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A level Chem OCR B > ES - redox *03 > Flashcards

ES - redox *03 Flashcards

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1
Q

how are halogens extracted?

A

by electrolysis of halide solutions

  1. when you electrolyse concentrated aqueous solutions containing halide ions
    - –> halogen element released at anode
  2. the halide ions lose electrons to the electrode and are oxidised to atoms which combine to form molecules
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2
Q

how is chlorine extracted from brine?

A

by electrolysis

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3
Q

what is brine?

A

a solution of water with high concentrations of salts

  • mainly sodium chloride
  • but also some bromine and iodine salts

occurs naturally in salt lakes and seawater, and can be made by dissolving rock salt in water

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4
Q

electrolysis of brine

A

Cathode:
- 2 H+ ions accept two electrons to become one hydrogen molecule
2H+ (aq) + 2e- —-> H2 (g)

Anode:
- 2 Cl- ions lose their electrons and become one chlorine molecule
2Cl- (aq) —-> Cl2 (g) + 2e-

sodium ions and hydroxide ions (from water) are left behind
- NaOH left in solution

electrolysis cell constantly fed with a fresh stream of brine

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5
Q

electrolysis of brine - what are electrode made of?

A

an inert material

  • carbon
  • platinum
  • titanium
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6
Q

electrolysis of brine - why does the NaCl solution need to be concentrated?

A

in dilute solution the chloride ions aren’t discharged —> they hang onto their electrons

the OH- ions lose electrons instead and the products at anode are oxygen and water, not chlorine
4OH- (aq) —-> 4e- + 2H2O (l) + O2 (g)

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7
Q

electrolysis of brine - bromine

A

the brine contains some bromide ions
chlorine is better at attracting electrons than bromine
- so when you bubble chlorine gas through the brine, the chlorine will displace the bromide

2Br- (aq) + Cl2 (g) —-> Br2 (g) + 2Cl- (aq)

the bromine is then collected, condensed into a liquid and purified

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8
Q

electrolysis of brine - iodine

A

the brine contains some iodide ions
chlorine is better at attracting electrons than iodine
- when you bubble chlorine gas through the brine, the chlorine will displace the iodide

2I- (aq) + Cl2 (g) —-> I2 (g) + 2Cl- (aq)

the iodine is then collected, purified and condensed into a grey solid

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9
Q

electrolysis - anode is….

A

positive

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10
Q

electrolysis - cathode is….

A

negative

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11
Q

electrolysis - what

A

breaking a substance down using electricity

  1. if you pass an electric current through an ionic substance that’s molten or in solution it breaks down into its elements
  2. requires a liquid to conduct the electricity
  3. electrolytes contain free ions —> usually the molten or dissolved ionic substances
  4. the free ions conduct electricity
  5. for the circuit to be complete needs to be a flow of electrons
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12
Q

electrolysis - at anode

A

negative anions move to the positive electrode, the anode, and lose electrons

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13
Q

electrolysis - at cathode

A

positive cations more to the negative electrode, the cathode, and gain electrons

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14
Q

electrolysis - of an aqueous solution, procedure

A
  1. use wires and clips to connect each electrode to the power supply
    - the electrode connected to the positive pole will be the anode
    - the electrode connected to the negative pole will be that cathode
  2. usually use inert electrodes (platinum or carbon)
    - so they don’t start reacting and interfering with the electrolysis
  3. place the electrodes in a beaker containing the electrolyte
    - make sure then don’t touch each other
  4. turn the power supply on
  5. depending on the electrolyte the products will form as: metals (thin layer on surface of cathode) or gases (as bubbles at cathode or anode)
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15
Q

electrolysis - what do half equations show

A

whats happening at each electrode
- show the movement of electrons during a reaction

anode: negative ion losing electrons to form atoms
cathode: positive ion gaining electrons to form atoms

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16
Q

electrolysis - in molten compounds

A

only one source of ions

the substance just breaks into its elements

the positive cation forms at the negative cathode

the negative anion forms at the positive anode

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17
Q

electrolysis - in aqueous solutions

A

have H+ and OH- ions from the water as well as the ions in the ionic compound

the products formed at each electrode depends on the reactivity of the ions, as well as the concentration of the salt

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18
Q

electrolysis - in aqueous solutions at CATHODE

A

positive cations move to the negative cathode

METAL LESS REACTIVE than HYDROGEN then the METAL WILL FORM
- silver and copper

METAL MORE REACTIVE than HYDROGEN then HYDROGEN GAS FORMS

  • group 1 and group 2 metals
  • aluminium
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19
Q

electrolysis - in aqueous solutions at ANODE

A

negative anions move towards positive anode

NO HALIDE = OXGEN FORMED
- from hydroxide ions in water
4OH- (aq) —-> O2 (g) + 2H2O (l) + 4e-

HALIDE SOLUTION CONCENTRATED = HALOGEN FORMED

HALIDE SOLUTION DILUTE = OXYGEN FORMED

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20
Q

electrolysis - metal electrodes

eg. purification of copper

A

metal ions can also be made at ANODE (apart from platinum ones)

in the purification of copper, anode = impure copper, cathode = pure copper

  • at ANODE: copper ions lose electrons and become copper ions, which enter the solution
  • these ions are then attracted to the CATHODE: gain electrons to become copper atoms again and place the pure copper cathode

ANODE: Cu (s) —> Cu2+ (aq) + 2e- CATHODE: Cu2+ (aq) + 2e- —-> Cu(s)

pure copper cathode increases in mass and the impure copper anode shrinks

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21
Q

oxidation states - oxygen

A

-2

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22
Q

oxidation states - hydrogen

A

+1

23
Q

oxidation states - group 1

A

+1

24
Q

oxidation states - group 2

A

+2

25
Q

oxidation states - group 3

A

+3

26
Q

oxidation states - group 4

A

+4

27
Q

oxidation states - group 5

A

-3

28
Q

oxidation states - group 6

A

-2

29
Q

oxidation states - group 7

A

-1

30
Q

oxidation states - d-block

A

variable oxidation states

31
Q

oxidation states - nitrate or nitrate(V)

A

NO3-

-1

32
Q

oxidation states - sulfate or sulfate(VI)

A

SO3 2-

-2

33
Q

oxidation states - carbonate

A

CO32-

-2

34
Q

oxidation states - manganate(VIII)

A

MnO4-

-1

35
Q

oxidation states - ammonium

A

NH4+

+1

36
Q

oxidation states - hydrogencarbonate

A

HCO3-

-1

37
Q

oxidation states - sulfide

A

S2-

-2

38
Q

oxidation states - what do they show?

A

how many electrons you have

when atoms react or bond with other atoms they lose or gain electrons

an atom’s oxidation state tells you have many electrons it has donated or accepted to form an ion or a bond

39
Q

oxidation states rules

A
  1. uncombined elements = 0
  2. neutral compound = 0
  3. compound ions = same as the overall charge
40
Q

redox reactions - what?

A

when electrons are transferred

41
Q

redox reactions - oxidation

A

loss of electrons

42
Q

redox reactions - reduction

A

gain of electrons

43
Q

redox reactions - oxidation states

A

goes up or down as electrons are lost or gained

  1. oxidation state increases +1 for every electron lost
  2. oxidation state decreases -1 for every electron gained
44
Q

redox reactions - half equations

A

show whats been reduced and whats been oxidised

45
Q

redox reactions - oxidising agents

A

accepts electrons and is reduced

46
Q

redox reactions - reducing agents

A

donates electrons and is oxidised

47
Q

redox reactions - balancing equations

A

need to balance charges as well as atoms

- do this by comparing the oxidation states of the reactants and products

48
Q

redox reactions

balance: Au3+ + I- —-> Au + I2

A
  1. first balance the atoms: Au3+ + 2I- —-> Au + I2
  2. check charges are balances: left side = +1 right side = 0, not balanced
  3. find the change in oxidation state of both elements
    - Au3+ —> Au = -3
    - 2I- —-> I2 = +2
  4. to balance the charges need to balance the change in oxidation states.
    - multiply the top and bottom rows of the table so the change cancels each other out
    - x2 for gold = -6
    - x3 for iodine = +6

now add these into the equation: 2Au3+ + 6I- —-> 2Au + 3I2

49
Q

Iodine-sodium thiosulfate titrations - what for?

A

to find the concentration of an oxidising agent

the more conc. an oxidising agent is the more ions will be oxidised by a certain vol. of it

50
Q

Iodine-sodium thiosulfate titrations - Potassium iodate(V), KIO3
STAGE 1: use a sample of oxidising agent to oxidise as much iodide as possible

A
  1. measure out a certain vol. of KIO3 - 25cm^3
  2. add this to an excess of acidic potassium iodide solution
    - the iodate(V) ions in the potassium iodate(V) solution will oxidise some of the iodide ions to iodine

IO3- (aq) + 5I- (aq) + 6H+ —-> 3I2 (aq) + 3H2O (l)

51
Q

Iodine-sodium thiosulfate titrations - Potassium iodate(V), KIO3
STAGE 2: find out how many moles of iodine have been produced

A

titrate the solution with sodium thiosulfate
- I2 (aq) + 2S2O32- (aq) —-> 2I- (aq) + S4O62- (aq)

  1. put all the solution from stage 1 in a flask
  2. add S2O3 from burette to solution in the flask
  3. when colour fades to a pale yellow add 2cm^3 of starch as an indicator —> solution goes dark blue
  4. add S2O3 one drop at a time until the blue colour disappears
  5. just as this happens means all the iodine has just reacted
  6. now can calculate no. of moles needed
52
Q

Iodine-sodium thiosulfate titrations - Potassium iodate(V), KIO3
STAGE 3: calculate the conc. of the oxidising agent

A

use the first equation and mole ratios

53
Q

titrations - need to be done accurately

A
  1. using contaminated apparatus could make results inaccurate
    - –> clean and rise it out with sodium thiosulfate before you start
  2. read burette correctly
    - –> bottom of the meniscus
  3. repeat to reduce random errors
    - —> concordant results, the take an average
  4. wash flask between repeats or use a new clean one
54
Q

Iodine-sodium thiosulfate titrations - specific problems

A
  1. solutions will react very slowly with oxygen from the air and should be made up as freshly as possible
  2. if you add starch indicator too soon the iodine will stick to the starch and won’t react as expected with the thiosulfate
    - –> making the results unreliable so only add when pale yellow

A level Chem OCR B (59 decks)

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