthalpy?

Answer 1

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

Question 2

What is the definition for standard enthalpy change of formation?

Answer 2

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

Question 3

What is the definition of the first ionisation energy?

Answer 3

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

Question 4

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

Answer 4

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.

Question 5

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

Answer 5

K^{<span>+ </span>}(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.

Question 6

What do exothermic lattice enthalpy values indicate?

Answer 6

• 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.

Question 7

Why is it not possible to measure lattic enethalpy directly?

Answer 7

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

Question 8

How can you calculate lattice enthalpy?

Answer 8

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

Question 9

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

Answer 9

• 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)

Question 10

Whta are key enthalpy changes?

Answer 10

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.

Question 11

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

Answer 11

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

Question 12

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

Answer 12

K_(s) + _1/2 Cl_2(g) → KCl_(s)

It is an exothermic process.

Question 13

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

Answer 13

Δ_aH^ø The enthalpy change that takes place when one mole of gaseous atoms forms from the element in it’s standard state.

Question 14

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

Answer 14

For potassium metallic bonds are broken;

K_(s) → K_(g)

For chlorine, covalent bonds are broken:

_1/2Cl_{2 (g)} → Cl_(g)

The process is always endothermic.

Question 15

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

Answer 15

Δ_I1H^ø 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.

Question 16

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

Answer 16

K⁺_(g) → K⁺_(g) + e^-

The process is endothermic.

Question 17

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

Answer 17

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.

Question 18

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

Answer 18

Ca⁺_(g) → Ca²⁺_(aq) + e^-

The process is endothermic.

Question 19

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

Answer 19

Δ_EA1H^ø 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.

Question 20

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

Answer 20

Cl_(g) + e^- → Cl^- _(g)

The process is exothermic.

Question 21

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

Answer 21

Δ_EA2H^{<strong>ø </strong>}

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

Question 22

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

Answer 22

• 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).

Question 23

What can Born-Haber cycles be used to calculate?

Answer 23

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

• Δ_fH^ø = 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)

Question 24

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

Answer 24

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

Question 25

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

Answer 25

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

Question 26

What happens when a solid dissolves?

Answer 26

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 (Δ_solH^ø).

Question 27

Is the dissolving of a substance exo or endothermic?

Answer 27

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).

Question 28

Why is the enthalpy change of ionic lattice breakdown equal to -Δ_LEH^ø?

Answer 28

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.

Question 30

The breakdown of an ionic lattice;

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

Answer 30

• 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.

Question 31

What dictates the size of lattice enthalpy?

Answer 31

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)

Question 32

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

Answer 32

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.

Question 33

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

Answer 33

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.

Question 34

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

Answer 34

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 Cl^- ions in the MgCl₂ lattice can pack closer together and exert a greater attraction on each other than the ions in MgBr₂ or MgI_2.​

(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.)

Question 35

Explain charge density.

Answer 35

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.

Question 36

What is hydration?

Answer 36

• 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.

Question 37

What happens during hydration?

Answer 37

• 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.

Question 39

What enthalpy change occurs when ions become hydrated?

Answer 39

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

Question 40

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

Answer 40

Hydration is an exothermic process.

Question 41

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

Answer 41

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

Question 42

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

Answer 42

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 Δ_hydH^ø becomes more exothermic.

Question 43

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

Answer 43

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.

Question 44

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

ionic solids go and gaseous ions go?

Answer 44

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

Question 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?

Answer 45

• 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)

Question 46

What is the definition of entropy?

Answer 46

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

Question 47

What is the definition of the standard entropy?

Answer 47

The standard entropy, S^ø, of a substance is the entropy content of one mole of the substance under standard conditions.

Question 48

What is the definition of standard entropy change of reaction?

Answer 48

Standard entropy change of reaction, ΔS^ø, 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.

Question 49

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

Answer 49

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

Question 50

What units do standard entropies have?

Answer 50

J K^-1 mol^-1

Question 51

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

Answer 51

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.

Question 52

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

Answer 52

thermodynamic

Question 53

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

Answer 53

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

Question 54

Does entropy tend to increase or decrease?

Answer 54

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.

Question 55

Can entropy decrease during a reaction?

Answer 55

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.

Question 56

How is entropy affected by temperature?

Answer 56

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

Question 57

Describe the entropy change when dissolving solids?

Answer 57

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.

Question 58

How do the number of gas molecules affect entropy?

Answer 58

• 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.

Question 59

How is the standard entropy change of a reaction calculated?

Answer 59

Using the entropies of the reactants and products.

Question 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.

Answer 60

Question 61

What is the definition of the free energy change?

Answer 61

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

Question 62

How do we calculate total change in entropy?

Answer 62

Question 63

What must happen for spontaneous changes to occur?

Answer 63

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.

Question 64

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

Answer 64

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.

Question 65

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

Answer 65

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.

Question 66

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

Answer 66

Question 67

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

Answer 67

• 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.

Question 68

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

Answer 68

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

Question 69

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

Answer 69

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.

Question 70

Define oxidation

Answer 70

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

Question 71

Define reduction

Answer 71

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

Question 72

Define an oxidising agent.

Answer 72

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

Question 73

Define a reducing agent.

Answer 73

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

Question 74

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

Describe oxidising agents.

Answer 74

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.

Question 75

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

Describe reducing agents.

Answer 75

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.

Question 76

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

Mg + 2HCl → MgCl₂ + H₂

Answer 76

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)

Question 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?

Answer 77

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)

Question 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?

Answer 78

• 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.

Question 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)

Answer 79

• 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.

Question 80

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

Answer 80

• 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.

Question 81

Why are titrations useful?

Answer 81

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.

Question 82

How do you carry out a redox titration?

Answer 82

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.

MnO₄^- is commonly used to oxidise solutions containing iron(II), it can be used as an oxidising agent in many other chemical reactions.

Question 83

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

Answer 83

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^-.)

Question 84

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

Answer 84

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

Question 85

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

Answer 85

The end point is seen when excess MnO₄^- ions are present - indicated by a faint pink colour appearing. This occurs because all the Fe²⁺ ions have reacted and the MnO₄^- can no longer be reduced to the colourless Mn²⁺ .

Question 86

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

Answer 86

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.

Question 87

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

Answer 87

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)

Question 88

Redox titrations between I₂ and S₂O₃ ;

Why is iodine a useful substance for a redox titration?

Answer 88

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

Question 89

The redox titration between I₂ and S₂O₃^2- ;

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

Answer 89

• This reduction of iodine to iodide ions will occur in the presence of thiosulphate ions, S₂O₃^2-.
• 2S₂O₃^2- + I₂ → S₂O₃^2- + 2I^- (thiosulfate to tetrathionate)
• We can use aqueous iodide ions and aqueous thiosulfate to determine the concentration of unknown reducible species.

Question 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.

Answer 90

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.

2S₂O₃^2- + I₂ → S₄O₆^2- + 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)

Question 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.

Answer 91

• 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)

Question 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?

Answer 92

1. iodide ions are oxidised to iodine, 2I^- → I₂ + 2e^-
2. copper(II) ions are reduced to copper(I) ions, Cu²⁺ + e^- → Cu^{<span>+</span>}

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

Question 93

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

Answer 93

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.

Question 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)

Answer 94

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)

Question 95

Define a standard electrode potential of a half-cell, E^{ø _.}

Answer 95

The standard electrode potential of a half-cell, E^ø , 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.

Question 96

Explain the electricity from redox reactions.

Answer 96

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.

Question 97

Describe the half cells used to create standard electrode potentials.

Answer 97

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.

Question 98

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

Answer 98

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.

Question 99

What is the hydrogen half cell?

Answer 99

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.

Question 100

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

Answer 100

A half cell of bromine and its ions;

Br₂ + 2e^- ⇔ 2Br^-​

Question 101

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

Answer 101

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.

Question 102

What does a standard hydrogen half-cell comprise of?

Answer 102

• 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.

Question 103

Describe a metal ion/metal ion half cell.

Answer 103

• 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.

Question 104

What is the standard electrode potential?

Answer 104

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.

Question 105

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

Answer 105

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.

Question 106

How are two cells joined and why?

Answer 106

• 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 .

Question 107

What do these half cells show?

Answer 107

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.

Question 108

What is the electrochemical series?

Answer 108

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

Question 109

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

Answer 109

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.

Question 110

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

Answer 110

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

Question 111

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

Answer 111

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

Fe ²⁺ + 2e^- ⇔ Fe

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

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

Question 112

How can a simple electrochemical cell be made?

Answer 112

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.

Question 113

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

1. Zn²⁺ + 2e^- ⇔ Zn E^ø = -0.76V
2. Cu²⁺ + 2e^- ⇔ Cu E^ø= +0.34V

Answer 113

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 ⇒

Question 114

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

Answer 114

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.

Question 115

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

Answer 115

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.

Question 116

Standard cell potentials;

How can the standard cell potential be calculated?

Answer 116

E^ø_cell = E^ø (positive terminal) - E^ø(negative terminal)

Question 117

Standard cell potentials;

What is the cell reaction?

Answer 117

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.

Question 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?

Answer 118

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.

Question 119

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

Answer 119

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.

Question 120

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

Use the examples of iron and copper.

Answer 120

• 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.

Question 121

Limitations of predictions of feasibility from cell potentials:

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

Answer 121

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.

Question 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?

Answer 122

• 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 E^ø 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.

Question 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?

Answer 123

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

Question 124

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

Answer 124

• 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.

Question 125

What are the risks of using electrochemical cells?

Answer 125

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.

Question 126

What are modern cells based on?

Answer 126

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

Question 127

How does a fuel cell create a voltage?

Answer 127

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.

Question 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^- E^ø = -0.83 V

1/2O₂ + H₂O +2e^- ⇔ 2OH^-E^ø = + 0.40 V

Answer 128

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

2H₂O + 2e^- ⇔ H₂ + 2OH^- E^ø = -0.83 V Negative terminal

1/2O₂ + H₂O +2e^- ⇔ 2OH^-E^ø = + 0.40 V Positive terminal

The more negative hydrogen system provides the electrons.

Question 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

Answer 129

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:

H₂ + 2OH^- ⇒ 2H₂O + 2e^-

1/2O₂ + H2O +2e- ⇔ 2OH-

Overall;

H₂ + 1/2O₂ → H₂O

E^ø_cell = E^ø (positive terminal) - E^ø (negative terminal)

= 0.4 - (-o.83)

=1.23 V

Question 130

You’re done!

Answer 130

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