Topic 14 Redox II

Question 1

Standard redox potential

Answer 1

Standard electrode potential

Question 2

Convention: how are half-equations written?

Answer 2

With the electrons on the left.

Question 3

Absolute potential difference

Answer 3

The potential difference between a metal electrode and its solution. We cannot measure this without using a reference electrode.

Question 4

What happens when metals such as Mg or Cu are placed in water?

Answer 4

Small tendency to lose electrons: M(s) –> M^n+ (aq) + ne^-
These electrons build up on the surface of the metal and attract positive ions. A layer of positive ions surrounds the metal. Some of the ions regain their electrons and become part of the metal surface.
M^n+ (aq) + ne^- –> M(s)
When these reactions occur at the same rate, a dynamic equilibrium is formed:
M^n+ (aq) + ne^- <–> M(s)

Question 5

What is the difference between Mg^2+ (aq) + 2e^- <–> Mg(s) and Cu^2+ (aq) + 2e^- <–> Cu(s)?

Answer 5

Mg has a greater tendency to lose e- and form Mg2+ ions in solution, so its equilibrium position will be shifted further left than for Cu.
The potential difference is greater with Mg than Cu.

Question 6

Electromotive force (emf)

Answer 6

The standard electrode potential of a half-cell connected to a standard hydrogen electrode under standard conditions: 298K, 100kPa and concentrations of 1 mol dm^-3).

Question 7

Why is the surface of a platinum foil electrode covered in porous platinum?

Answer 7

It has a large surface area, so allows an equilibrium to be established quickly.

Question 8

The standard hydrogen electrode

Answer 8

H+ (aq) + e- <–> 1/2H2(g)
Hydrogen gas is bubbled over a piece of Pt foil dipped in a solution of HCl (or H2SO4). Under standard conditions.

Question 9

Standard conditions (for half-cells)

Answer 9

Gas pressure: 100kPa.
Temperature: 298K.
Concentration of ions in solution: 1 mol dm^-3.

Question 10

Why is a salt bridge needed?

Answer 10

To complete the circuit.

Question 11

How does a salt bridge work?

Answer 11

It contains a concentration solution of KNO3 as a liquid or a gel that allows the movement of ions. The ions present should not interfere with the components of the half-cells.

Question 12

Why is a high-resistance voltmeter used in electrochemical cells?

Answer 12

Ideally infinite resistance, so there would be no flow of electrons around the external circuit (no current flowing). This would mean the voltmeter reading represents the potential difference between the half-cells when both are in equilibrium.
(Even better would be to use a potentiometer, but that’s not practical.)

Question 13

Negative standard electrode potential

Answer 13

The metal electrode is negative w.r.t. the SHE.

Question 14

Positive standard electrode potential

Answer 14

The metal electrode is positive w.r.t. the SHE.

Question 15

Standard electrode potential is a sign invariant quantity.

Answer 15

However the reaction is represented, the sign doesn’t change.

Question 16

The more negative the standard electrode potential of a half-cell…

Answer 16

… the further the equilibrium lies to the left relative to the SHE, i.e., the more readily the metal loses e- to form ions and the better reducing agent it is.

Question 17

The more positive/less negative the standard electrode potential of a half-cell…

Answer 17

… the further the equilibrium lies to the right relative to the SHE, the less readily the metal loses e- to form ions and the worse a reducing agent it is.

Question 18

Standard electrode potential of a half-cell

Answer 18

The emf of a half-cell containing the half-cell connected to the standard hydrogen electrode under standard conditions of 298K, 100kPa and 1 mol dm^-3 solutions.

Question 19

Why can’t the standard electrode potentials of Li, Ca, K, Na or F be determined experimentally?

Answer 19

The react with water to completion, i.e., they don’t form an equilibrium, so E values are calculated using thermodynamic data.

Question 20

How can we measure the standard electrode potential of a half-cell involving gases? e.g., for Cl-|1/2Cl2(g)?

Answer 20

Bubble the chlorine gas into a solution containing Cl- ions and use a Pt electrode (comprised of a Pt wire and Pt foil).

Question 21

How can we measure the standard electrode potential of a half-cell involving a non-metal element and their ions in solution? e.g., for 1/2Br2|Br^-?

Answer 21

A half-cell containing a solution of Br2 & Br- ions each of concentration 1 mol dm^-3 is connected to a SHE.

Question 22

How can we measure the standard electrode potential of a half-cell involving ions of the same element with different oxidation numbers, e.g., Fe3+ (aq) | Fe2+ (aq)?

Answer 22

Pt electrode in a beaker of Fe2+ & Fe3+ ions connected to a SHE under standard conditions.

Question 23

In the electrochemical series, species on the RHS of the half-cell equations:

Answer 23

Reducing agents as they can lose electrons.

Question 24

In the electrochemical series, what is the most powerful reducing agent and why?

Answer 24

Species in the top (right). It’s redox system has the most negative electrode potential, so the equilibrium position of its half-cell reaction is furthest left.

Question 25

In the electrochemical series, what is the least powerful reducing agent and why?

Answer 25

Bottom (right) species as its redox system has the least negative/most positive standard electrode potential value. The position of equilibrium of the half-cell is furthest to the right.

Question 26

In the electrochemical series, species on the LHS of the half-cell equations:

Answer 26

Capable of acting as oxidising agents as they can gain electrons.

Question 27

An electrochemical cell

Answer 27

A device for producing an electric current from chemical reactions, constructed from two half-cells.

Question 28

E cell equation

Answer 28

E cell = E right - E left
Under standard conditions.
Don’t change the sign of the left-hand half-cell, even when the reaction is written as an oxidation.

Question 29

Consider a zinc-copper electrochemical cell.

Answer 29

The Zn2+| Zn half-cell has a more negative E, so the Zn will be the negative electrode and electrons will flow from the Zn to the Cu.

Question 30

In a cell diagram, what does | represent?

Answer 30

A phase boundary, e.g., Al3+ (aq) | Al (s) or Pt(s) | Fe3+ (aq), Fe2+ (aq) or H+(aq) | 1/2H2(g) | Pt (s).

Question 31

|| in a cell diagram represents…

Answer 31

…the salt bridge.

Question 32

Rules of a cell diagram

Answer 32

• The positive electrode is shown on the RHS.
• The 2 reduced forms of the species are shown on the outside of the cell diagram.
• Commas separate species where there is no phase boundary.
  Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)

Question 33

When do we break cell diagram convention?

Answer 33

When measuring standard electrode potential. The SHE is always on the RHS.
Pt (s) | 1/2H2 (g) | H+ (aq) || Zn2+ (aq) | Zn (s)

Question 34

Even if standard electrode potentials indicate a reaction is thermodynamically feasible, why might it not take place?

Answer 34

The reactants may be kinetically stable because the activation energy for the reaction is very large and the reaction may not be taking place under standard conditions.

Question 35

Will zinc react with dilute sulfuric acid?

Answer 35

Yes!
Zn (s) + 2H+ –> Zn2+ (aq) + H2(g)
Zn2+ + 2e- <–> Zn(s) E= -0.76V
2H+ + 2e- <–> H2(g) E= 0.00V
The Zn half equation has a more negative E value, so the equilibrium moves to the left. The H2 half-equation has a less negative E value, so the equilibrium moves to the right. This means the reaction is thermodynamically feasible as zinc will release electrons to the H+ ions.

Question 36

Will manganese (IV) oxide react with hydrochloric acid?
MnO2 (s) + 4HCl (aq) –> Mn^2+ (aq) + 2Cl^-(aq) + 2H2O(l) + Cl2(g)

Answer 36

MnO2 (s) + 4H+ (aq) + 2e- <–> Mn2+ (aq) + 2H2O (l) +1.23V
Cl2(g) + 2e- <–> 2Cl-(aq) +1.36V
The 2nd E value is more positive, so Cl- ions cannot release e- to MnO2. The reaction is not feasible under standard conditions.
But…
If 10 mol dm^-3 HCl is used, the concentration of H+ & Cl- ions increases. This shifts the top equilibrium right and the bottom equilibrium left, so that the Cl- ions can release e- to MnO2.
The reaction is feasible under non-standard conditions.

Question 37

Shifting an equilibrium right… (Redox II context.)

Answer 37

The electrode potential becomes more positive because the redox system is a better electron acceptor.

Question 38

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

Answer 38

Changing the conditions may alter the electrode potential of a half-cell by changing the position of equilibrium.

Question 39

Disproportionation of copper: 2Cu^+ (aq) –> Cu^2+ (aq) + Cu(s).

Answer 39

Cu2+ + e- <–> Cu+ +0.15V
Cu+ + e- <–> Cu(s) +0.52V
The electrode potential for the top equilibrium is more negative than for the bottom equilibrium, so the equilibrium position moves left.
Cu+ ions both release and accept electrons to form Cu2+ + Cu species. Thus, Cu+ ions disproportionate into Cu atoms & Cu2+ ions.

Question 40

Relationship between Gibbs & E cell:

Answer 40

ΔG° = -nFE°cell
where n is the number of moles of electrons involved in the cell reaction and F is the Faraday constant.

Question 41

ΔG° = -TΔS°total
How?

Answer 41

ΔG° = ΔH° - TΔS°
ΔS°surroundings = -ΔH° / T
ΔH° = -TΔS°surroundings
ΔG° = -TΔS°surroundings -TΔS°system= -TΔS°total

Question 42

How does E cell relate to total entropy?

Answer 42

TΔS°total = nFE°cell
ΔS°total is proportional to E°cell.

Question 43

How can E°cell be used to predict the direction of a cell reaction (using knowledge of entropy)?

Answer 43

If E°cell is positive, the reaction will occur as written from left to right in the cell diagram. It is thermodynamically feasible as ΔS°total will be positive.
If E°cell is negative, the reaction will occur from right to left in the half-cell.

Question 44

How can E°cell be used to calculate the equilibrium constant, K, for a reaction?

Answer 44

ΔG° = -RTlnK
ΔG° = -nFE°cell
RTlnK= nFE°cell
ln K = nFE°cell /RT
n, F & R are constants, so at a given temperature, ln K is proportional to E°cell.

Question 45

Storage cell/secondary cell

Answer 45

A cell that can be recharged by passing a current through it in the opposite direction to the flow of current generated by the cell.

Question 46

How is a storage cell recharged?

Answer 46

An external potential difference is applied that reverses the reactions.

Question 47

Common storage cells:

Answer 47

Ag-Zn cell or nickel metal hydride cells (NiMH).

Question 48

How do fuel cells produce a voltage?

Answer 48

The reaction of a fuel with oxygen. Fuels tend to be hydrogen, methanol or ethanol.

Question 49

Example of a storage cell: NiCd.
NiO(OH) is the anode, & Cd is the cathode.
KOH is the electrolyte.

Answer 49

Cd(OH)2 (s) + 2e- <–> Cd(s) + 2OH-(aq) -0.88V
NiO(OH) (s) + H2O (l) + e- <–> Ni(OH)2 (s) + OH- (aq) +0.52V
The upper equilibrium shifts to the left as it is more negative.
Overall: Cd(s) + 2NiO(OH) (s) + 2H2O(l) –> Cd(OH)2 (s) + Ni(OH)2 (s)

Question 50

How does a hydrogen-oxygen fuel cell work?

Answer 50

The metal electrodes are coated in a Pt catalyst. Acidic electrolyte.
At the cathode: H2(g) –> 2H+(aq) + 2e-
At the anode: 1/2O2(g) + 2H+(aq) + 2e- –> H2O(l)
The protons pass through the proton exchange membrane, so they can react with the oxygen. The H+ ions move across a polymer electrolyte membrane.
Overall: H2(g) + 1/2O2(g) –> H2O (l)

Question 51

How are the reactions different in an alkaline hydrogen-oxygen fuel cell?

Answer 51

At the cathode, hydrogen is fed in: H2(g) + 2OH-(aq) –> 2H2O(l) + 2e-
At the anode, oxygen is fed in: O2(g) + 2H2O(aq) + 4e- –> 4OH-(l)
The electrolyte is KOH(aq). Ion exchange membrane line the Pt(s) electrodes and allow these OH- ions to pass through, but not H2(g) or O2(g).
The overall reaction is the same as in acidic conditions: H2(g) + 1/2O2(g) –> H2O (l)

Question 52

Advantages of hydrogen-oxygen fuel cells.

Answer 52

• An alternative to the direct use of fossil fuels, e.g., petrol.
• Products don’t include pollutants.
• Engines are lighter and more efficient than those that use fossil fuels.

Question 53

Disadvantages of hydrogen oxygen fuel cells.

Answer 53

• Supply of hydrogen: most is produced from methane– a finite resource. Producing hydrogen using renewables tends to be expensive, but with electrolysis it is becoming increasingly cheaper.
• Hydrogen is explosive, so hard to transport.

Question 54

Transporting hydrogen by compressing it as a gas.

Answer 54

Safety issues, especially when transporting under pressure.

Question 55

Transporting hydrogen by adsorbing it onto the surface of a solid.

Answer 55

Many metals adsorb onto hydrogen. Carbon nanotubes could be used to adsorb hydrogen.

Question 56

Transporting hydrogen by absorbing it into a material.

Answer 56

Many metal hydrides absorb hydrogen, which can be released under the correct conditions. High temperatures are needed to release hydrogen, however.

Question 57

Potassium manganate(VII) in redox titrations.

Answer 57

Used for the quantitative estimation of many reducing agents, such as iron(II) compounds & ethanedioic acid & its salts, as it is a powerful oxidising agent.

Question 58

Why is KMnO4 used under acidic conditions in redox titrations?

Answer 58

To prevent the formation of the manganese (IV) oxide brown precipitate that forms under alkaline conditions. This would interfere with the endpoint colour.

Question 59

When is the endpoint in a redox titration involving KMnO4?

Answer 59

The first permanent pink colour. Colourless to pink. KMnO4 acts as its own indicator. Occurs as soon as the KMnO4 is in excess.

Question 60

Titration of KMnO4 with Fe2+ ions:

Answer 60

Fe2+ ions are oxidised, & manganate(VII) ions are reduced.
Overall equation: MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) –> Mn2+ (aq) + 4H2O(l) + 5Fe3+(aq).
Pipette 25 cm^3 of Fe2+ into a conical flask. Add a small volume of dilute sulfuric acid.
Slowly add KMnO4 solution of a known concentration from a burette, and swirl.
Endpoint: colourless to pale pink.

Question 61

Fe2+ (aq) = pale green.
Fe3+ (aq) = yellow.
Mn2+ (aq) = pale pink.
Why in redox titrations do these solutions appear colourless?

Answer 61

The solutions in titrations are very dilute.

Question 62

Why is dilute sulfuric acid boiled beforehand in a redox titration with KMnO4 + Fe2+?

Answer 62

This removes dissolved oxygen that might oxidise Fe2+ ions to Fe3+ ions.

Question 63

Titration of ethanedioic acid with KMnO4.

Answer 63

Overall: 2MnO4- (aq) + 6H+ (aq) + 5H2C2O4 (aq) –> 2Mn2+ (aq) + 8H2O(l) + 10CO2(g)
Aq ethanedioic acid is pipetted into a conical flask with a pipette filler. Then acidified with dilute sulfuric acid. Aq KMnO4 is added from the burette. The reaction is autocatalysed by the Mn2+ ions, so it starts slowly, hence heat the reaction to 60°C first.

Question 64

Redox titrations with iodine + sodium thiosulfate: how does it work?

Answer 64

Overall: 2S2O3^2- + I2 –> S4O6^2- + 2I- (All aqueous.)
- Add sodium thiosulfate 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 to produce a blue/black colour.
- Add the sodium thiosulfate solution drop by drop until the blue-black solution turns colourless.

Question 65

Redox titrations with iodine + sodium thiosulfate: what happens if starch is added too early?

Answer 65

It adsorbs some of the iodine, and reduces the accuracy of the titration.

Question 66

Redox titrations with iodine + sodium thiosulfate: why do we need the starch?

Answer 66

To see the endpoint more clearly. the colour of the iodine becomes faint towards the endpoint, so it’s difficult to accurately assess when the endpoint has been obtained.

Question 67

The larger the cell potential, the larger the equilibrium constant.

Answer 67

E 0 is directly proportional to delta S total, so E0 is also directly proportional to ln K.

Question 68

How would you determine the mass of copper in an alloy?

Answer 68

1. Dissolve a known mass of alloy in conc. HNO3 (aq), and pour into a 250 cm^3 conical flask. Make up to 250 cm^3 with deionized water.
2. Pipette 25 cm^3 of the solution into a flask. Add Na2CO3 (aq) to neutralize any residual nitric acid. Keep adding until you just start to see a precipitate form. Then add several drops of ethanoic acid to remove this precipitate.
3. Add excess acidified KI(aq), which will react with the Cu2+ ions.
4. Titrate the mixture against Na2S2O3 (aq).

Question 69

Titration errors: Cu2+ + KI, then the resulting mixture is titrated with Na2S2O3 (aq).

Answer 69

If most of the I2 has not yet reacted when the starch is added, it will take a long time for the blue colour to disappear.
The CuI precipitate makes it difficult to see the colour of the solution.
Keep the solution as cool as possible. I2 produced in the reaction can evaporate readily at RTP, which could lead to a false titre and the final copper % will be too low.
Make the starch solution only when you’re ready to use it.

Question 70

Electrons flow from…

Answer 70

…a more reactive metal to a less reactive metal.

Question 71

If there is an oxidising agent in the half-cell containing oxygen…

Answer 71

Add acid too!

Question 72

To get 1 mol dm^-3 of H+ ions in the half-cell, what could you do?

Answer 72

Use 1 mol dm^-3 of HCl (aq) or 0.5 mol dm^-3 of H2SO4 (diprotic).

Question 73

Ethanol or methanol fuel cell

Answer 73

The alcohol is oxidised at the anode in the presence of water:
CH3OH + H2O –> CO2 + 6e- + 6H+
H+ ions pass through the electrolyte and are oxidised to water:
6H+ + 6e- + 3/2O2 –> 3H2O