chapter 23

Question 1

carrying out redox titrations- common titrations

Answer 1

• potassium manganate (VII) (KMnO4(aq)) under acidic conditions
• sodium thiosulfate (Na2S2O3(aq)) for determination of iodine

Question 2

Manganate (VIII) titration procedure

Answer 2

1) a standard solution of KMnO4 added to the burette
2)using a pipette, add a measured volume of the solution being analysed to a conical flask. Excess of dilute sulfuric acid to provide the H+ ions fro reduction of MnO4-
3)manganate solution is decolourised as it is added, end point= first permanent pink colour
4) repeat until you obtain concordant titres
read from top of meniscus

Question 3

iodine/ thiosulfate titration equations

Answer 3

thiosulfate ions are oxidised and Iodine is reduced

oxidation: 2S2O3 2- (aq) –> S4O6 2- + 2e-
reduction: I2 (aq) + 2e- –> 2I-

overall: 2S2O3 2- (aq) + I2 (aq) + S4O6 2-(aq)

starch can be used to find the end point- blue/black colour disappears

Question 4

half cells

Answer 4

contains the chemical species present in a redox half equation. voltaic cell can be made by joining two half cells.

Question 5

Metal/metal ion half-cells

Answer 5

consists of a metal rod dipped into a solution of its aq metal ion- represented by vertical line eg Zn2+(aq)|Zn(s)

Question 6

Ion/ion half cells

Answer 6

contains ions of the same element in different oxidation states eg Fe3+(aq) + e- –> Fe2+ (aq)
platinum metal electrode

Question 7

how do you know which electrode has a greater tendency to gain or lose electrons?

Answer 7

in an operating cell:

• the electrode w/ more reactive metal loses electrons and is oxidised- negative electrode
• the electrode w/ the less reactive metal gains electrons and is reduced- positive electrode

Question 8

standard electrode potential

Answer 8

• the tendency to be reduced and gain electrons is measured as standard electrode potential.
• the standard is a half-cell containing hydrogen gas and a solution containing H+ ions. platinum electrode is used to allow electrons in/out the half cell
• 298K, 100KPa (1 bar), conc 1 moldm-3
• the standard potential of hydrogen electrode is 0V

Question 9

measuring a standard electrode potential

Answer 9

• to measure a standard electrode potential the half-cell is connected to a standard hydrogen electrode
  -2 electrodes are connected by a wire to allow a controlled flow of electrons
• 2 solutions connected by a salt bridge which allows ions to flow. typically contains a conc solution of an electrolyte that does not react w/ either solution. eg strip of filter paper soaked in aq KNO3
  equilibrium always shows the forward reaction is reduction

Question 10

the more negative the E value

Answer 10

• greater the tendency to lose electrons and undergo oxidation
• the less the tendency to gain electrons and undergo reduction
• reducing agent

Question 11

the more positive the E value

Answer 11

• greater the tendency to gain electrons and undergo reduction
• less tendency to lose electrons and undergo oxidation
• oxidising agent

Question 12

e value summary

Answer 12

• the more negative the E value , the greater the reactivity of a metal in losing electrons
• the more positive the E value, the greater the reactivity of a non-metal on gaining electrons

Question 13

E cell =

Answer 13

E(positive electrode) - E(negative electrode)

Question 14

predicting redox reactions

Answer 14

• the most negative system has the greatest tendency to be oxidised and lose electrons
• the most positive system has the greatest tendency to be reduced and gain electrons
  you can predict the feasibility of a reaction from E values

Question 15

limitations of predictions using E values- reaction rate

Answer 15

• reactions w/ v large Ea- slow rate

- electrode potentials give no indication of rate of a reaction

Question 16

limitations of predictions using E values- concentration

Answer 16

• if the conc of a solution is not 1 moldm-3 then the value will be different from the standard electrode potential
• eg Zn2+(aq) + 2e- –> Zn(s), if the conc of Zn2+ is greater than 1 moldm-3, the equilibrium will shift to the right, removing electrons from the system and making the potential less negative
• any change to electrode potential will affect the value of the overall cell potential

Question 17

limitations of using E values- other factors

Answer 17

• conditions may be different from standard

- the electrode potentials apply to aq equilibria, many reactions are not aqueous

Question 18

primary cells

Answer 18

• non-rechargeable, designed to be used once only
• electrical energy is produced by oxidation and reduction at the electrodes
• reactions cannot be reversed, chemicals will be used up & battery will go flat
• eg wall clocks and smoke detectors- low-current, long-storage devices

Question 19

secondary cells

Answer 19

-rechargeable
-cell reaction producing electrical energy can be reversed during recharging, chemicals are regenerated
eg lead-acid car batteries, nickel-cadmium NiCd cells, lithium ion cells.

Question 20

fuel cells

Answer 20

• uses the energy from the reaction w/ oxygen to create a voltage
• fuel & oxygen flow in and products flow out. electrolyte remains in the cell
• do not have to be recharged

Question 21

hydrogen fuel cells- alkali

Answer 21

oxidation: H2(g) + 2OH-(aq) –> 2H2O(l) + 2e-
reduction: 1/2O2(g) + H2O(l) + 2e- –> 2OH-(aq)
overall: H2(g) + 1/2 O2(g) –> H20(l) Ecell= 1.23V

Question 22

hydrogen fuel cells- acid

Answer 22

oxidation: H2(g) –> 2H+(aq) + 2e-
reduction: 1/2O2(g) + 2H+(aq) + 2e- –> H2O(l)
overall: H2(g) + 1/2O2 –> H2O(l) Ecell= 1.23V