9/19 Flashcards
(71 cards)
1
Q
oxidation is the —– of oxygen
reduction is the —– of oxygen
A
oxidation is gain of oxygen
reduction is loss of oxygen
2
Q
oxidation is the —– of hydrogen
reduction if the —— of hydrogen
A
oxidation is loss of hydrogen
reduction is gain of hydrogen
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
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)
5
Q
group 1 metals always have an oxidation number of ?
A
+1
6
Q
group 2 metals always have an oxidation number of ?
A
+2
7
Q
aluminium always has an oxidation number of ?
A
+3
8
Q
fluorine always has an oxidation number of ?
A
-1
9
Q
oxygen is always …
A
-2 except when in peroxides, where it is -1
9
Q
A
10
Q
KMnO4 is potassium manganate (VII)- what does the (VII) refer to?
A
the oxidation number on the manganese ion
11
Q
define an oxidation number
A
reflects the no of electrons the atom uses in bonding to another element
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
13
Q
hydrogen is always…
A
+1 except when in metal hydrides, where it is -1
14
Q
what does the activity series do?
A
it ranks metals according to the ease with which they undergo oxidation
15
Q
describe how you would write a redox reaction equation
A
- write down word equation as a symbol equation and balance it
- work out the oxidation numbers of the elements that change
- work out the increase and decreases in oxidation number and balance them
- add H2O and H+ as needed to balance O and H
16
Q
state and explain 2 different redox titrations
A
- 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 - 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.
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
18
Q
in the case of solubility of oxygen in water, we calculate the amount dissolved in …
A
1dm3
19
Q
high concentration of dissolved oxygen =
A
low level of pollution
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
21
Q
give 3 sources of organic matter in a body of water
A
- untreated sewage
- brewery waste
- abattoirs
22
Q
describe the Winkler Method
A
- a precipitate of manganese (II) hydroxide is made
Mn2+ (aq) + 2OH- (aq) -> Mn(OH)2 (s) - 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) - The brown precipitate is react with an excess of iodide ions, creating iodine
MnO(OH)2 (s) + 4H+ (aq) + 2I- (aq) -> Mn2+ + I2 + 3H2O - The amount of iodine formed is determined by titrating the sample with sodium thiosulphate, Na2S2O3 (redox titration)
I2 + 2S2O32- -> 2I- + S4O62-
23
Q
give the Winkler method ratios
A
4:2:2:1
S2O32-: I2: MnO(OH)2: O2
24
what takes place in electrochemical cells?
chemical energy - electrical energy conversions
25
state and describe the two types of electrochemical cell
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
26
define an electrode
a conductor of electricity used to make contact with a non-metallic part of a circuit, such as the solution in a cell
27
similarity between voltaic and electrolytic cells
oxidation always takes place at the anode
reduction always takes place at the cathode
28
difference between voltaic and electrolytic cells
voltaic cell:
- cathode is positive electrode
- anode is negative electrode
electrolytic cell (CNAP):
- cathode is negative electrode
- anode is positive electrode
29
a voltaic cell consists of
two half-cells
30
state the 3 different types of electrode used in voltaic cells
- metal/metal-ion electrode
- metal ions in two different oxidation states
- gas-ion electrode
31
define electrodes and half cells
electrodes: the two metal strips
half cells: beaker containing strip of metal atoms in equilibrium with aqueous solution of its ions
32
what is the function of the salt bridge?
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
what does a salt bridge usually consist of and why?
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
explain the use of electrochemical series to create voltaic cells
- 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
draw a metal/metal-ion electrode
36
draw a metal ions in two different oxidation states electrode
37
draw a gas-ion electrode
38
the more negative the value of E^0 the ----- the reducing agent
stronger
39
use zinc and copper half cells as examples of the flow of electrons/charge
- 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
by convention the anode is always written on the ----- and the cathode on the ------
left; right
41
draw a zinc-copper voltaic cell
page 228
42
EMF (electromotive force) and standard cell potential
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
what is the result of the generation of an EMF?
the movement of electrons from the anode to the cathode via the external circuit
44
EMF is termed
the cell potential (Ecell^⊖)
45
define standard hydrogen electrode (SHE)
an inert platinum electrode in contact with 1mol/dm3 hydrogen ion (H+) and hydrogen gas at 100kPa and 298K
46
what is the standard electrode potential (E^⊖) of a substance?
the potential (voltage) of the reduction half equation under standard conditions measured relative to the SHE
47
standard conditions
solute concentration of 1 mol/dm3 or 100kPa for gases, 298k
48
standard electrode potential of the SHE is
0V
49
if a reaction is to happen spontaneously, E^0cell must be
positive
50
E^0cell =
E(red) - E(ox)
E(+ve) - E(-ve)
51
give the limitations of using standard electrode potentials
electrode potentials are affected by cell concentration, cell temperature, and cell pressure
52
For M2+ + 2e- <-> M
describe the effects of changing concentration and temperature
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
a negative standard electrode potential signifies
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
a positive standard electrode potential signifies
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
how do you calculate standard free-energy changes (∆Gº)?
using the expression ∆Gº=-nFEcell
n is amount, in mol, of electrons transferred in the balanced equation
F is Faraday's constant
56
when Ecell is positive, ∆Gº is
negative and the process is spontaneous
57
when Ecell is negative, ∆Gº is
positive and the process is non-spontaneous
58
draw a diagram for electrolysis of PbBr2
59
when aqueous solutions are electrolysed, what two things happen to water?
it can be oxidised to oxygen at the anode and reduced to hydrogen at the cathode
60
water being reduced at the cathode
water being oxidised at anode
H2O (l) + e- <-> 1/2H2 (g) + OH- (aq)
H2O (l) <-> 1/2O2 (g) + 2H+ (aq) + 2e-
61
Which species is discharged depends on three things:
The relative values of Eθ
The concentration of the ions present
The identity of the electrode
62
Relative values of Eθ
cathode- reduction with the more positive Eθ value will be favoured
anode- oxidation with the more negative Eθ value will be favoured
63
concentration of the ions present
an anion in higher concentration is always being preferentially discharged
64
active electrodes
Electrodes that take part in the redox processes
65
passive electrodes
inert electrodes such as platinum and graphite
66
how to calculate the amount of product formed at the electrodes during electrolysis
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
define electroplating and draw a diagram
electrolytic coating of an object with a very thin metallic layer
68
anode is usually made from the same metal to
replenish the loss of the metal during electrolysis and maintain a constant concentration of the electrolyte
69
equation for EMF
70
define the Faraday constant
the amount of electric charge carried by one mol of electrons
