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IB chemistry > 9/19 > Flashcards

9/19 Flashcards

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1
Q

oxidation is the —– of oxygen
reduction is the —– of oxygen

A

oxidation is gain of oxygen
reduction is loss of oxygen

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2
Q

oxidation is the —– of hydrogen
reduction if the —— of hydrogen

A

oxidation is loss of hydrogen
reduction is gain of hydrogen

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3
Q

oxidation is the —– of electrons
reduction is the ——- of electrons

A

oxidation is the loss of electrons
reduction is the gain of electrons

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4
Q

what happens to oxidising/reducing agents in redox reactions?

A

the oxidising agent is reduced (gains electrons), the reducing agent is oxidised (loses electrons)

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5
Q

group 1 metals always have an oxidation number of ?

A

+1

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6
Q

group 2 metals always have an oxidation number of ?

A

+2

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7
Q

aluminium always has an oxidation number of ?

A

+3

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8
Q

fluorine always has an oxidation number of ?

A

-1

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9
Q

oxygen is always …

A

-2 except when in peroxides, where it is -1

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9
Q
A
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10
Q

KMnO4 is potassium manganate (VII)- what does the (VII) refer to?

A

the oxidation number on the manganese ion

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11
Q

define an oxidation number

A

reflects the no of electrons the atom uses in bonding to another element

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12
Q

how can you tell if an element has been oxidised or reduced by its oxidation numbers?

A

oxidised if its oxidation number increases
reduced if its oxidation number decreases

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13
Q

hydrogen is always…

A

+1 except when in metal hydrides, where it is -1

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14
Q

what does the activity series do?

A

it ranks metals according to the ease with which they undergo oxidation

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15
Q

describe how you would write a redox reaction equation

A
  1. write down word equation as a symbol equation and balance it
  2. work out the oxidation numbers of the elements that change
  3. work out the increase and decreases in oxidation number and balance them
  4. add H2O and H+ as needed to balance O and H
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16
Q

state and explain 2 different redox titrations

A
  1. acidified manganate (VII) ions and iron (II) ions;
    MnO4– (aq) + 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l)
    - This reaction needs no indicator as the manganate (VII) is a strong purple colour which disappears at the end point, so the titration is self-indicating
  2. iodine and thiosulfate ions:
    2S2O32– (aq) + I2 (aq) → 2I–(aq) + S4O62– (aq)
    - The light brown/yellow colour of the iodine turns paler as it is converted to colourless iodide ions
    - When the solution is a straw colour, starch is added to clarify the end point
    - The solution turns blue/black until all the iodine reacts, at which point the colour disappears.
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17
Q

give the 2 equations for concentration in parts per million

A

mass of component in solution/total mass of solution x10^6

or

mass of solute in mg/volume of solution in dm3

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18
Q

in the case of solubility of oxygen in water, we calculate the amount dissolved in …

A

1dm3

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19
Q

high concentration of dissolved oxygen =

A

low level of pollution

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20
Q

what does BOD stand for and what does it measure?

A

biochemical oxygen demand- the amount of oxygen used (for bacteria) to decompose the organic matter in a sample of water over a specified time period, usually 5 days, at a specified temperature- in ppm

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21
Q

give 3 sources of organic matter in a body of water

A
  • untreated sewage
  • brewery waste
  • abattoirs
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22
Q

describe the Winkler Method

A
  1. a precipitate of manganese (II) hydroxide is made
    Mn2+ (aq) + 2OH- (aq) -> Mn(OH)2 (s)
  2. this precipitate will react with any oxygen present in the water sample to form a brown precipitate of MnO(OH)2
    2Mn(OH)2 (s) + O2(g) -> 2MnO(OH)2 (s)
  3. The brown precipitate is react with an excess of iodide ions, creating iodine
    MnO(OH)2 (s) + 4H+ (aq) + 2I- (aq) -> Mn2+ + I2 + 3H2O
  4. The amount of iodine formed is determined by titrating the sample with sodium thiosulphate, Na2S2O3 (redox titration)
    I2 + 2S2O32- -> 2I- + S4O62-
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23
Q

give the Winkler method ratios

A

4:2:2:1
S2O32-: I2: MnO(OH)2: O2

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24
Q

what takes place in electrochemical cells?

A

chemical energy - electrical energy conversions

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25
Q

state and describe the two types of electrochemical cell

A
  1. voltaic (galvanic) cells- convert chemical energy to electrical energy; convert energy from spontaneous, exothermic chemical processes to electrical energy
  2. electrolytic cells- convert electrical energy to chemical energy, bringing about non-spontaneous processes
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26
Q

define an electrode

A

a conductor of electricity used to make contact with a non-metallic part of a circuit, such as the solution in a cell

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27
Q

similarity between voltaic and electrolytic cells

A

oxidation always takes place at the anode
reduction always takes place at the cathode

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28
Q

difference between voltaic and electrolytic cells

A

voltaic cell:
- cathode is positive electrode
- anode is negative electrode

electrolytic cell (CNAP):
- cathode is negative electrode
- anode is positive electrode

29
Q

a voltaic cell consists of

A

two half-cells

30
Q

state the 3 different types of electrode used in voltaic cells

A
  • metal/metal-ion electrode
  • metal ions in two different oxidation states
  • gas-ion electrode
31
Q

define electrodes and half cells

A

electrodes: the two metal strips
half cells: beaker containing strip of metal atoms in equilibrium with aqueous solution of its ions

32
Q

what is the function of the salt bridge?

A

to allow ions to flow between the solutions to complete the circuit, keeping the balance of positive and negative ions in each half cell without them mixing together

33
Q

what does a salt bridge usually consist of and why?

A

it is often a piece of filter paper or agar gel soaked in a compound that will not react with either of the solutions in the half cells

34
Q

explain the use of electrochemical series to create voltaic cells

A
  • the more negative the value of E⊖, the more to the left the equilibrium lies, releasing electrons = reducing agent
  • undergoes oxidation at the anode
35
Q

draw a metal/metal-ion electrode

A
36
Q

draw a metal ions in two different oxidation states electrode

A
37
Q

draw a gas-ion electrode

A
38
Q

the more negative the value of E^0 the —– the reducing agent

A

stronger

39
Q

use zinc and copper half cells as examples of the flow of electrons/charge

A
  • zinc is higher up in the activity/has more negative E cell so undergoes oxidation more easily than copper; Zn undergoes oxidation at anode
  • metal zinc atoms on the strip form Zn2+ ions and join the solution, leaving electrons on the metal strip
  • electrons flow from the anode (-ve electrode) to the cathode (+ electrode)
  • solution at zinc half cell is positively charged as it has an excess Zn2+ ins
  • copper is lower in the activity series/has less negative E cell value so undergoes oxidation less easily; Cu2+ ions in the solution undergo reduction at the anode
  • take electrons from the metal strip and so are discharged as copper metal on the strip
  • solution at copper half cell is negatively charged as there is a deficit of Cu2+ ions
40
Q

by convention the anode is always written on the —– and the cathode on the ——

A

left; right

41
Q

draw a zinc-copper voltaic cell

A

page 228

42
Q

EMF (electromotive force) and standard cell potential

A

EMF is the energy supplied by a source divided by the electric charge transported through the source

In a voltaic cell, EMF = electric potential difference for zero current through the cell = maximum voltage that can be delivered by the cell

43
Q

what is the result of the generation of an EMF?

A

the movement of electrons from the anode to the cathode via the external circuit

44
Q

EMF is termed

A

the cell potential (Ecell^⊖)

45
Q

define standard hydrogen electrode (SHE)

A

an inert platinum electrode in contact with 1mol/dm3 hydrogen ion (H+) and hydrogen gas at 100kPa and 298K

46
Q

what is the standard electrode potential (E^⊖) of a substance?

A

the potential (voltage) of the reduction half equation under standard conditions measured relative to the SHE

47
Q

standard conditions

A

solute concentration of 1 mol/dm3 or 100kPa for gases, 298k

48
Q

standard electrode potential of the SHE is

A

0V

49
Q

if a reaction is to happen spontaneously, E^0cell must be

A

positive

50
Q

E^0cell =

A

E(red) - E(ox)
E(+ve) - E(-ve)

51
Q

give the limitations of using standard electrode potentials

A

electrode potentials are affected by cell concentration, cell temperature, and cell pressure

52
Q

For M2+ + 2e- <-> M

describe the effects of changing concentration and temperature

A

if the concentration of M2+ is increased, the equilibrium will shift to the right and the electrode potential will become less negative

if the temperature increases there is an increased tendency for metals to dissolve and form M2+, the equilibrium will shift left and the electrode potential will become more negative

53
Q

a negative standard electrode potential signifies

A

the potential on the metal electrode is more negative compared to the hydrogen half cell. the equilibrium lies to the left as electrons are liberated and oxidation is occurring

54
Q

a positive standard electrode potential signifies

A

the potential on the metal electrode is more positive compared to the hydrogen half cell. the equilibrium lies to the left as electrons are gained and reduction is occurring

55
Q

how do you calculate standard free-energy changes (∆Gº)?

A

using the expression ∆Gº=-nFEcell

n is amount, in mol, of electrons transferred in the balanced equation
F is Faraday’s constant

56
Q

when Ecell is positive, ∆Gº is

A

negative and the process is spontaneous

57
Q

when Ecell is negative, ∆Gº is

A

positive and the process is non-spontaneous

58
Q

draw a diagram for electrolysis of PbBr2

A
59
Q

when aqueous solutions are electrolysed, what two things happen to water?

A

it can be oxidised to oxygen at the anode and reduced to hydrogen at the cathode

60
Q

water being reduced at the cathode

water being oxidised at anode

A

H2O (l) + e- <-> 1/2H2 (g) + OH- (aq)

H2O (l) <-> 1/2O2 (g) + 2H+ (aq) + 2e-

61
Q

Which species is discharged depends on three things:

A

The relative values of Eθ
The concentration of the ions present
The identity of the electrode

62
Q

Relative values of Eθ

A

cathode- reduction with the more positive Eθ value will be favoured

anode- oxidation with the more negative Eθ value will be favoured

63
Q

concentration of the ions present

A

an anion in higher concentration is always being preferentially discharged

64
Q

active electrodes

A

Electrodes that take part in the redox processes

65
Q

passive electrodes

A

inert electrodes such as platinum and graphite

66
Q

how to calculate the amount of product formed at the electrodes during electrolysis

A
  1. CALCULATE THE CHARGE PASSED, Q
    - Q = I x t (Charge = current x time)
    - C = A x s (Coulombs = Amps x seconds)
  2. CALCULATE THE NUMBER OF MOLES OF PRODUCT
    - n = Q/(moles of electron needed to form one mol of product) x F (faraday constant)
  3. FIND THE MASS OF PRODUCT, m, USING
    THE MOLAR MASS, M, or volume
67
Q

define electroplating and draw a diagram

A

electrolytic coating of an object with a very thin metallic layer

68
Q

anode is usually made from the same metal to

A

replenish the loss of the metal during electrolysis and maintain a constant concentration of the electrolyte

69
Q

equation for EMF

A
70
Q

define the Faraday constant

A

the amount of electric charge carried by one mol of electrons

IB chemistry (29 decks)

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