5.2 Flashcards

Question

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

Answer 1

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

Answer 2

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_sol*H*^ø).

Answer 3

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).

Answer 4

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.

Answer 5

* 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.

Answer 6

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)

Answer 7

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.

Answer 8

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.

Answer 9

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. ## Footnote (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 MgBr₂you 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.)

Answer 10

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.

Answer 11

* 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.

Answer 12

* 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.

Answer 13

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

Answer 14

Hydration is an exothermic process.

Answer 15

* the size of the ions involved * the charges on the ions

Answer 16

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_hyd*H*^øbecomes more exothermic.

Answer 17

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.

Answer 18

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

Answer 19

* 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)

Answer 20

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

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

J K^-1 mol^-1

Answer 25

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.

Answer 26

thermodynamic

Answer 27

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

Answer 28

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.

Answer 29

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 Δ*S* . 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.

Answer 30

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

Answer 31

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.

Answer 32

* 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.

Answer 33

_Using the entropies of the reactants and products._

Answer 34

* ΔG, is the balance between enthalpy, entropy and temperature for a process: * Δ*G* = Δ*H - T*Δ*S* * A process can take place spontaneously when ΔG \< 0.

Answer 35

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.

Answer 36

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.

Answer 37

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.

Answer 38

* 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.

Answer 39

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^-1mol^-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

Answer 40

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.

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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**.

Answer 46

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**.

Answer 47

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)

Answer 48

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)

Answer 49

* 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.

Answer 50

* 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.

Answer 51

* 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.

Answer 52

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.

Answer 53

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.

Answer 54

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^-.)

Answer 55

MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²⁺ +5Fe³⁺ + 4H₂O

Answer 56

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⁺ .

Answer 57

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.

Answer 58

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)

Answer 59

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

Answer 60

* 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.

Answer 61

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)

Answer 62

* 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)

Answer 63

1. iodide ions are oxidised to iodine, 2I^- → I₂ + 2e^- 2. copper(II) ions are reduced to copper(I) ions, Cu²⁺ + e^- → Cu⁺ So the overall ionic equation is: 2Cu²⁺ + 4I^- → 2CuI + I₂

Answer 64

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.

Answer 65

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)

Answer 66

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.

Answer 67

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.

Answer 68

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.

Answer 69

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.

Answer 70

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.

Answer 71

A half cell of bromine and its ions; Br₂ + 2e^- ⇔ 2Br^-

Answer 72

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.

Answer 73

* 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.

Answer 74

* 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.

Answer 75

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.

Answer 76

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.

Answer 77

* 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 .

Answer 78

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.

Answer 79

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

Answer 80

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.

Answer 81

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.

Answer 82

From this small series, we could say that the half cell Fe²⁺ + 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^øvalues for standard electrode potentials can be used to predict what will happen when half cells are connected together to form a cell.

Answer 83

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.

Answer 84

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 ⇒

Answer 85

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.

Answer 86

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.

Answer 87

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

Answer 88

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.

Answer 89

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.

Answer 90

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.

Answer 91

* 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.

Answer 92

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.

Answer 93

* 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.

Answer 94

* 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.

Answer 95

* 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.

Answer 96

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.

Answer 97

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

Answer 98

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.

Answer 99

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.

Answer 100

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

Answer 101

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