5.2

Question 1

5.2.1 a) Explain what lattice enthalpy is, and how it can be used as a measure of the strength of ionic bonding in a giant ionic lattice?

Answer 1

• Lattice enthalpy, ∆LEH⊖, is the enthalpy change that accompanies the formation of one mole of an ionic lattice from its gaseous ions under standard conditions.
• It is an exothermic change and the value for lattice enthalpy will always be negative (i.e. when gaseous ions combine to make a solid lattice, energy is given out).
• Lattice enthalpy can be used as a measure of the strength of ionic bonding in a giant ionic lattice. The more exothermic ( or the more negative) the lattice enthalpy, the stronger the ionic bonding, and the higher the melting and boiling point (as more energy is required to overcome the stronger electrostatic attractions).
  e.g. K+ (g) + Cl- (g) → KCl (s)
  ∆LEH⊖ = -711 kJ mol-1
  e.g. Mg(2+) (g) + O(2-) (g) → MgO (s)
  ∆LEH⊖ = -3291 kJ mol-1
  The most exothermic lattice enthalpies arise when ions are small (allowing them to get closer together) and have large charges (resulting in large electrostatic forces).

Question 2

5.2.1 b) i) Explain how a Born-Haber cycle (a type of Hess’ cycle) is constructed (using MgO as an example).

Answer 2

Draw a line near the bottom of your cycle, to the left-hand side (called a datum line). On this line, write the elements in their standard states.
Mg (s) + 1/2O2 (g)
Under this line, draw another, longer line. Place the iconic compound (as a solid ionic lattice) on this line.
MgO (s)
Join the two lines with a downward arrow, to represent an exothermic change. This is the enthalpy change of formation of the iconic compound.
Draw a line near the top of your cycle, to the right-hand side. On this line, write the gaseous ions needed to form the iconic compound.
Mg(2+) (g) + O(2-) g)
Join this line with the line showing the ionic compound (MgO (s)) as well. Ensure you use a downward arrow to represent an exothermic change. This is the lattice enthalpy of the ionic compound from its gaseous ions.

Draw 3 or 4 lines on the left-hand side (above the datum line), joining each successive line with an upward, endothermic arrow:
1) On the first line above the datum line, write one of the elements in its gaseous state. This will represent the enthalpy change of atomisation for the first element.
Mg (g) + 1/2O2 (g)
2) On the second line above the datum line, write the other element in its gaseous state. This will represent the enthalpy change of atomisation for the second element.
Mg (g) + O (g)
3) On the third line above the datum line, write one of the elements as a 1+ ion. This represents the first ionisation energy, forming gaseous 1+ ions.
Mg(+) (g) + O (g)
4) If necessary, repeat this for a second ionisation energy of that element.
Mg(2+) (g) + O (g)

For an element that only requires one electron affinity, simply connect the previous line (with the ionisation energy) to the one on the right-hand side (with the gaseous ions) using a downward, exothermic arrow.
If you require two electron affinities for an element, instead draw a line near the centre of the cycle, just below the previous line (with the ionisation energy). On this line, write one of the elements as a 1- ion.
Mg(2+) (g) + O(-) (g)
Join the two lines with a downward, exothermic arrow (to represent the first electron affinity). and then connect this line to the one on the right-hand side (with the gaseous ions), using an upward endothermic arrow (this represents the second electron affinity). Remember that second electron affinities are endothermic.

Question 3

5.2.1 b) ii) Explain how a Born-Haber cycle can be used to calculate various enthalpy changes.

Answer 3

When you draw a Born-Haber cycle, you place the enthalpies provided alongside the corresponding arrows. You can calculate various enthalpies using Hess’s Law:
The enthalpy of formation of the ionic compound (i.e. the sum of the anti-clockwise enthalpy changes) = the sum of all other enthalpy changes (i.e. the sum of all the clockwise enthalpy changes), including the lattice enthalpy of course
Rearrange this to calculate the unknown quantity.
[Remember to multiply the enthalpies if an element has more than one atom in the ionic compound, e.g. for chlorine in MgCl2, remember to multiply any enthalpy involving Cl and Cl- by 2]

Question 4

5.2.1 b) Explain what the standard enthalpy change of formation is.

Answer 4

The standard enthalpy change of formation, ∆fH⊖, is the enthalpy change that takes place when 1 mole of a compound is formed from its constituent elements in their standard states, and under standard conditions. This is usually an exothermic process for an ionic compound, as bonds are being formed.
e.g. Na (g) + 1/2Cl2 (g) → NaCl (s)

Question 5

5.2.1 b) Explain what the standard enthalpy change of atomisation is.

Answer 5

The standard enthalpy change of atomisation, ∆atH⊖, is the enthalpy change that takes place when an element in its standard state forms 1 mole of gaseous atoms, under standard conditions. This is always an endothermic process because bonds are broken.

e. g. K (s) → K (g)
e. g. 1/2 Cl2 (g) → Cl (g)

Question 6

5.2.1 b) Explain what first, and second, ionisation energy is.

Answer 6

The first ionisation energy, ∆IEH⊖, is the energy required to remove 1 mole of electrons from 1 mole gaseous atoms, forming 1 mole of gaseous 1+ ions.
e.g. Ca (g) → Ca(+) (g)
The second ionisation energy, ∆IE2H⊖, is the energy required to remove 1 mole of electrons from 1 mole of gaseous 1+ ions, forming 1 mole of gaseous 2+ ions.
e.g. Ca(+) (g) → Ca(2+) (g)
Ionisation energies are endothermic: energy is required to overcome the electrostatic attraction between a negative electron and the positive nucleus.

Question 7

5.2.1 b) Explain what first, and second, electron affinity is.

Answer 7

The first electron affinity, ∆EAH⊖, is the enthalpy change that takes place when 1 mole of electrons is added to 1 mole of gaseous atoms, forming 1 mole of gaseous 1- ions.
e.g. O (g) + e- → O(-) (g)
First electron affinities are exothermic, because the electron being added is attracted in towards the nucleus.
The second electron affinity, ∆EA2H⊖, is the enthalpy change that takes place when 1 mole of electrons is added to 1 mole of gaseous 1-ions, forming 1 mole of gaseous 2- ions.
e.g. O(-) (g) + e- → O(2-) (g)
Second electron affinities are endothermic: a second electron is being gained by a negative ion, which repels the electron away, so energy must be put in to force the negatively-charged electron onto the negative ion.

Question 8

5.2.1 c) i) Explain what the standard enthalpy change of solution is.

Answer 8

The standard enthalpy change of solution, ∆solH⊖, is the enthalpy change that takes place when 1 mole of a solute is completely dissolved in a solvent (e.g, water) under standard conditions.

Question 9

5.2.1 c) i) Explain what the enthalpy change of hydration is.

Answer 9

The enthalpy change of hydration, ∆hydH⊖, is the enthalpy change that takes place when 1 mole of gaseous ions is dissolved in water to form 1 mole of aqueous ions.

Question 10

5.2.1 i) Explain how enthalpy change of solution can be calculated experimentally.

Answer 10

The enthalpy change of solution can be calculated using coffee cup (polystyrene) calorimetery:
1) Using a measuring cylinder, add 25 cm3 of distilled water into the plastic cup, and measure the temperature of the water using a thermometer.
2) Add a weighed sample of the ionic compound to the water. Stir the mixture with the thermometer until all of the ionic compound has dissolved, and the temperature no longer changes.
3) Calculate the temperature change of solution (initial temperature of water - final temperature of solution), and the mass that is changing the temperature (i.e the mass of solution = mass of water + mass of the ionic compound).
4) Calculate the energy change, q, in solution:
q = mc∆t (then divide by 1000 for kJ).
5) Calculate the amount, in moles, of the ionic compound that dissolved:
n(ionic compound) = m/M
6) Calculate ∆solH⊖ in kJ mol-1:
∆H = q/n

Question 11

5.2.1 d) Explain the process that takes place when a solid ionic compound dissolves in water, and the enthalpy changes that accompany with each step.

Answer 11

When a solid ionic compound dissolves in water, two processes take place:
1) The ionic lattice breaks up. If lattice enthalpy is the enthalpy change that accompanies the formation of 1 mole of an ionic compound from its gaseous ions, then the reverse of this occurs when the lattice breaks down (we imagine the lattice becoming gaseous ions). Therefore the enthalpy change of ionic lattice breakdown has the same value of lattice enthalpy, but with a different sign (the bonds between the ions break to give gaseous ions - this is an endothermic process).
Lattice enthalpy is exothermic, e.g, when the ionic lattice of potassium chloride forms:
K(+) (g) + Cl(-) (g) → KCl (s)
∆LEH⊖ = -711 kJ mol-1
The breakdown of the ionic lattice is endothermic, e.g, when the ionic lattice of potassium chloride breaks down after dissolving:
KCl (s) → K(+) (g) + Cl(-) (g)
∆H = +711 kJ mol-1
Multiply lattice enthalpy by -1 to get the enthalpy change of the breakdown of the ionic lattice.
2) The free ions then become part of the solution. During hydration, the positive ions will be attracted to the slightly negative oxygen in the water molecules, and the negative ions will be attracted to the slightly positive hydrogens in the water molecules. The water molecules will completely surround the ions.
When the separate gaseous ions interact with polar water molecules, hydrated aqueous ions are formed. An enthalpy change occurs when ions become hydrated (energy is released when new bonds are formed between ions and water molecules). The energy change involved is called the enthalpy change of hydration, and is an exothermic process.
e.g. K(+) (g) + aq → K(+) (aq)
∆hydH⊖ = -322 kJ mol-1
e.g. Cl(-) (g) + aq → Cl(-) (aq)
∆hydH⊖ = -363 kJ mole -1

The overall enthalpy change of this process is known as the enthalpy change of solution. The enthalpy change of solution can be either exothermic or endothermic, depending on the relative sizes of the lattice enthalpy and the enthalpy changes of hydration.

Question 12

5.2.1 d) Outline how enthalpy cycles can be constructed using enthalpy change of solution, enthalpy change of hydration and lattice enthalpy.

Answer 12

A
XY (s) → X+ (aq) + Y- (aq)
B ↘ ↗ C
X+ (g) + Y (g)

A = enthalpy of solution
B = (-) lattice enthalpy
C = enthalpies of hydration

OR:

          A XY (s)   →   X+ (aq) + Y- (aq)
 B ↖       ↗ C
X+ (g) + Y (g)

A = enthalpy of solution
B = lattice enthalpy
C = enthalpies of hydration

1) Enthalpy change of solution = (-) lattice enthalpy + enthalpies of hydration
2) Enthalpies of hydration = lattice enthalpy + enthalpy of solution
3) Lattice enthalpy = enthalpy change of solution - enthalpies of hydration

[Remember to multiply where necessary, e.g, for CaBr2, there are 2 Br- ions involved, so the enthalpy of hydration for Br- should be multiplied by 2]

Question 13

5.2.1 e) Explain the effect of ionic charge and ionic radius on the exothermic value of a lattice enthalpy.

Answer 13

• The higher the ionic charge, the stronger the electrostatic forces of attraction between ions in a lattice. This results in stronger ionic bonds being made, with more energy being released when these bonds are made. This, in turn, gives rise to more exothermic lattice enthalpy values (i.e. more negative values).
• Ions with a smaller ionic radius sit closer together in a lattice. This increases the electrostatic attraction between oppositely charged ions, resulting stronger ionic bonds being made (and more energy being released when these bonds are made). This, in turn, gives rise to more exothermic lattice enthalpy values (i.e. more negative values).
• The most exothermic lattice enthalpy values arise from small, highly charged ions. These ions have a higher charge density.

Question 14

5.2.1 e) Explain the effect of ionic charge and ionic radius on the exothermic value of enthalpy of hydration.

Answer 14

• Ions with a higher charge experience stronger electrostatic forces of attraction with surrounding water molecules. This means that on hydration, more energy is released when bonds are made, and the value for ∆hydH⊖ is more exothermic (i.e. more negative).
• Ions with smaller ionic radii can get closer to water molecules than those with a larger ionic radii. This increases the electrostatic forces of attraction between ions and water molecules, so that on hydration, more energy is released when bonds are made, and the value for ∆hydH⊖ is more exothermic (i.e. more negative).
• The most exothermic ∆hydH⊖ values arise from small, highly charged ions. These ions have a higher charge density.

Question 15

5.2.2 a) Explain what entropy is?

Answer 15

Entropy, S, is a measure of the dispersal of energy in a system.
- A system which is more disordered in space will tend to have more disorder in the way the energy is arranged as well. This is because the more random the arrangement of particles, the more random the distribution of energy.
- A system becomes more stable when its energy is spread out in a more disordered state. So entropy is higher when the number of ways the particles can be arranged, and the number of ways that the energy can be shared out between the particles, is greater.
= Therefore, the more disordered the particles are in a system, the greater the dispersal of energy, and the higher the entropy.

Question 16

5.2.2 b) i) Explain the difference in magnitude of the entropy of a system of solids, liquids and gases.

Answer 16

Entropy increases during changes in state that give a more random arrangement of particles: solid → liquid → gas
In general, solids have the smallest entropies and gases have the greatest entropies. e.g. Consider H2O:
- ice, H2O (S) = +48
- liquid water, H2O (l) = +69.9
- steam, H2O (g) = +189
The particles in a system, and the dispersal of energy, become more disordered. This random arrangement of particles, and random distribution of energy, causes the entropy to increase.
A similar process takes place when a highly ordered solid ionic lattice dissolves in water: the ions spread out, and their arrangement becomes more random. Because this enables more ways for energy to be arranged or dispersed, entropy has increased.

Question 17

5.2.2 b i) Explain the effect an increase in temperature will have on entropy.

Answer 17

Entropy of pure substances increases with increasing temperature. One of the reasons is because an increase in temperature can cause a change in physical state (from solid to liquid, or liquid to gas, depending on the substance’s melting and boiling point). However, even if a substances remains the same physical state, particles at higher temperatures have more energy, and therefore move around more. The arrangement of particles, and distribution of energy, becomes more random/more disordered, and the entropy increases.

Question 18

5.2.2 b) ii) Explain the difference in magnitude of the entropy of a system for a reaction in which there is a change in the number of gaseous molecules.

Answer 18

If the number of gas molecules changes during a reaction, entropy changes.
- 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.
You can predict the sign of the entropy change for reactions where the reactants and products have different numbers of gas molecules.
e.g. N2 (g) + 3H2 (g) → 2NH3 (g)
4 moles of gas → 2 moles of gas
There is a decrease in the number of gas molecules, and so a decrease in the randomness of particles. The energy is less dispersed, and ∆S will be negative.

Question 19

5.2.2 c) Outline why the entropy value of a substance is always positive, and explain what standard entropy is.

Answer 19

At 0K, there would be no energy and all substances would have an entropy value of 0. Above 0K, energy becomes dispersed amongst the particles, and all substances will possess a certain degree of disorder because they are in constant motion.

• This means that entropy, S, is always a positive number above 0. Systems that are more disordered have a higher entropy value.
• The standard entropy, S⊖, is the entropy content of one mole of the substance under standard conditions. Standard entropies have units of J K-1 mol-1, and are always positive.

Question 20

5.2.2 c) Outline the ways in which entropy can change during the course of a reaction, and entropy change, ∆S, can be calculated.

Answer 20

Entropy can change during the course of a chemical process or reaction.
- If a system changes to become more disordered/more random, energy will be more dispersed. The entropy change, ∆S, will be positive.
There is always a tendency towards higher entropy, however it is possible for entropy to decrease as well:
- If a system changes to become more ordered/less random, energy will become more concentrated. The entropy change, ∆S, will be negative.
Standard entropies can be used to calculate the entropy change, ∆S, of a reaction:
∆S = ∑S⊖(products) - ∑S⊖(reactants)

Question 21

5. 2.2 d) What two things does the feasibility if a process depends upon?
6. 2.2 d) Explain how the feasibility of a process can be calculated.

Answer 21

The feasibility of a process depends upon two things:

• the entropy change and temperature in the system, T∆S, and
• the enthalpy change of the system, ∆H

The relationship between these two components can be expressed using Gibbs’ equation - this allows you to calculate the free energy change, ∆G, which in turn allows you to predict whether a reaction is feasible:
∆G = ∆H - T∆S
For a reaction to be feasible, there must be a decrease in free energy, i.e. ∆G must be negative.
[Note: ∆H is usually given in kJ mol-1, and ∆S is usually given in J K-1 mole-1. You should divide the units for ∆S by 1000 to convert them into kJ K mole-1.]
You may have to calculate the minimum temperature that a reaction is feasible at - if this is the case, rearrange Gibbs’ equation:
T = ∆H/∆S

Question 22

5.2.2 e) Outline how the feasibility of a process can depends on enthalpy, entropy and temperature.

Answer 22

A feasible process is one which occurs without a continuous supply of energy. While there is still an activation energy barrier to be overcome, once the minimum energy has been provided to start the reaction, the process will be self-sufficient until the all of the reactants have been used up.
Some reactions are spontaneous because they give off energy in the form of heat, however this isn’t always the case: reactions can be spontaneous because they lead to an increase in the disorder of the system.
1) If a reaction is exothermic, energy will be released - this energy can help to maintain the reaction so that it is self-sufficient. If the entropy of this reaction is also positive, then the free energy change will always be negative - this is because the ∆H value tends have a larger magnitude than ∆S, and so ∆G will mainly be dependent on the negative ∆H . Therefore, the reaction will be feasible at both low and high temperatures.
2) However, an exothermic reaction is not always feasible: the entropy might be negative. If this is the case, the free energy change will be positive at high temperatures (when you subtract a negative T∆S value from ∆H, you are adding it, and if the magnitude of T∆S is larger than ∆H, ∆G will be positive) - but ∆G can be negative at low temperatures (when the magnitude of T∆S has decreased). This means that the reaction will only be feasible at low temperatures.
3) Endothermic reactions (despite absorbing energy from the surroundings) can be feasible as well: i.e. the entropy might be positive and act as a driving force. However, even then the free energy change is only negative at high temperatures (when you subtract T∆S from ∆H, and the value of T∆S is larger than ∆H, ∆G will be negative) - but ∆G will be positive at low temperatures (as T∆S will be smaller than ∆H). This means that the reaction would only be feasible at high temperature.
4) If the reaction is endothermic but the entropy is negative, then the free energy will always be positive (this is because you subtract a negative T∆S value, which will only add it to the already positive ∆H value)i. This means that the reaction is never feasible, whatever the temperature.

Question 23

5.2.2 e) Summarise how feasibility can vary with different signs and temperatures.

Answer 23

∆H: Negative, ∆S: Positive
= ∆G is negative [at low or high T
and = feasible [at low or high T]

∆H: Negative, ∆S: Negative
= ∆G is negative [at low T]
and = feasible [at low T] 
OR
= ∆G is positive [at high T]
and = NOT feasible [at high T] 

∆H: Positive, ∆S: Positive
= ∆G is negative [at high T]
and = feasible [at high T] 
OR
= ∆G is positive [at low T]
and = NOT feasible [at low T] 

∆H: Positive, ∆S: Negative
= ∆G is positive [at low or high T
and = NOT feasible [at low or high T]

This is why the feasibility of a process depends upon the balance between ∆H and T∆S.

Question 24

5.2.2 f) Explain the limitations of predictions made by ∆S

Answer 24

Many reactions that have a negative ∆G value don’t take place. This is because the reaction may have a high activation energy - energy needs to be supplied initially, in order to overcome this - or the rate of reaction may be extremely slow.

Question 25

5.2.3 a) Explain the following terms: an oxidising agent and a reducing agent.

Answer 25

[Remember: Oxidation is the loss of electrons - there is an increase in oxidation number. Reduction is the gain of electrons - there is a decrease in oxidation number.]

1) In a redox reaction, one of the species loses electrons, and is oxidised - this species is the reducing agent: it provides the electrons in the reaction, causing the other species to be reduced.
2) In a redox reaction, one of the species gains electrons, and is reduced - this species is the oxidising agent: it gains the electrons in the reaction, causing the other species to be oxidised.

Question 26

5.2.3 b) Explain how a redox reaction can be broken down into two half-equations, using the following example:
Magnesium reacts with hydrochloric acid:
Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

Answer 26

Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)
1) Consider Mg: it forms a Mg2+ ion. This means magnesium is oxidised, and loses 2 electrons:
Mg → Mg(2+) + 2e-
2) Consider the two H+ ions in HCl: H2 is formed from them. This means that hydrogen is reduced, and each of the 2 hydrogen ions gain 1 electron:
2H(+) + 2e- → H2
Make sure that the total charges for each side of a half-equation are equal - in this case, the charge is 0 on each side, for both half-equations.
Also ensure that the number of electrons lost in the oxidation half-equation is the same as the number of electrons gained in the reduction half-equation.
[Notice that Cl- isn’t included in the half-equations. It is a spectator ion: it isn’t involved in the redox reaction and its oxidation number doesn’t change.]

Question 27

5.2.3 b) Explain how a half-equation can be constructed for reactions containing oxygen or hydrogen in the reducing/oxidising agent. Use the following example:
As part of a redox reaction, acidified dichromate (VI) ions, Cr2O7(2-), are reduced to Cr(3+) ions. Write a half equation for this reaction.

Answer 27

Cr2O7(2-) → Cr(3+)
1) Start by balancing all the elements in the reaction except for O and H. In this case, you just balance the Cr atoms:
Cr2O7(2-) → 2Cr(3+)
2) Next, balance the oxygen atoms : add 7 H2O molecules to the right-hand side:
Cr2O7(2-) → 2Cr(3+) + 7H2O
3) Next, balance the hydrogen atoms: add 14 H+ ions (which are present due to the acidic condition) to the left-hand side:
Cr2O7(2-) + 14H(+) → 2Cr(3+) + 7H2O
4) Finally, add electrons to balance the charges. The total charge of the left-hand side is 12+ , and the total charge on the right-hand side is 6+. Add 6 negative electrons to the left-hand side, balancing the half-equation so that the charge on each side is 6+.
Cr2O7(2-) + 14H(+) + 6e-→ 2Cr(3+) + 7H2O

[Note: electrons (e-), protons (H+), water (H2O), and very occasionally OH- ions, are the only things that you’re allowed to add to balance half-equations.]

Question 28

5.2.3 b) Outline how to construct redox equations using half-equations, with the following example:
Iron reacts with aqueous copper(II) ions to form iron(III) ions and copper metal. The half equations are as follows:
Fe → Fe(3+) + 3e-
Cu(2+) + 2e- → Cu

Answer 28

1) Multiply the half equations so that the electrons are balanced:
Fe → Fe(3+) + 3e-
(x2) = 2Fe → 2Fe(3+) + 6e-
Cu(2+) + 2e- → Cu
(x3) = 3Cu(2+) + 6e- → 3Cu
2) Add the two half-equations together and cancel out where appropriate:
2Fe + 3Cu(2+) + 6e- → 2Fe(3+) + 6e- + 3Cu
= 2Fe + 3Cu(2+) → 2Fe(3+) + 3Cu

Question 29

5.2.3 b) Outline how to construct redox equations using half-equations, with the following example:
Zinc metal displaces silver ions from silver nitrate solution to form zinc nitrate and a deposit of silver metal. The half equations are as follows:
Zn (s) → Zn(2+) (aq) + 2e-
Ag(+) (aq) + e- → Ag (s)

Answer 29

1) Multiply the half equations so that the electrons are balanced:
Zn (s) → Zn(2+) (aq) + 2e- (doesn’t need to be multiplied)
Ag(+) (aq) + e- → Ag (s)
(x2) = 2Ag(+) (aq) + 2e- → 2Ag (s)
2) Add the two half-equations together and cancel out where appropriate:
Zn (s) + 2Ag(+) (aq) + 2e- → Zn(2+) (aq) + 2e- + 2Ag (s)
= Zn (s) + 2Ag(+) (aq) → Zn(2+) (aq) + 2Ag (s)

Question 30

5.2.3 b) Outline how to construct redox equations using half-equations, with the following example:
The redox reaction between hydrogen peroxide, H2O2, and alkaline Cr3+ ions has the following half-equations:
H2O2 + 2e- → 2OH-
Cr(3+) + 8OH- → CrO4(2-) + 4H2O + 3e-

Answer 30

1) Multiply the half equations so that the electrons are balanced:
H2O2 + 2e- → 2OH-
(x3) = 3H2O2 + 6e- → 6OH-
Cr(3+) + 8OH- → CrO4(2-) + 4H2O + 3e-
(x2) = 2Cr(3+) + 16OH- → 2CrO4(2-) + 8H2O + 6e-
2) Add the two half-equations together and cancel out where appropriate:
3H2O2 + 6e- + 2Cr(3+) + 16OH- → 6OH- + 2CrO4(2-) + 8H2O + 6e-
= 3H2O2 + 2Cr(3+) + 16OH- → 6OH- + 2CrO4(2-) + 8H2O
= 3H2O2 + 2Cr(3+) + 10OH- → 2CrO4(2-) + 8H2O

Question 31

5.2.3 b) Outline how to construct redox equations using half-equations, with the following example
Acidified manganate (VII) ion, MnO4(-), can be reduced to Mn(2+), by Fe(2+) ions.

Answer 31

Write the two half equations for this reaction. Then construct the redox equation.
1) Iron is being oxidised, and manganate is being reduced:
Fe(2+) → Fe(3+) + e-
MnO4(-) → Mn(2+)
You need to balance the elements in the manganate half-equation.
2) First balance the oxygen atoms : add 4 H2O molecules to the right-hand side:
MnO4(-) → Mn(2+) + 4H2O
3) Next, balance the hydrogen atoms: add 8 H+ ions (which are present due to the acidic condition) to the left-hand side:
MnO4(-) + 8H+ → Mn(2+) + 4H2O
4) Finally, add electrons to balance the charges. The total charge of the left-hand side is 7+ , and the total charge on the right-hand side is 2+. Add 5 negative electrons to the left-hand side, balancing the half-equation so that the charge on each side is 2+.
MnO4(-) + 8H+ + 5e- → Mn(2+) + 4H2O

Now construct the redox equation.
1) Multiply the half equations so that the electrons are balanced:
MnO4(-) + 8H+ + 5e- → Mn(2+) + 4H2O (doesn’t need to be multiplied)
Fe(2+) → Fe(3+) + e-
(x5) = 5Fe(2+) → 5Fe(3+) + 5e-
2) 2) Add the two half-equations together and cancel out where appropriate:
MnO4(-) + 8H+ + 5e- + 5Fe(2+) → Mn(2+) + 4H2O + 5Fe(3+) + 5e-
= MnO4(-) + 8H+ + 5Fe(2+) → Mn(2+) + 4H2O + 5Fe(3+)

Question 32

5.2.3 b) Outline how to construct redox equations using oxidation numbers, with the following example:
Sulfur, S, reacts with concentrated nitric acid, HNO3, to form sulfuric acid, H2SO4, nitrogen dioxide, NO2, nad water, H2O.

Answer 32

1) Construct an incomplete draft equation (using the information provided), and assign oxidation numbers:
S + HNO3 → H2SO4 + NO2 + H2O
0 +1/+5/-2 +1/+6/-2 +4/-2 +1/-2
2) Identify which atoms have been reduced (with a decrease in oxidation numbers), and which atoms have been oxidised (with an increase in oxidation numbers):
S + HNO3 → H2SO4 + NO2 + H2O
0 +6 Oxidised (S) : +6
+5 +4 Reduced (N) : -1
3) Balance the species that contain these elements. The total increase in oxidation number should equal the total decrease in oxidation number. You need to multiply HNO3 and NO2 by 6, so that there is a decrease of -6 from nitrogen, matching the increase of +6 for sulfur:
S + 6HNO3 → H2SO4 + 6NO2 + H2O
0 +6 Oxidised (S) : +6
+5(x6) +4(x6) Reduced (N) : -1
4) Balance any remaining atoms.
S + 6HNO3 → H2SO4 + 6NO2 + 2H2O

Question 33

5.2.3 b) Outline how to construct redox equations using oxidation numbers, with the following example:
Hydrogen iodide, HI, is oxidised to iodine, I2, by concentrated sulfuric acid, H2SO4, which is reduced to hydrogen sulfide, H2S.

Answer 33

1) Construct an incomplete draft equation (using the information provided), and assign oxidation numbers:
2HI + H2SO4 → H2S + I2
+1/-1 +1/+6/-2 +1/-2 0
2) Identify which atoms have been reduced (with a decrease in oxidation numbers), and which atoms have been oxidised (with an increase in oxidation numbers):
2HI + H2SO4 → H2S + I2
+6 -2 Reduced (S) : -8
-1(x2) 0(x2) Oxisdised (I) : +1
3) Balance the species that contain these elements. The total increase in oxidation number should equal the total decrease in oxidation number. You need to multiply 2HI and I2 by 4, so that there is an increase of +8 for iodine, matching the decrease of -8 for sulfur:
8HI + H2SO4 → H2S + 4I2
+6 -2 Reduced (S) : -8
-1(x8) 0(x8) Oxisdised (I) : +8
4) Balance any remaining atoms - in this case, add 2 H2O molecules to the right-hand side:
8HI + H2SO4 → H2S + 4I2 + 4H2O

Question 34

5.2.3 b) Outline how to construct redox equations using oxidation numbers, with the following example:
Acidified manganate(VII) ions, MnO4(-), can be reduced to manganese(II) ions by iron(II) ions. The iron(II) ions are oxidised to iron(III) ions. Write an overall redox equation for this reaction.

Answer 34

1) Iron changed from iron (II) to iron (III) - its oxidation number has increased by 1, so it must have lost 1 electron:
Fe(2+) → Fe(3+) + e-
2) Manganate(VII) is reduced to manganese(II) - its oxidation number has decreased by 5, so it must have gained 5 electrons:
MnO4(-) + 5e- → Mn(2+)
Balance this half-equation by adding 4 H2O molecules to the right-hand side, and 8H+ ions to the left-hand side:
MnO4(-) + 5e- + 8H+ → Mn(2+) + 4H2O
3) Multiply the half equations so that the electrons are balanced:
MnO4(-) + 8H+ + 5e- → Mn(2+) + 4H2O (doesn’t need to be multiplied)
Fe(2+) → Fe(3+) + e-
(x5) = 5Fe(2+) → 5Fe(3+) + 5e-
4) 2) Add the two half-equations together and cancel out where appropriate:
MnO4(-) + 8H+ + 5e- + 5Fe(2+) → Mn(2+) + 4H2O + 5Fe(3+) + 5e-
= MnO4(-) + 8H+ + 5Fe(2+) → Mn(2+) + 4H2O + 5Fe(3+)

Question 35

5.2.3 d) Outline the techniques and procedures used when carrying out redox titrations.

Answer 35

• Titrations are a way of determining amounts of substances, e.g. 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: a known concentration of either a reducing agent or an oxidising agent is placed in a burette and titrated against an unknown concentration of the chemical being oxidised or reduced.
• Whilst for acid-base reactions, an indicator is used to identify the end point, this isn’t necessary for redox titrations: many of the species are self-indicating, and change colour with different oxidation states.
  To carry out a redox titration:
  1) Measure out a known quantity of the reducing agent (using a pipette) and place into a conical flask.
  2) Add a known concentration of the oxidising agent to the burette (which is set up in a clamp-stand).
  3) Gradually add the oxidising agent to the reducing agent, swirling the conical flask as you do so. Stop at the end point (indicated by a colour change) and record the volume of oxidising agent added.
  4) Repeat the titration until you obtain concordant results, i.e. two titres that agree within +0.10 cm3. Calculate a mean from the concordant results.

Question 36

5.2.3 d) Outline the techniques and procedures used when carrying out a redox titration for Fe2+/MnO4(-)

Answer 36

Manganate(VII), MnO4(-), is an oxidising agent, usually obtained from potassium permanganate(VII), KMnO4. It has a deep purple colour but becomes colourless when reduced from +7 oxidation state to +2 oxidation state This occurs in the presence of H+ ions - so the reaction must take place under acidic conditions (e.g. in dilute sulfuric acid).
MnO4(-) + 8H+ + 5e- → Mn(2+) + 4H2O
MnO4- can be used to oxidise solutions containing iron(II):
MnO4(-) + 8H+ + 5Fe(2+) → Mn(2+) + 4H2O + 5Fe(3+)
1) Add a standard solution (with known concentration) of potassium permanganate(VII), KMnO4, to a burette.
2) Using a pipette, add a measured volume of the solution containing iron(II) (e.g. a solution of iron sulfate(II), FeSO4, which is also FeSO4•7H2O dissolved in solution) to the conical flask. Add an excess of dilute sulfuric acid as well.
3) Gradually add the permanganate(VII) to the iron(II) solution, swirling the conical flask as you do so: you will see the purple colour of the MnO4- disappearing as it reacts with the acidified Fe2+.
4) The end point is seen when excess MnO4- ions are present - indicated by a permanent pale pink colour appearing. This occurs when all the Fe2+ ions have reacted, and the Mn04- can no longer be reduced to the colourless Mn2+.
5) Repeat the titration until you obtain concordant results. Calculate a mean from the concordant results.

Question 37

5.2.3 d) Outline the techniques and procedures used when carrying out a redox titration for I2/S2O2(2-). How can this titration be used to determine the concentration of different oxidising agents?

Answer 37

Consider an iodine-thiosulfate titration:
The thiosulfate ions S2O3(2-), are oxidised:
2S2O3(2-) (aq) → S4O6(2-) (aq) + 2e-
The iodine, I2, is reduced:
I2 (aq) + 2e- → 2I- (aq)
So the overall redox reaction is:
2S2O3(2-) (aq) + I2 (aq) → S4O6(2-) (aq) + 2I- (aq)
The concentration of aqueous iodine can be determine by titration with a standard solution of thiosulfate. However, this reaction can also be used to determine the concentration of an unknown reducible species, i.e. an oxidising agent. This is done by first reacting a solution containing iodine ions, I-, with the oxidising agent - this will form iodine, I2. The resulting solution, containing iodine, is then titrated with sodium thiosulfate (as shown above).
Titration calculations can be used to calculate the concentration of the oxidising agent (this is because the more concentrated an oxidising agent is, the more ions will be oxidised by a certain volume of it).
To carry out this procedure:
1) Measure out a certain volume of the oxidising agent, and add this to an excess of potassium iodide (KI) solution - this is the source of iodide ions. The oxidising agent will be reduced, and the iodide ions will be oxidised to iodine - consider the following examples of different oxidising agents:
e.g. The concentration of a solution of chlorine (Cl2) could be determined:
Cl2 (aq) + 2I- (aq) → 2Cl- (aq) + I2 (aq)
e.g. The concentration of a solution containing Cu2+ ions could be determined:
2CU(2+) (aq) + 4I- (aq) → 2CuI (s) + I2 (aq)
e.g. The concentration of an (acidified) potassium iodate(V) solution could be determined:
IO3- (aq) + 5I- (aq) + 6H+ (aq) → 3H2O (l) + 3I2 (aq)
e.g. The concentration of ClO- ions in (acidified) bleach could be determined:
ClO- (aq) + 2I- (aq) + 2H+ (aq) → Cl- (aq) + I2 (aq) + H2O (l)
In each reaction, iodine (I2) is formed, and the solution will be yellow-brown colour.
2) Add a standard solution of sodium thiosulfate, Na2S2O3 (aq), to the burette.
3) Titrate the yellow-brown solution formed in step 1 with sodium thiosulfate - add it drop by drop, swirling the solution as you do so. During the titration, the iodine reacts with thiosulfate ions, and is reduced back to iodide ions.
2S2O3(2-) (aq) + I2 (aq) → S4O6(2-) (aq) + 2I- (aq)
The yellow-brown colour fades gradually as the iodine reacts, making it difficult to determine the exact end point of the reaction. This problem is solved by adding starch solution. When the end point is being approached, and the iodine colour has faded into a pale yellow, add the starch solution. A deep blue-black colour will form (due to the presence of any remaining iodine).
4) Carry on with the titration: as more sodium thiosulfate is added, reacting with the iodine, the blue-black colour fades. At the end-point, all of the iodine will have just reacted and the blue-back colour will have disappeared.
5) Now you can calculate the number of moles of thiosulfate ions that reacted, and use the stoichiometry of the reactions to determine:
i) the number of moles of iodine that reacted
i) the number of moles of the oxidising agent that reacted
iii) the concentration of the oxidising agent.

Question 38

5.2.3 g) Explain what a half-cell is.

Answer 38

• In an electrochemical cell, chemical energy is converted into electrical energy. This electrical energy is a result of the movement of electrons in redox reactions.
• When we consider redox reactions that are part of a cell, each half-equation involved is equivalent to a half-cell.
• So a half-cell will contain the chemical species present in a redox half-equation.
• A half-cell comprises of an element in two oxidation states. The equilibrium in a half-cell is written so that the forward reaction shows reduction (the gain of electrons), and the reverse reaction shows oxidation (the loss of electrons), i.e. the electrons are always on the left:
  e. g. X+ (aq) + e- ⇄ X (s)
  e. g. Y(3+) (aq) + 2e- ⇄ Y(2+) (aq)
• In an isolated half-cell there is no net transfer of electrons either into or out of the half-cell.
• When two half-cells are connected, the direction of electron flow depends upon the relative tendency of each electrode to release electrons.

Question 39

5. 2.3 g) Explain what the following half-cells consist of:
   i) A metal/metal ion half-cell
   ii) An ion/ion half-cell
   iii) A non-metal/non-metal ion half-cell

Answer 39

1) A metal/metal ion half-cell consists of a metal placed in an aqueous solution of its ions. At the phase boundary where the metal is in contact with its ions, an equilibrium will be set up.
The piece of solid metal serves as an electrode when the half-cell is connected to another half-cell. It allows the transport of electrons either into or out of the half-cell.
e.g. A copper half-cell comprises of a strip of copper metal (oxidation state: 0) placed in an aqueous solution of Cu(2+) ions (oxidation state: +2). The following equilibrium exists at the surface of the solid Cu:
Cu(2+) (aq) + e- ⇄ Cu (s)

2) An ion/ion half-cell contains ions of the same element in different oxidation states. The different ions should be the same concentrations, i.e. be equimolar (the standard concentration is 1 mol dm-3).
In this type of half-cell, there is no solid piece of metal that can act as the electrode. This is overcome using an inert platinum electrode: it doesn’t react with any elements, but is in contact with both ions, allowing the transfer of electrons into or out of the half-cell, via a connecting wire.
e.g. A standard Fe(3+) (aq)/ Fe(2+) (aq) half-cell is made up of a solution containing equimolar aqueous iron(II) and iron (III) ions. The redox equilibrium is:
Fe(3+) (aq) + e- ⇄ Fe(2+) (aq)

3) A non-metal/non-metal ion half-cell consists of a gas being released into a glass tube, with this glass tube being placed in a solution containing ions of that element. Holes in the glass tube allow bubbles of the gas to escape, so that an equilibrium can be set up between the two oxidation states.
An inert platinum electrode allows electrons to pass into or out of the half-cell, via a connecting wire.
e.g. A standard hydrogen half-cell consists of hydrogen gas, H2 (at 298k temperature, and 100kPa pressure), and a solution containing 1 mol dm-3 of H+ ions (e.g. from 1 mol dm-3 of HCl).
The redox equilibrium is:
2H+ (aq) + 2e- ⇄ H2 (g)

Question 40

5.2.3 f) Outline what standard electrode

potential, E⊖, is, and how it is measured.

Answer 40

The standard electrode potential, E⊖, of a half-cell can be determined by connecting it to a hydrogen electrode. The tendency for different half-cells to accept or release electrons is measured as a voltage (otherwise known as electromotive force, e.m.f) with the unit: volts, V.
To measure a standard electrode potential, the half-cell is connected to a standard hydrogen electrode (i.e. a H2(g)/H+(aq) half-cell), under the following standard conditions:
1) a temperature of 298K
2) a pressure of 100kPa
3) any solution with a concentration of 1 mol dm-1
- The two electrodes from each half-cell are connected by a wire to allow a controlled flow of electrons.
- The two solutions are connected with a salt bridge which allows ions to flow. The salt bridge typically contains a concentrated solution of an electrolyte that does not react with either solution. An example of a salt bridge is a strip of filter paper soaked in an aqueous solution of an ionic substance, e.g. KNO3 (aq).

The reading on the voltmeter (the e.m.f value) measures the difference between the electrode potentials of the half-cells. However, by definition the hydrogen half-cell has a standard electrode potential of 0V (which is why it is used as a reference to measure other standard electrode potentials by). So any reading on the voltmeter gives the standard electrode potential of the connected half-cell.

Half-cells can be listed in order of their standard electrode potentials (with the most negative value at the top, and the most positive value at the bottom) - this is known as the electrochemical series.
The more reactive a metal is, the more it wants to lose electrons to form a positive ion. So more reactive metals have more negative standard electrode potentials.
The more reaction a non-metal is, the more it wants to gain electrons to form a negation ion. So more reaction non-metals have more positive standard electrode potentials.

Question 41

5.2.3 g) Outline the techniques and procedures used for the measurement of cell potentials.

Answer 41

• An electrochemical cell can be made by connecting together two different half-cells. The electrode potential of each half-cell indicates its tendency to lose or gain electrons. Each cell has an overall cell potential - this is the difference in electrode potentials between the two half-cells.
  The cell potential can be measured experimentally:
  1) Prepare two standard half-cells. Connect the electrode of the half-cell to a voltmeter using wires.
  2) Prepare a salt bridge by soaking a strip of filter paper in a saturated aqueous solution of potassium nitrate, KNO3. Connect the two solutions of the half-cells with a salt bridge.
  3) Record the cell potential from the voltmeter.
  The reading on the voltmeter (the e.m.f value) measures the difference between the electrode potentials of the half-cells. This means that the e.m.f value can also be taken as the cell potential.
  [Note: if the half-cells were prepared under standard conditions, then voltmeter will give the standard cell potential]

Question 42

5.2.3 h) Explain how the standard cell potential can be calculated by combining two standard electrode potentials.

Answer 42

While you can work out standard cell potential experimentally, it can also be calculated from standard electrode potentials, using the following equation:
E⊖cell = E⊖(positive electrode) - E⊖(negative electrode)
[Note: even if both half-cells have a positive standard electrode potential, +E⊖, or both have a negative standard electrode potential, -E⊖, one of them will act as a positive electrode, and the other as a negative electrode.]

e.g. Calculate the standard cell potential of a silver-copper cell. The standard electrode potentials are:
Ag+ (aq) + e- ⇄ Ag (g)
E⊖ = +0.80 v
Cu(2+) (aq) + e- ⇄ Cu (s)
E⊖ = +0.34 v
Remember that the more negative of the two systems is the negative terminal of the cell.
E⊖cell = E⊖(positive electrode) - E⊖(negative
E⊖ = +0.80 - +0.34
E⊖ = 0.46 V

Question 43

5.2.3 h) Outline how to calculate the redox equation for a cell, given the standard electrode potentials of the two half-cells.

Answer 43

The overall redox reaction of a cell can be seen as two separate reactions: oxidation, which takes place in one half cell, and reduction, which takes place in the other one. While the reversible arrow shows that both reactions can go in either direction, one half-cell will favor the forward reaction (reduction), and the other half-cell will favor the reverse reaction (oxidation) - which direction each reaction goes in depends on the standard electrode potential.

e.g. Y(+) + e- ⇄ Y
e.g. Z(+) + e- ⇄ Z
- The more negative the E⊖ value, the more the equilibrium lies to the left of the half equation: this is because there is a greater tendency to lose electrons and undergo oxidation. So when half-cells are in combination, the one with the more negative value will favor the reducing agent:
Y → Y(+) + e-
- The more positive the E⊖ value, the more the equilibrium lies to the right: this is because there is a greater tendency to gain electrons and undergo reduction. So when half-cells are in combination, the one with the more positive value will favor the oxidising agent:
Z(+) + e- → Z
The overall cell equation is obtained by combining the reduction and oxidation half-equations:
Y + Z(+) + e- → Y(+) + e- + Z
Y + Z(+) → Y(+) + Z
This means there is a flow of electrons from the negative electrode, where oxidation takes place, to the positive electrode, where reduction takes place.

e.g. Consider a zinc-copper cell:
Zn(2+) (aq) + e- ⇄ Zn (s)
E⊖ = - 0.76 V
Cu(2+) (aq) + e- ⇄ Cu (s)
E⊖ = + 0.34 V

Reduction (positive electrode): Cu(2+) (aq) + e- → Cu (s)
Oxidation (negative electrode): Zn (s) → Zn(2+) (aq) + e-
Overall cell reaction:
Cu(2+) (aq) + Zn (s) → Cu (s) + Zn(2+) (aq)

Question 44

5.2.3 i) Explain how you can predict of the feasibility of a reaction using standard cell potentials, with the following examples:
1) Predict whether zinc reacts with aqueous copper ions.
Zn(2+) (aq) + e- ⇄ Zn (s)
E⊖ = - 0.76 V
Cu(2+) (aq) + e- ⇄ Cu (s)
E⊖ = + 0.34 V
2) Predict whether copper reacts with dilute sulfuric acid.
2H+ (aq) + 2e- ⇄ H2 (g)
E⊖ = 0.00 V
Cu(2+) (aq) + e- ⇄ Cu (s)
E⊖ = + 0.34 V

Answer 44

Consider the half-equation that accompanies a half-cell - the reducing agents are on the left, and the oxidising agents are on the right:
e.g. X(+) + e- ⇄ X, E⊖ = - 0.76 v
e.g. Y(+) + e- ⇄ Y, E⊖ = - 0.45
You may have to determine if a reaction between one of the oxidising agents, e.g. X, and one of the reducing agents, e.g. Y(+), is feasible. It may not be: instead, the reaction that occurs could be between the oxidising agent Y and the reducing agent X(+):
Y + X(+) → Y(+) + X
This is because a redox reaction will take place between an oxidising agent and a reducing agent only if the redox system of the oxidising agent has a more positive E⊖ than the redox system of the reducing agent.

e.g. Consider a zinc and aqueous copper ions.
1) The half-cell with the more negative E⊖ will react from right to left, and lose electrons:
Zn (s) → Zn(2+) (aq) + e-
2) The half-cell with the more positive E⊖ will react from left to right, and gain electrons:
Cu(2+) (aq) + e- → Cu (s)
The overall, feasible redox reaction is:
Cu(2+) (aq) + Zn (s) → Cu (s) + Zn(2+) (aq)
So zinc does react with aqueous copper ions.

e.g. Consider copper and dilute sulfuric acid.
1) The half-cell with the more negative E⊖ will react from right to left, and lose electrons:
H2 (g) → 2H(+) (aq) + 2e-
2) The half-cell with the more positive E⊖ will react from left to right, and gain electrons:
Cu(2+) (aq) + e- → Cu (s)
The overall, feasible redox reaction is:
Cu(2+) (aq) + H2 (g) → Cu (s) + 2H(+) (aq)
This feasible reaction that occurs is between aqueous copper ions and gaseous hydrogen. This means that copper and sulfuric acid (i.e. the H+ ions) won’t react, and isn’t feasible.

Question 45

5.2.3 i) What are the limitations of predicting the feasibility of a reaction using standard cell potentials?

Answer 45

1) A prediction using E⊖ only states if a reaction is possible under standard condition - but non-standard conditions alter the value of an electrode potential:
e.g. Standard electrode potentials are measured using concentrations of 1 mol dm-3.
Consider Zn (s) → Zn(2+) (aq) + e-:
If the concentration of Zn(2+) (aq) is greater than 1 mol dm-3, the equilibrium will shift to the right, removing the electrons from the system and making the electrode potential less negative. If the concentration of Zn(2+) (aq) is less than 1 mol dm-3, the equilibrium will shift to the left, increasing the electrons in the system and making the electrode potential more negative.
[Note, even if the concentration is not 1 mol dm-3 for ion/ion half-cells, the e.m.f could still remain the same: as long as the concentrations are equimolar]
2) Some reactions also have a very high activation energy - this could result in an extremely slow rate of reaction.

Question 46

5.2.3 j) A nickel-iron cell has a nickle oxide hydroxide (NiO(OH)) cathode and an iron (Fe) anode. Using the half-equations below, write the full equation for the reaction.
Fe + 2OH- ⇄ Fe(OH)2 + 2e-
E⊖ = - 0.44 V
NiO(OH) + H2O + e- ⇄ Ni(OH)2 + OH-
E⊖ = + 0.76 V

Answer 46

Oxidation: Fe + 2OH- → Fe(OH)2 + 2e-
Reduction: NiO(OH) + H2O + e- → Ni(OH)2 + OH-
(x2) : 2NiO(OH) + 2H2O + 2e- → 2Ni(OH)2 + 2OH-
The overall reaction is:
= Fe + 2OH- + 2NiO(OH) + 2H2O + 2e- → Fe(OH)2 + 2e- + 2Ni(OH)2 + 2OH-
= Fe + 2NiO(OH) + 2H2O → Fe(OH)2 + 2Ni(OH)2
E⊖cell = + 1.2 V

Question 47

5.2.3 k) Explain how a fuel cell works.

Answer 47

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

• It is an electrochemical cell, where the fuel flows into one half-cell, and the oxygen flows into the other half-cell (any products flow out of the cell). An electrolyte, i.e. a conductive substance containing free ions, is also present. And while a membrane keeps the fuel and oxygen apart, it does allow the electrolyte to move through.
• The fuel loses electrons at one electrode (the anode) - and is oxidised.
• The oxygen gains electrons at the other electrode (the cathode) - and is reduced.
• This flow of electrons produces a voltage.

Many different fuels can be used, with hydrogen (and hydrogen-rich fuels, such as methanol, CH2OH) being the most common. Hydrogen fuel cells produce no carbon dioxide during combustion, with water being the only product.

Question 48

5.2.3 k) Explain the two different versions of a hydrogen-oxygen fuel cell.

Answer 48

There are two versions of the hydrogen-oxygen fuel cell: one with an acid electrolyte, and one with an alkali electrolyte. 
1) An acid (H+) hydrogen fuel cell:
The two half cells are: 
- 2H+ (aq) + 2e- ⇄ H2 (g)  
E⊖ = 0.00 V
- 1/2O2 (g) + 2H+ (aq) + 2e- ⇄ H2O (l)  
E⊖ = + 1.23 V
- Oxidation takes place at the anode:  
H2 (g) → 2H+ (aq) + 2e-
- Reduction takes place at the cathode: 
1/2O2 (g) + 2H+ (aq) + 2e- → H2O (l)
- The overal redox reaction:
1/2O2 (g) + 2H+ (aq) + 2e- + H2 (g) → H2O (l) + 2H+ (aq) + 2e-
=   1/2O2 (g) + H2 (g) → H2O (l)
E⊖cell = + 1.23 V

2) An alkali (OH-) hydrogen fuel cell:
The two half-cells are:
- 2H2O + 2e- ⇄ H2 (g) + 2OH- (aq)
E⊖ = - 0.83 V
- 1/2O2 (g) + 2H2O(l) + 2e- ⇄ 2OH- (aq)  
E⊖ = + 0.40 V
- Oxidation takes place at the anode:
H2 (g) + 2OH- (aq) → 2H2O + 2e-
- Reduction takes place at the cathode:
1/2O2 (g) + H2O(l) + 2e- ⇄ 2OH- (aq)
- The overal redox reaction:
H2 (g) + 2OH- (aq) + 1/2O2 (g) + H2O(l) + 2e- → 2H2O + 2e- + 2OH- (aq)
=   1/2O2 (g) + H2 (g) → H2O (l)
E⊖cell = + 1.23 V