Chapter 23 - Redox and electrode potentials

Question 1

What species will always be in a redox reaction?

Answer 1

In a redox reaction, there will always be an oxidising agent and a reducing agent.

Question 2

What is an oxidising agent and a reducing agent?

Answer 2

a. An oxidising agent takes electrons from the species being oxidised and the oxidising agent contains the species that is reduced.
b. A reducing agent adds electrons to the species being reduced and the reducing agent contains the species that is oxidised.

Question 3

How would you write a half equation in alkaline conditions?

Answer 3

a. Firstly, work out the change in oxidation number which is equal to the number of electrons to be placed on the side of the equation with the greatest oxidation state.
b. After the electrons have been placed, balance the charge of the equation using OH- ions.
c. Then add H20 molecules to balance the oxygen atoms in the remaining equation.

Question 4

How would you write a half equation in acidic conditions?

Answer 4

a. Balance the atoms on either side of the equation using H20 molecules and H+ ions only.
b. Then balance the overall charges on either side of the equation by using electrons.

Question 5

How do Manganate (VII) titrations work?

Answer 5

• In Manganate (VII) [permanganate] titrations, MnO4- (aq) ions are reduced to Mn2+ ions and so the other chemical used must be a reducing agent that is oxidised.
  a. A standard solution of known concentration of potassium manganate (VII) is added to the burette.
  b. Using a pipette, add a measured volume of the solution being analysed to the conical flask. An excess of dilute sulfuric acid is also added to provide the H+(aq) ions required for the reduction of MnO4- (aq) ions. You do not need an indicator, as the reaction is self-indicating.
  c. During the titration, the manganate (VII) solution reacts and is decolourised as it is being added. The end point of the titration is judged by the first permanent pink colour, indicating when there is an excess of MnO4- ions present.
  d. Repeat the titration until you obtain concordant titres (two titres that agree within 0.10 cm^3)

Question 6

How do you read the meniscus on Manganate (VII) titrations?

Answer 6

KMnO4 (aq) is a deep purple colour and is very difficult to see the bottom of the meniscus through the intense colour. So, burette reading are read from the top rather than the bottom.

Question 7

How do you calculate the percentage purity of a compound?

Answer 7

Percentage purity = (mass worked out from the experiment [pure sample])/(mass of impure sample) x 100

Question 8

How can the permanganate titration be extended to non-familiar redox titrations?

Answer 8

a. Permanganate titrations can be used to analyse other reducing agents that reduce MnO4- to Mn2+.
b. KMnO4 can be replaced with other oxidising agents, the commonest used being acidified dichromate (VI), H+ / Cr2072-.

Question 9

How do iodine-thiosulfate titrations work and what can they be used to determine?

Answer 9

• Thiosulfate is oxidised from S2O32- to S4O62- and Iodine is reduced from I2 to I-.
• Iodine/thiosulfate titrations can be used to determine: The ClO- (oxidising agent) content in household beach, the Cu2+ (oxidising agent) content in copper (II) compounds and the Cu (oxidising agent) content in copper alloys.
  a. Add a standard solution of Na2S2O3 to the burette.
  b. Prepare a solution of the oxidising agent to be analysed. Using a pipette, add this solution to a conical flask. Then add an excess of potassium iodide. The oxidising agent reacts with iodide ions to produce iodine, which turns the solution a yellow-brown colour.
  c. Titrate the solution with the Na2S2O3 (aq). During the titration, the iodine is reduced back to I- ions and the brown colour fades quite gradually, when the iodine colour has faded enough to become a pale straw colour, add a small amount of starch indicator. A deep blue-black colour forms to assist with the identification of the end point. As more sodium thiosulfate is added, the blue-black colour fades. At the end point, all the iodine will have just reacted, and the blue-black colour disappears.

Question 10

What does it mean when the sulfur in the tetrathionate ion has a decimal oxidation state?

Answer 10

It just means that the sulfurs within the tetrathionate ion have different oxidation states.

Question 11

In the iodine/thiosulfate reaction why is it important to only add the starch indicator towards the end of an iodine-thiosulfate titration?

Answer 11

a. The blue-black colour is a complex of starch and iodine. If added too early, the starch-iodine complex may precipitate out of solution, preventing some of the iodine reacting with the thiosulfate.
b. Close to the end point, the iodine concentration is low enough for the complex not to precipitate out.

Question 12

Can the iodine-thiosulfate titration be extended to other non-familiar redox reactions?

Answer 12

The same principles can be applied for the analysis of other oxidising agents, provided that they are capable of oxidising I- ions to I2.

Question 13

What are the ClO- ions commonly known as?

Answer 13

They are commonly known as hypochlorite.

Question 14

How is the Cu2+ content in copper (II) compounds determined?

Answer 14

a. For copper (II) salts, Cu2+ (aq) ions are produced simply by dissolving the compound in water.
b. Insoluble copper (II) compounds can be reacted with acids to form Cu2+(aq) ions.
c. Cu2+ ions react with I-(aq) to form a solution of Iodine, I2(aq) and a white precipitate of Copper(I) iodide, CuI(s). The mixture appears as a brown colour.
d. The iodine in the brown mixture is then titrated with a standard solution of sodium thiosulfate.

Question 15

How is the Cu content in copper alloys determined?

Answer 15

a. For copper alloys, such as brass or bronze, the alloy is reacted and dissolved in concentrated nitric acid, followed by neutralisation to form Cu2+(aq) ions.
b. Cu2+ ions react with I-(aq) to form a solution of Iodine, I2(aq) and a white precipitate of Copper(I) iodide, CuI(s). The mixture appears as a brown colour.
c. The iodine in the brown mixture is then titrated with a standard solution of sodium thiosulfate.

Question 16

How do you calculate the percentage composition in an alloy?

Answer 16

% composition = mass of element/mass of alloy x 100

Question 17

What is a voltaic cell?

Answer 17

a. A voltaic cell is a type of electrochemical cell which converts chemical energy into electrical energy.
b. As electrical energy results from movement of electrons, you need chemical reactions that transfer electrons from one species to another known as redox reactions.
c. A voltaic cell can be made by connecting together two different half cells which allow electrons to flow. A half-cell contains the chemical species present in a redox half-equation. The two half cells must be kept apart as if allowed to mix, electrons would flow in an uncontrolled way and heat energy would be released rather than electrical energy.

Question 18

What is a metal/metal ion half-cell?

Answer 18

a. A metal/metal ion half-cell consists of a metal rod dipped into a solution of its aqueous metal ion. This is represented using a vertical line for the phase boundary between the aqueous solution and the metal.

b. At the phase boundary where the metal is in contact with its ions, an equilibrium will be set up.
i. By convention, the equilibrium in a half cell is written so that the forward reaction shows reduction and the reverse reaction shows oxidation.

c. When two half cells are connected, the direction of electron flow depends upon the relative tendency of each electrode to release electrons.

Question 19

What is an ion/ion half-cell?

Answer 19

a. An ion/ion half-cell contains ions of the same element in different oxidation states.
b. In this type of half-cell there is no metal to transport electrons either into or out of the half-cell, so an inert metal electrode made out of platinum is used.

Question 20

What is known as the standard electrode potential?

Answer 20

a. The standard electrode potential is the e.m.f. of a half-cell connected to a standard hydrogen half-cell under standard conditions of 298K, solution concentrations of 1 mol dm^-3, and a pressure of 100kPa.
b. The tendency to be reduced and gain electrons is measured as a standard electrode potential.

Question 21

What is a standard hydrogen electrode?

Answer 21

a. A half-cell containing hydrogen gas, H2(g), and a solution containing H+ ions. An inert platinum electrode is used to allow electrons into and out of the half-cell. This is known as a standard hydrogen electrode.
b. By definition, the standard electrode potential of a standard hydrogen electrode is 0 V.
c. The sign of a standard electrode potential shows the sign of the half-cell connected to the standard hydrogen electrode and shows the relative tendency to gain electrons compared with the hydrogen half-cell.

Question 22

What are the standard conditions used in a standard hydrogen half-cell?

Answer 22

a. Solutions have a concentration of exactly 1 mol dm^-3.
b. The temperature is 298K (25 degrees celsius).
c. The pressure is 100kPa (1 bar).

Question 23

How do you measure a standard electrode potential?

Answer 23

• To measure a standard electrode potential, the half-cell is connected to a standard hydrogen electrode.
  a. The two electrodes are connected by a wire to allow a controlled flow of electrons.
  b. 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.
  c. One such example of a salt bridge is a strip of filter paper soaked in aqueous potassium nitrate, KNO3 (aq)

Question 24

Within a voltaic cell, what does the standard electrode potential tell us?

Answer 24

a. The more negative the E value:
i. The greater the tendency to lose electrons and undergo oxidation.
ii. The less the tendency to gain electrons and undergo reduction.
iii. In general, the more negative the E value, the greater the reactivity of a metal in losing electrons.

b. The more positive the E value:
i. The greater the tendency to gain electrons and undergo reduction.
ii. The less the tendency to lose electrons and undergo oxidation.
iii. In general, the more positive the E value, the greater the reactivity of a non-metal in gaining electrons

Question 25

How can a voltaic cell be set up to measure standard cell potentials?

Answer 25

a. Prepare two standard half cells.
i. For a metal/metal ion half-cell, the metal ion must have a concentration of 1 mol dm^-3.

ii. For an ion/ion half-cell, both metal ions present in the solution must have the same concentration. There must be an inert electrode, usually platinum. (In an ion/ion half-cell, both solutions can be 1 mol dm^-3, but it may be difficult to dissolve enough solute to get these concentrations. However equal ion concentrations give the same e.m.f.).
iii. For a half-cell containing gases (e.g. a hydrogen half-cell), the gas must be at 100kPa pressure, in contact with a solution with an ionic concentration of 1 mol dm^-3. There must be an inert electrode, usually platinum.
iv. For all half-cells, the temperature must be at 298 K.
b. Connect the metal electrodes of the half-cells to a voltmeter using wires.
c. Prepare a salt bridge by soaking a strip of filter paper in a saturated aqueous solution of potassium nitrate, KNO3.
d. Connect the two solutions of the half-cells with a salt bridge.
e. Record the standard cell potential from the voltmeter.
f. The cell that has the more positive E, has a greater tendency to undergo reduction and gain electrons, so is the positive electrode.
g. The cell that has the more negative E, has a greater tendency to undergo oxidation and lose electrons, so is the negative electrode.

Question 26

How can you calculate the standard cell potential?

Answer 26

a. Ecell = E (positive electrode) – E (negative electrode).
b. A standard cell potential can be calculated directly from standard electrode potentials using the standard electrode potentials given on a data sheet.

Question 27

In a redox system which are the oxidising agents and reducing agents?

Answer 27

a. An oxidising agent takes electrons away from the species being oxidised. So oxidising agents are reduced and are on the left.
b. A reducing agent adds electrons to the species being reduced. So, reducing agents are oxidised and are on the right.

Question 28

How can you predict the feasibility of redox reactions within redox systems using the E values?

Answer 28

a. A reaction should take place between an oxidising agent on the left and a reducing agent on the right, provided that the redox system of the oxidising agent has a more positive E value than the redox system of the reducing agent.
i. The strongest reducing agent is at the top on the right of a redox system.
ii. The strongest oxidising agent is at the bottom on the left of a redox system.

b. For example, redox system C has a more positive E value and will have a greater tendency to be reduced than redox system A and B. So, the oxidising agent Ag+(aq) of redox system C should react with reducing agents on the right in redox systems A and B, that is, with Cr(s) and Cu(s).
c. Another example, redox system B has a more positive E value and will have a greater tendency to be reduced than redox system A. So, the oxidising agent Cu2+(aq) of redox system B should react with reducing agents on the right in redox system A, that is, only with Cr(s).

Question 29

What are the limitations of predicting the feasibility of redox reactions using E values?

Answer 29

• Reaction Rates:
  o One limitation of predictions for feasibility based on ΔG lies with reactions that have a very large reaction rate, resulting in a very slow rate. This same limitation is applied to predictions made based on E values. So, electrode potentials may indicate thermodynamic feasibility but give no indication on the rate of a reaction.
• Concentration:
  o Standard electrode potentials are measured using concentrations of 1 mol dm^-3. Many reactions take place using concentrated or dilute solutions. If the concentration of a solution is not 1 mol dm^-3, then the value of the electrode potential will be different from the standard value.
  o For example, for Zn2+(aq) + 2e- <=> Zn (s) E= -0.76 V
   If the concentration of Zn2+(aq) is greater than 1 mol dm^-3, the equilibrium will shift to the right, removing electrons from the system and making the electrode potential less negative.
   If the concentration of Zn2+(aq) is less than 1 mol dm^-3, the equilibrium will shift to the left increasing electrons in the system and making the electrode potential more negative.
   Any change to the electrode potential will affect the value of the overall cell potential.
• Other factors:
  o The actual conditions used for the reaction may be different from the standard conditions used to record E values. This will affect the value of the electrode potential.

o Standard electrode potentials apply to aqueous equilibria. Many reactions take place that are not aqueous.

Question 30

What are the three types of voltaic cell?

Answer 30

Primary cells
Secondary cells
Fuel cells

Question 31

What are primary cells?

Answer 31

a. Primary cells are non-rechargeable and are designed to be used only once.
b. Electrical energy is produced by oxidation and reduction reactions at the electrodes. However, the reactions cannot be reversed.
c. Primary cells find use for low-current, long-storage devices such as wall clocks and smoke detectors.
d. Most modern primary cells are alkaline based on zinc and manganese dioxide Zn/MnO2, and a potassium hydroxide alkaline electrolyte, Zinc is oxidised to ZnO and manganese in MnO2 is reduced to manganese in Mn2O3.

Question 32

What are secondary cells?

Answer 32

a. Secondary cells are rechargeable.
b. The cell reaction producing electrical energy can be reversed during recharging.

c. Examples of secondary cells include:
i. Lead-acid batteries used in car batteries

ii. nickel-cadmium, NiCd, cells and nickel-metal hydride NiMH – the cylindrical batteries used in radios, torches and so on.
iii. Lithium-ion and lithium-ion polymer cells used in our modern appliances – laptops, tablets, cameras, mobile phones – are also being developed for cars.

Question 33

What are fuel cells?

Answer 33

a. Fuel cells use the energy from the reaction of a fuel with oxygen to create a voltage.
i. The fuel and oxygen flow into the fuel cell and the products flow out. The electrolyte remains in the cell.

ii. Fuel cells can operate continuously provided that the fuel and oxygen are supplied into the cell.
iii. Fuel cells do not have to be recharged.
b. Various fuels can be used, but hydrogen is the most common.
c. Hydrogen fuel cells produce no carbon dioxide during combustion, with water being the only combustion product.
d. Fuel cells using other hydrogen-rich fuels, such as methanol, are also being developed.

Question 34

What are lithium-ion polymer cells?

Answer 34

a. Lithium is a light metal so translates into a very high energy density when used in lithium ion batteries.
b. Cells can be a regular shape with an internal salt bridge made of a micro porous polymer covered in electrolytic gel.
c. When a lithium-ion cell charges and discharges, electrons move through the connecting wires to power the appliance, whilst Li+ ions move between the electrodes in the cell.

d. Negative electrode made of graphite coated with lithium metal.
i. Li=Li+ + e-

e. Positive electrode made out of a metal oxide, typically CoO2.
i. Li+ + CoO2 + e- = LiCoO2.

f. When fully charged a lithium-ion cell has a voltage of 4.2 V but typical operating voltage is about 3.7-3.8 V.
g. Limitations of lithium-ion cell include becoming unstable at high temperatures and on very rare occasions they have ignited laptops and mobile phones, care must be taken when igniting as lithium is a very reactive metal.

Question 35

How does alkali hydrogen fuel cells work and what redox systems are present?

Answer 35

a. Hydrogen gas fuel comes through left and oxygen gas comes into cell on right, and water leaves on hydrogen side.
b. Electrodes are made of platinum which also acts as catalysts.
c. The redox systems include: 2H20 + 2e- = H2 + 2OH- E=-0.83 V
i. 0.5O2 + H20 + 2e- = 2OH- E=+0.40 V

Question 36

How does acid hydrogen fuel cells work and what redox systems are present?

Answer 36

a. Hydrogen gas fuel comes through left and oxygen gas comes into cell on right, and water leaves on oxygen side.
b. Electrodes are made of platinum which also acts as catalysts.
c. The redox systems include: 2H+ + 2e- = H2 E= 0.00 V
i. 0.5O2 + 2H+ + 2e- = H20 E=1.23 V

Question 37

What is the cell voltage of hydrogen fuel cells?

Answer 37

1.23 V

Question 38

What is the colour of Cr3+ ions?

Answer 38

green

Question 39

What is the colour of Cr6+ ions or CrO42-?

Answer 39

Yellow

Question 40

What are the advantages of using a methanol fuel as opposed to using hydrogen as the fuel?

Answer 40

Methanol is a liquid and methanol is easier to store/transport.

Question 41

What are the charge carriers that transfer current through the wire and through the solution?

Answer 41

Electrons in the wire and ions in the salt bridge.