Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

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

DM - redox *03 Flashcards

You may prefer our related Brainscape-certified flashcards:
Periodic Table flashcards Chemistry 101 flashcards AP® Chemistry flashcards Organic Chemistry 101 flashcards
1
Q

what is oxidation?

A

a loss of electrons

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What is reduction?

A

A gain of electrons

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is a redox reaction?

A

when reduction and oxidation happens simultaneously

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

What is a reducing agent?

A

donates its electrons to something

(oxidised itself)

the more powerful a reducing agent the more easily it can be oxidised

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What is an oxidising agent?

A

takes electrons away from something

(reduced itself)

the more powerful the oxidising agent the more easily it can be reduced

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What do half equations show?

A

show whats being reduced and whats been oxidised

eg
Na —-> Na+ + e-
Cl2 + 2e- —-> 2Cl-

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Balancing half equations

A

sometimes cannot balance the equation by just multiplying up the reactants and products and adding electrons

  1. If oxidising agent contains OXYGEN, need to add some H+ ions (reactants side) and H2O (products side) to make half equations balance
  2. also need to balance charges
    - might need to multiply up and add electrons
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Balancing half equations - MnO4- —> Mn2+ by Fe2+ ions

A

iron is oxidised = Fe2+ (aq) —> Fe3+ (aq) + e-

manganate reduced = MnO4- (aq) —> Mn2+ (aq)

to balance oxygens = MnO4- (aq) —> Mn2+ (aq) + 4H2O (l)

then need to balance hydrogens = MnO4- (aq) + 8H+ (aq) —-> Mn2+ (aq) + 4H2O (l)

finally need to balance charges by adding electrons =
MnO4- (aq) + 8H+ (aq) + 5e- —-> Mn2+ (aq) + 4H2O (l)

overall equation
MnO4- (aq) + 5Fe2+ + 8H+ —-> Mn2+ + 5Fe3+ + 4H2O

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Acid-Base titrations - what do they find?

A

how much acid is needed to neutralise a base

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

acid-base titrations - how to set up?

A
  1. measure out alkali using a pipette, put it in a flask with some indicator, eg. phenolphthalein
  2. add acid to the burette and do a rough titration to get an idea of where the end point is
    - swirl each time acid is added
  3. now do accurate titrations, run the acid within 2cm^3 of the end point then add the acid dropwise
  4. record the vol. of acid used to neutralise the alkali
    - repeat until concordant results (within 0.1cm^3)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Redox titrations - what do they show?

A

how much oxidising agent is needed to exactly react with a quantity of reducing agent

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

redox titration - what do you need to know?

A

concentration of either the oxidising or reducing agent

can use titration results to find the concentration of the other

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

redox titrations - Manganese (VII) ions, MnO4- method

A
  1. measure quantity of reducing agent eg. aq Fe2+ ions using a pipette and put it into a conical flask
  2. add some dilute sulfuric acid to the flask (in excess)
    - so plenty of H+ ions to allow the oxidising agent to be reduced
  3. Add aq MnO4- (oxidising agent) to the reducing agent using a burette, swirling the flask as you do so
  4. stop when the mixture in the flask becomes tainted with the colour of MnO4- (purple) this is the end point - a rough titration
  5. then do accurate titrations until concordant results (within 0.1cm^3)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

redox titrations - Manganese (VII) ions, MnO4- - colour change

A

MnO4- in aq potassium manganate (VII), KMnO4, are purple

when added to the reducing agent, they react with it are are reduced to colourless Mn2+ ions

the reaction continues until all the reducing agent has reacted
the next drop into the flask will give the mixture the pink colour of the oxidising agent

need to spot exactly where this happens

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

redox titrations - what can results be used for?

A

to find the conc. of the reagent

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

redox titrations - Manganese (VII) ions, MnO4- and Fe2+ equation

A

MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) —–> Mn2+ (aq) + 2H2O (l) + 5Fe3+ (aq)

5 moles of Fe2+ react with 1 mole of MnO4-

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

electrochemical cells - what are they?

A

made from two different metals dipped in salt solution of their own ions and connected by a wire

always two reactions - redox (one reduction and one oxidation)
- oxidation at anode (-ve electrode)
- reduction at cathode (+ve electrode)
CHARGES ARE THE OTHER WAY ROUND

so electrons flow through the wire from the most reactive metal to the least

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

electrochemical cells - zinc/copper

A

zinc loses electrons more easily than copper, so in the half cell zinc (s) (from the zinc electrode) loses electrons and is oxidised to Zn2+ (aq) ions
Zn (s) —-> Zn2+ (aq) + 2e-

this releases electrons into the external circuit

in the other half cell the same number of electrons are taken from the external circuit, reducing the Cu2+ ions to copper atoms
Cu2+ (aq) + 2e- ——> Cu (s)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

electrochemical cells - how can you measure the cell potential, Ecell?

A

put a voltmeter between the two half cells to measure the voltage between the two half cells

this is the cell potential

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

electrochemical cells - half cells involving solutions of two aqueous ions of the same element

eg. Fe2+ (aq) / Fe3+ (aq)

A

the conversion of Fe2+ Fe3+ happens on the surface of the electrode

neither the reactants nor products are solids need something else for the electrode

it needs to conduct electricity and be very inert so that it won’t react with anything in the half cell
- Platinum excellent but v. expense so graphite often used instead.

21
Q

electrochemical cells - drawing them, anode left or right

A

the half cell where oxidation happens (anode) always on the left

the half cell where reduction happens (cathode) always on the right

22
Q

electrochemical cells - how to set up?

A
  1. get a strip of each of the metals you’re investigating (these are the electrodes)
    - clean the surface of the metals using emery paper (or sandpaper)
  2. clean any grease or oil from the electrodes using propanone
    - don’t touch the surface with your hands and could transfer grease back onto the electrode
  3. place electrode in beaker filled with a solution containing ions of that metal
    - sometimes also have to add acid
  4. create a salt bridge to link the two solutions together
    - soak a piece of filter paper in salt solution, KCl (aq) or KNO3 (aq)
    - drape between the two beakers, the ends should be immersed in the salt solution
  5. connect the electrodes to a voltmeter using crocodile clips and wires
    - if connected correctly should get a reading on your voltmeter
23
Q

electrochemical cells - are reactions at each electrode reversible?

A

yes

the reactions can go in either direction

24
Q

electrochemical cells - equations for zinc/copper

A

Zn2+ (aq) + 2e- Zn (s)

Cu2+ (aq) + 2e- Cu (s)

25
Q

electrochemical cells - how are half equations written?

A

the reduction reaction always go forwards

electons always added on left side

26
Q

electrode potential

A

the volatage

  • between or half cells
  • each half cell has its own potential (hydrogen electrode)
27
Q

why does each half cell have an electrode potential

A

there is a potential difference between electrode and solution

eg. zinc half cell - Zn electrode -vely charged as electrons left behind when Zn2+ ions form, Zn2+ ions in solution +vely charged

28
Q

elecrode potential - combining electrode potentials

A

the more +ve standard electrode potential goes forward (reduced)

the more -ve standard electrode potential goes backwards (oxidised)

29
Q

elecrode potential - combining electrode potentials

eg. zinc-copper cell

A

Zn2+(aq)/Zn(s) electrode potential = -0.76V
Cu2+(aq)/Cu(s) electrode potenial = +0.34V

zinc half cell more -ve so is oxidised, goes backwards
copper half cell more +ve so is reduced, goes forwards

overall reaction = Cu2+ (aq) + Zn (s) Cu(s) + Zn2+ (aq)

reducing agent = zinc
oxiding agent = copper

30
Q

how do you measure the electrode potential of a half cell?

A

against a standard hydrogen electrode

31
Q

what is a standard hydrogen electrode?

A

H2 gas bubbles into H+ solution

Platinum foil surface on electrode

salt bridge

voltmeter

always drawn on the left

32
Q

what are the standard conditions for measuring the electrode potential of a half cell using a hydrogen electrode?

A

any solution must have a conc. of 1.00 mol dm^03

temperature must be 298 K (25degrees)

pressure must be 100kPa

33
Q

half equation for the reaction at the hydrogen electrode?

A

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

34
Q

what is the standard electrode potential of a hydrogen half cell?

A

0.00V

35
Q

Draw the standard hydrogen electrode half cell

A

H2 gas bubbles in at 100kPa

tube thing surrounding it

H+ ions in solution at 1 mol dm^-3

Solid Pt foil surface on electrode

wire connecting it to metal electrode

voltmeter

salt bridge

36
Q

How do you work out Ecell?

A

use electrode potentials

Ecell = Emore +ve - Emore -ve

this value will always be a +ve Volatge

37
Q

why do reactive metals have large negative electrode potentials?

A

the more reactive a metal is the more it wants to LOSE ELECTRONS to form a +ve ion

(its a stronger reducing agent)

38
Q

why do more reactive non-metals have more positive standard electrode potentials

A

the more reactive a non-metal is the more it wants to GAIN ELECTRONS, to form a -ve ion

(its a stronger oxidising agent)

39
Q

what is the electrochemical series?

A

a list a redox equilibria

arranged in order of electrode potential
- most -ve at top

40
Q

what does the electrochemical series show?

A

what is reactive

41
Q

electrochemical series - more negative electrode potentials mean that:

A
  1. the right hand substances are more easily oxidised
  2. the left hand substances are more stable

decreases down series (obvs) as getting less -ve

42
Q

electrochemical series - more positive electrode potentials mean that:

A
  1. the left hand substances are more easily reduced
  2. the right hand substances are more stable

increases down series (obvs) as getting more +ve

43
Q

how can you use electrode potentials to predict whether a reaction will happen?

eg. predict whether zinc metal reacts with aqueous copper ions

A

need half equations and electrode potentials:
Zn2+ (aq) + 2e- Zn (s) E= -0.76 V
Cu2+ (aq) + 2e- Cu (s) E = +0.34 V

put the most -ve on top
Zn2+ (aq) + 2e- —-> Zn (s)
Cu2+ (aq) + 2e- —–> Cu (s)
read anti clockwise

Zn (s) + Cu2+ (aq) —> Zn2+ (aq) + Cu (s)

zinc metal WILL react with copper ions

ONLY UNDER STANDARD CONDITIONS

44
Q

predicitng feasible reactions - why could predicition be wrong?

A

kinetics not favourable:

  1. rate of reaction may be so slow that the reaction might not appear to happen
  2. if a reaction has a high activation energy, this may stop it happening

conditions not standard

  1. changing conc. or temp. changes electrode potential
    - its shifts the equilibrium
45
Q

rusting - what is it?

A

when iron is exposed to oxygen and water it will turn into rust

46
Q

rusting - how?

A

two half equations:
Fe2+ (aq) + 2e- < —> Fe (s) E= -0.44 V
2H2O (l) + O2 (g) + 4e- < —-> 4OH- (aq) E = +1.23 V

overall: 2H2O (l) + O2 (g) + 2Fe (s) —-> 2Fe2+ (aq) + 4OH- (aq)
E = +1.67 V

Fe2+ and OH- combine to form a percipiate of iron (II) hydroxide
Fe2+ (aq) + 2OH- (aq) —-> Fe(OH)2 (s)

Fe(OH)2 if further oxidised to Fe(OH)3 by oxygen
2H2O (l) + O2 (g) + 4Fe(OH)2 (s) —–> 4Fe(OH)3 (s)

Iron (III) hydroxide will eventually turn into hydrated iron (III) oxide, Fe2O3.xH2O (this is rust)

47
Q

rusting - why is rust less likely to form in alkaline conditions?

A

would shift equilibrium of 2H2O (l) + O2 (g) + 4e- 4OH- (aq) would shift left to get rid of extra OH-

as a result, the equilibrium of Fe2+ (aq) + 2e- < —>Fe (s) is pushed right to get rid of extra electrons this produces

so less Fe2+ to take part in Fe2+ (aq) + 2OH- (aq) —-> Fe(OH)2 (s) so LESS rust

48
Q

rusting - how to prevent? - COATING

A

coat iron with barrier to keep out either O2 or H2O or both

  1. painting/coating with a polymer - idea for big for small structures can also be decorative
  2. oiling/greasing - this has to be using when moving parts are involves, eg. bike chains

A level Chem OCR B (59 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions