Unit 5- Topic 16

Question 1

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

Answer 1

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

Question 2

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

Answer 2

Reduction is the gain of electrons

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

Question 3

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

Answer 3

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

Question 4

Definition of standard electrode potential

Answer 4

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

Question 5

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

Answer 5

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

These standard conditions apply to all equilibria

Question 6

What reactions are involved in the measurement of electrode potential

Answer 6

All redox reactions and therefore equilibria

Question 7

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

Answer 7

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)

Question 8

Why is the standard hydrogen electrode necessary

Answer 8

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.

Question 9

The conditions for only the SHE

Answer 9

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

Question 10

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

Answer 10

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

Question 11

Why is the salt bridge needed in electrochemical cells

Answer 11

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

Question 12

Why is the platinum foil covered in porous platinum

Answer 12

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

Question 13

Why is a high-resistance voltmeter used

Answer 13

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

Question 14

In electrochemical cells, inn which electrode does oxidation occur

Answer 14

Anode

Question 15

Equation for emf (E cell)

Answer 15

E cell = E rhs - E lhs

Question 16

How to measure E wit a systems involving gases

Answer 16

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

Question 17

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

Answer 17

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.

Question 18

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

Answer 18

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

Question 19

What half cell goes to the left of the electrochemical cells

Answer 19

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

Question 20

What does the vertical solid line mean in a cell diagram

Answer 20

A phase boundary

Question 21

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

Answer 21

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

Question 22

What does double vertical lines represent in cell diagram

Answer 22

Salt bridge

Question 23

Draw the standard hydrogen electrode diagram

Answer 23

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

Question 24

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

Answer 24

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

Question 25

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

Answer 25

Always written in the left-hand side

Question 26

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

Answer 26

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

Question 27

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

Answer 27

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

Question 28

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

Answer 28

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

Question 29

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

Answer 29

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.

Question 30

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

Answer 30

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

Question 31

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

Answer 31

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

Question 32

What is the relationship between the equilibrium constant and Ecell

Answer 32

Since change Stotal = RlnK and change in Stotal= nFEcell / T nFEcell lnK=____________ RT

Question 33

What other name can standard electrode potentials be referred as

Answer 33

Standard reduction potentials

Question 34

How is the electrochemical series ordered

Answer 34

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

Question 35

Where are reducing agents located in the electrochemical series

Answer 35

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

Question 36

Where are oxidising agents located in the electrochemical series

Answer 36

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

Question 37

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

Answer 37

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

Question 38

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

Answer 38

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

Question 39

What is a disproportionation reaction

Answer 39

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

Question 40

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

Answer 40

Equilibrium will move to the left, releasing electrons

Question 41

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

Answer 41

Equilibrium will move to the right, accepting electrons

Question 42

How to calculate percentage uncertainty

Answer 42

uncertainty Percentage uncertainty =_______________ x 100 measured value

Question 43

How to reduce percentage uncertainty in brunette readings

Answer 43

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

Question 44

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

Answer 44

A powerful oxidising agent

Question 45

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

Answer 45

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

Question 46

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

Answer 46

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

Question 47

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

Answer 47

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

Question 48

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

Answer 48

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

Question 49

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

Answer 49

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

Question 50

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

Answer 50

-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

Question 51

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

Answer 51

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

Question 52

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

Answer 52

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

Question 53

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

Answer 53

-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.

Question 54

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

Answer 54

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

Question 55

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

Answer 55

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

Question 56

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

Answer 56

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

Question 57

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

Answer 57

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

Question 58

Overall equation for the redox titration of iodine and sodium thiosulfate

Answer 58

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

Question 59

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

Answer 59

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

Question 60

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

Answer 60

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

Question 61

Method of the redox titration of sodium thiosulfate and iodine

Answer 61

-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

Question 62

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

Answer 62

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

Question 63

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

Answer 63

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

Question 64

What effect does increasing the concentration have in the emf

Answer 64

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

Question 65

What effect does increasing the pressure have in the emf

Answer 65

More negative as more electrons are produced

Question 66

Which electrode is the cathode in a fuel cell

Answer 66

The positive electrode

Question 67

Which electrode is the anode in a fuel cell

Answer 67

The negative electrode

Question 68

What is a fuel cell

Answer 68

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.

Question 69

What fuels are used in fuel cells

Answer 69

Hydrogen, methanol/ethanol, hydrogen-rich methanol

Question 70

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

Answer 70

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

Question 71

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

Answer 71

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

Question 72

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

Answer 72

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

Question 73

Overall reaction for fuel cells in both conditions

Answer 73

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

Question 74

Overall reaction for fuel cells in both conditions

Answer 74

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

Question 75

What happens in a hydrogen-oxygen fuel cell

Answer 75

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.

Question 76

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

Answer 76

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

Question 77

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

Answer 77

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

Question 78

Advantages of hydrogen-oxygen fuel cells

Answer 78

-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

Question 79

Disadvantages of hydrogen-oxygen fuel cells

Answer 79

-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.