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Chemistry IAL EDEXCEL A-Level > Unit 5- Topic 16 > Flashcards

Unit 5- Topic 16 Flashcards

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1
Q

Describe the term oxidation in terms of electron transfer and oxidation number

A

Oxidation is the loss of electrons
An element in a species is oxidised when its oxidation number increases

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2
Q

Describe the term reduction in terms of electron transfer and oxidation number

A

Reduction is the gain of electrons

An element in a species is reduced when its oxidation number decreases

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3
Q

To what elements of the Periodic table does the definitions of oxidation and reduction apply to

A

They apply to all elements whether they are in the s-, p- or d- block of the Periodic Table

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4
Q

Definition of standard electrode potential

A

The measured potential difference of a cell when no current is flowing (emf) when a half-cell is connected to a standard hydrogen electrode under standard conditions

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5
Q

What are the standard conditions that the standard electrode potential are measured by

A

298K temperature
Gas pressure: 100kPa
Concentration of ions: 1 mol dm-3

These standard conditions apply to all equilibria

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6
Q

What reactions are involved in the measurement of electrode potential

A

All redox reactions and therefore equilibria

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7
Q

What is the dynamic equilibrium formed of a metal immersed in water

A

The rate at which ions are leaving the surface of the metal to go into solution is the same as the rate at which they are joining it from the solution so:

M x+ (aq) + xe- <——> M (s)

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8
Q

Why is the standard hydrogen electrode necessary

A

Because it is not possible to measure the absolute potential difference between a metal electrode and its solution since to establish a electrical connection with the solution a metal would be needed therefore you would be ensuring the potential difference between the two pieces of metal.

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9
Q

The conditions for only the SHE

A

Hydrogen gas at pressure: 100kPa
Platinum foil dipped in solution
Solution with concentration of [H+] = 1 mol dm-3
At 298K temperature

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10
Q

What is needed to connect a SHE to a metal half-cell

A

A high resistant voltmeter connected with platinum wire to metal and platinum foil of SHE
A salt bridge

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11
Q

Why is the salt bridge needed in electrochemical cells

A

Allows movement of ions
The ions in the salt ridge should not interfere with the ions in both half-cells

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12
Q

Why is the platinum foil covered in porous platinum

A

Porous platinum has a large surface area and allows an equilibrium between hydrogen ions in solution and hydrogen gas to be established quickly

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13
Q

Why is a high-resistance voltmeter used

A

To achieve no flow of electrons in the external circuit so that the voltmeter represents the difference in potential between the two half-cells when both reactions are in equilibrium

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14
Q

In electrochemical cells, inn which electrode does oxidation occur

A

Anode

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15
Q

Equation for emf (E cell)

A

E cell = E rhs - E lhs

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16
Q

How to measure E wit a systems involving gases

A

Gas is bubbled into a solution containing its ions. A piece of platinum covered in porous platinum is dipped into the solution to connect the gas molecules and its ions to the externa circuit through an electrical connection

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17
Q

How to calculate E cell of two non-metal elements and their ions

A

Each non- metal is compared to a SHE. A half-cell containing a solution of both ions (of one element) with each of concentration 1 mol dm-3 is connected to the SHE and E annotated. Same is done with the other non-metal, and redox reaction calculated.

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18
Q

Components needed to measure the E of ions of same element but different oxidation numbers

A

-platinum wire
-platinum foil covered in porous platinum
-solution with both ions at a concentration of 1 mol dm-3

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19
Q

What half cell goes to the left of the electrochemical cells

A

The negative electrode, the one with the most negative E value

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20
Q

What does the vertical solid line mean in a cell diagram

A

A phase boundary

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21
Q

What are the two rules to follow when drawing a cell diagram

A

1-the two reduced forms of the species are shown on the outside of the cell diagram
2-the positive electrode is shown on the right-hand side of the cell diagram

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22
Q

What does double vertical lines represent in cell diagram

A

Salt bridge

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23
Q

Draw the standard hydrogen electrode diagram

A

H+(aq) l 1/2 H2 (g) l Pt(s)

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24
Q

Draw the cell diagram of Fe 2+ and Fe 3+ half cell

A

Pt(s) l Fe 3+ (aq), Fe 2+ (aq)

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25
Q

Where does the S.H.E always go in cell diagram

A

Always written in the left-hand side

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26
Q

Write the cell diagram for the S.H.E and zinc metal

A

Pt(s) l 1/2H2 (g) l H+ (aq) ll Zn 2+ (aq) l Zn (s)

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27
Q

What may be the reasons for a thermodynamic feasible reaction to not take place

A

The reactants may be kinetically stable because the activation energy of the reaction is very large
The reaction may not be taking place under standard conditions since:

A reaction that is not thermodynamically feasible under standard conditions may become feasible when the conditions are altered

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28
Q

Why is the electrode potential altered by a change in the conditions of the reaction

A

Because the position of equilibrium of the half-cell reaction may change

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29
Q

How can you use standard electrode potentials to predict thermodynamic feasibility of a reaction

A

Change the sign of the reversed reaction and calculate the new Ecell for the whole reaction. If the Ecell is positive then th reaction is likely to be thermodynamically feasible if not then it is not thermodynamically feasible.

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30
Q

What is the relationship between total entropy (Stotal) and emf of a cell (Ecell)

A

T change in Stotal = nFEcell

n and F are both constants therefore at a given temperature, the total entropy change is proportional to the emf of the cell

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31
Q

How can entropy change and Ecell be used to predict direction of reaction

A

If Ecell is positive, change S total is also positive and therefore reaction will occur.

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32
Q

What is the relationship between the equilibrium constant and Ecell

A

Since change Stotal = RlnK and change in Stotal= nFEcell / T

      nFEcell lnK=\_\_\_\_\_\_\_\_\_\_\_\_
          RT
33
Q

What other name can standard electrode potentials be referred as

A

Standard reduction potentials

34
Q

How is the electrochemical series ordered

A

The most negative E values are placed at the top and most positive at the bottom

35
Q

Where are reducing agents located in the electrochemical series

A

The species on the right-hand side of the half-equations are all capable of behaving as reducing agents as they can lose electrons

36
Q

Where are oxidising agents located in the electrochemical series

A

The species on the left-hand side of the half-cell equations are all capable of acting as oxidising agents as they can gain electrons

37
Q

What does a very negative E value mean in the electrochemical series

A

The more negative the E value, the more the equilibrium lies towards the left- and the more readily the species on the right loses electrons therefore the more powerful the reducing agent

38
Q

What does a very positive E value mean in the electrochemical series

A

The more positive the E value, the more the equilibrium lies towards the right - and the more readily the species on the left loses electrons therefore the more powerful the oxidising agent

39
Q

What is a disproportionation reaction

A

A reaction in which an element in a species is both oxidised and reduced in the same reaction at the same time

40
Q

If the E value is the most negative of the two reactions proposed, to which side will the equilibrium fall

A

Equilibrium will move to the left, releasing electrons

41
Q

If the E value is the most positive of the two reactions proposed, to which side will the equilibrium fall

A

Equilibrium will move to the right, accepting electrons

42
Q

How to calculate percentage uncertainty

A

uncertainty
Percentage uncertainty =_______________ x 100
measured value

43
Q

How to reduce percentage uncertainty in brunette readings

A

Dilute the chemical in the burette
This means that more will be required for the titration
Therefore, the overall uncertainty in the burette readings is reduced

44
Q

Is potassium manganate (VII) a powerful reducing or oxidising agent?

A

A powerful oxidising agent

45
Q

Give the half equation for the reduction of manganate (VII) ions in acidic solution

A

MnO4 - (aq) + 8H+ (aq) + 5e- -> Mn2+ (aq) + 4H2O (l)

46
Q

Why is potassium manganate(VII) not used in redox titrations in alkaline solutions

A

Because it yields manganese (IV) oxide as a brown precipitate which interferes with te end point colour of the titration. So it is used in acidic solutions to prevent the formation of manganese (IV) oxide

47
Q

Describe the behaviour (colour) of potassium manganate (VII) in a redox titration

A

As the titration proceeds, manganese (II) ions, Mn2+, accumulate but, at the dilution used, they give a colourless solution. As soon as the potassium manganate (VII) is in excess, the solution becomes pink. It therefore acts as its own indicator- the end point is the first permanent pink colour

48
Q

Give the ionic half-equations of the titration of potassium manganate (VII) and iron (II) ions

A

MnO4- (aq) + 8H+ (aq) + 5e- -> Mn2+ (aq) + 4H2O (l)
Fe2+ (aq) -> Fe3+ + e-

49
Q

Give the overall equation for the titration of potassium manganate (VII) and iron (II) ions

A

MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) -> Mn2+ (aq) +4H2O (l) +5Fe3+ (aq)

50
Q

Method for performing the titration of potassium manganate (VII) and iron (II) ions

A

-Pipette an accurately measured volume, usually 25.0cm3, of the iron (II) solution into a conical flask and then add a small volume of dilute sulfuric acid
-Slowly add potassium manganate(VII) solution of accurately known concentration from a burette and swirl the mixture
-The potassium manganate(VII) solution will turn colourless until all of the iron (II) ions have been oxidised
-The addition of one more drop of potassium manganate(VII) solution will turn the mixture pale pink

51
Q

What occurs to ethanedioic acid in the titration of this with potassium manganate (VII), give the ionic half-equation

A

The ethanedioic acid is reduced to carbon dioxide
H2C2O4 (aq) -> 2H+ (aq) + 2CO2 (g) +2e-

52
Q

Give the overall equation for the titration of ethanedioic acid and potassium manganate (VII)

A

2MnO4- (aq) + 16H+(aq) + 5C2O4 2+ (aq) -> 2Mn2+ (aq) + 8H2O (l) + 10CO2 (g)

53
Q

Method for the titration of ethanedioic acid and potassium manganate (VII)

A

-Aqueous ethanedioic acid heated to 60C is pipetted into the conical flask. (The acid is highly poisonous so a pipette filler must be used) and is the acidified with dilute sulfuric acid.
-Aqueous potassium manganate (VII) is added from the burette.
-Since the reaction is slow, the pink does not immediately disappear.
-After the initial sample is added, the Mn2+ ions produced act as a catalyst and the reaction speeds up, allowing you to titrate normally and obtain an accurate end point.

54
Q

Why is the aqueous ethanedioic acid heated to around 60 C before its titration with potassium manganate(VII)

A

Because the reaction is very slow, help overcome the activation energy.

55
Q

How is the sulfuric acid treated when it is going to be used in a titration with iron (II)

A

The aqueous sulfuric acid is boiled to remove dissolved oxygen that might otherwise oxidise some of the iron(II) ions to iron (III)

56
Q

State the ionic half-equations involved in the titration of iodine and sodium thiosulfate

A

Thiosulfate ions reduce iodine to iodide ions.
2S2O3 2- (aq) -> S4O6 2- (aq) + 2e-
I2 (aq) + 2e- -> 2I- (aq)

57
Q

State the ionic half-equations involved in the titration of iodine and sodium thiosulfate

A

Thiosulfate ions reduce iodine to iodide ions.
2S2O3 2- (aq) -> S4O6 2- (aq) + 2e-
I2 (aq) + 2e- -> 2I- (aq)

58
Q

Overall equation for the redox titration of iodine and sodium thiosulfate

A

2S2O3 2- (aq) + I2 (aq) -> S4O6 2- (aq) + 2I- (aq)

59
Q

What is the indicator used non the redox titration of iodine and sodium thiosulfate

A

The indicator is starch solution. With free iodine it produces a deep blue-black colour. The blue-black colour disappears as soon as sufficient sodium thiosulfate has been added to react with all of the iodine

60
Q

What can the redox titration of sodium thiosulfate and iodine be used for

A

-the direct estimation of iodine
-the estimation of a substance that can take part in a reaction that produces iodine

61
Q

Method of the redox titration of sodium thiosulfate and iodine

A

-Add sodium thiosulfate solution from the burette to the iodine solution until the original brown colour of the iodine changes to pale yellow
-Add a few drops of starch solution to produce a blue colouration
-Add the sodium thiosulfate solution drop by drop until the blue-black solution turn colourless

62
Q

Why is starch added to the redox titration of sodium thiosulfate and iodine

A

Because if not, it would be very difficult to detect the end point. The colour of the iodine solution becomes very faint towards the end of the reaction, and this makes it difficult t accurately assess when the end point has been reached

63
Q

Why do you need to start the redox titration of sodium thiosulfate and iodine without the starch solution and add it later

A

Because if starch is added too early, it adsorbs some of the iodine and reduces the accuracy of the titration.

64
Q

What effect does increasing the concentration have in the emf

A

Makes it more positive as fewer electrons are made in the reaction

65
Q

What effect does increasing the pressure have in the emf

A

More negative as more electrons are produced

66
Q

Which electrode is the cathode in a fuel cell

A

The positive electrode

67
Q

Which electrode is the anode in a fuel cell

A

The negative electrode

68
Q

What is a fuel cell

A

A fuel cell produces a voltage from the chemical reaction of a fuel with oxygen. In the hydrogen-oxygen fuel cell, hydrogen is supplied externally as a gas, and the cell can operate as long as the fuel supply is maintained.

69
Q

What fuels are used in fuel cells

A

Hydrogen, methanol/ethanol, hydrogen-rich methanol

70
Q

Why are both metal electrodes coated with platinum in the hydrogen-oxygen fuel cell

A

Because it catalyses the reactions that take place at the electrodes.

71
Q

Reaction in negative electrode [anode] in acidic conditions in fuel cells

A

H2 (g) -> 2H+ (aq) + 2e-

72
Q

Reaction in positive electrode [cathode] in acidic conditions in fuel cells

A

1/2 O2 (g) + 2H+ (aq) + 2e- -> H2O (l)

73
Q

Overall reaction for fuel cells in both conditions

A

H2 (g) + 1/2 O2 (g) -> H2O

74
Q

Overall reaction for fuel cells in both conditions

A

H2 (g) + 1/2 O2 (g) -> H2O

75
Q

What happens in a hydrogen-oxygen fuel cell

A

The hydrogen ion (protons) pass through the proton exchange membrane, which allows them to enter the compartment containing the positive electrode, where they can react with oxygen.

76
Q

Reaction in negative electrode [anode] in alkaline conditions in fuel cells

A

H2 (g) + 2OH- (aq) -> H2O (l) + 2e-

77
Q

Reaction in positive electrode [cathode] in alkaline conditions in fuel cells

A

1/2 O2 (g) + H2O (l) + 2e- -> 2OH- (aq)

78
Q

Advantages of hydrogen-oxygen fuel cells

A

-offer alternative to fossil fuels
-the only waste product is water, so no pollutants like CO or CO2 are produced
-they are lighter and more efficient than engines that use fossil fuels

79
Q

Disadvantages of hydrogen-oxygen fuel cells

A

-hydrogen explodes very easily so great care is needed when transporting it. You need to either: compress the gas, adsorb it onto the surface of a suitable solid material, absorb it into a suitable material
-compressing the gas is dangerous as transporting hydrogen under pressure is hazardous
-there is no firm conclusion as to what metal should be used to adsorb hydrogen
-metal hydride absorb hydrogen but high temperatures are required to release the hydrogen.
-supply of hydrogen is a problem since we are now extracting it from methane which is a limited source. Extracting it from renewable sources is possible but expensive.

Chemistry IAL EDEXCEL A-Level (7 decks)

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