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Aqa a level chemistry > Electrochemical cells + period 3 > Flashcards

Electrochemical cells + period 3 Flashcards

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1
Q

identify which species can be used to reduce VO2- ions to VO2+

A

reagent: Fe2+

justification: EƟ VO2+(/ VO2+) > EƟ Fe3+(/Fe2+) > EƟ VO2+(/V3+)

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2
Q

ionisation energy across period 3

A

Atomic radius across period 3 remains similar as shielding is similar. However nuclear charge increases slightly as proton number increases so stronger electrostatic force of attraction.

Across period 3 there is an overall increase in ionisation energy from sodium to argon. However Al has smaller ionisation energy as it enters a subshell higher in energy. Distance between nucleus and outer electron shell increases. There is also spin paired repulsion in sulfur = unstable so requires less energy to overcome.

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3
Q

melting point across period 3

A

From Na to Al melting point increases. Increase in attraction between delocalised electrons and positive lattice = increase in metallic bonding.

Silicon has the highest melting point because it is giant covalent structure

P to Ar = simple van der waal forces

Sulfur = different shape = higher melting point

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4
Q

reaction of sodium and magnesium in water

A

-Sodium has a very exothermic reaction with cold water producing hydrogen and a colourless solution of sodium hydroxide.

-Magnesium has a very slight reaction with cold water, but burns in steam.

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5
Q

period 3 oxides

A

Na2O

MgO

Al2O3

SiO2

P4O10

SO2

SO3

Cl2O

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6
Q

electrical conductivity of period 3 elements

A

Electrical conductivity of period 3 elements –> None of these oxides has any free or mobile electrons. That means that none of them will conduct electricity when they are solid.

The ionic oxides can, however, undergo electrolysis when they are molten. They can conduct electricity because of the movement of the ions towards the electrodes and the discharge of the ions when they get there.

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7
Q

sodium + oxygen

A

Sodium + oxygen –> 4Na + O2 –> 2Na2O (oxidation state of Na = +1)

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8
Q

magnesium + oxygen

A

Magnesium + oxygen –> 2Mg + O2 –> 2MgO (oxidation state of Mg = +2)

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9
Q

aluminium + oxygen

A

4Al + 3O2 –> 2Al2O3 (Al = +3 oxidation state)

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10
Q

silicon + oxygen

A

Si + O2 –> SiO2 (Si = +4 oxidation state)

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11
Q

phosphorus + oxygen

A

P4 + 5O2 –> P4O10 (P = +5 oxidation state)

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12
Q

sulfur + oxygen

A

S + O2 –> SO2 (+4 oxidation state)4 oxidation state_

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13
Q

sodium + water

A

Sodium + water –> 2Na + 2H2O –> 2NaOH + H2

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14
Q

Na2O

A

-dissolves and reacts to form a solution
-basic oxide
-pH 14

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15
Q

MgO

A

-ionic
-slightly soluble
-less soluble than Na2O due to higher lattice enthalpy

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16
Q

Al2O3
SO2

A

insoluble
high lattic energy (Al)
strong covalent bonds (SO2)

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17
Q

chlorine + water

A

Cl2(g) + H2O(l) ⇌ HCl(aq) + HClO(aq)

Chlorine + water = dissolves in water to give green solution

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18
Q

amphoteric

A

Amphoteric –> (of a compound, especially a metal oxide or hydroxide) able to react both as a base and as an acid

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19
Q

H3PO4

A

H3PO4 = tetrahedral shape (double bond for first oxygen) and ionic is –3 charge

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20
Q

element X

A

Element X forms an oxide that has a low melting point and dissolves to form an acidic solution

-Type of bonding = covalent

-Element X = Phosphorous

-Equation for P with water = P4O10 + 6H2O –> 4H3PO4

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21
Q

element Y

A

Element Y reacts vigorously with water and dissolves in water to form a solution with pH 14

-bonding = ionic

-element = sodium

-2Na + 2H2O –> 2Na+ + 2OH- + H2

-Na2O + 2HCl –> 2NaCl + H2O

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22
Q

element Z

A

Element Z forms an amphoteric oxide that has a high boiling point:

-bonding = ionic

-element = Al2O3

-amphoteric = reacts to acids and bases

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23
Q

burning of period 3 elements

A

-sodium burns with a yellow flame to produce a white solid

-Mg, Al, Si and P burn with a white flame to give white solid smoke.

  • S burns with a blue flame to form an acidic choking gas.
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24
Q

summary of equations period 3

A

4 Na (s) + O2 (g)

2 Na2O (s) 2Mg (s) + O2 (g)

2MgO (s) 4Al (s) + 3O2 (g)

2Al2O3 (s) Si (s) + O2 (g)

SiO2 (s) 4P (s) + 5O2 (g)

P4O10 (s) S (s) + O2 (g)

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25
Q

metal vs non-metal oxides

A

Metal ionic oxides tend to react with water to form hydroxides which are alkaline (exothermic)

Non-metal simple covalent molecules react with water to give acids

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26
Q

electrochemical cells

A

-cells are used to measure electrode potentials by reference to the standard hydrogen electrode

-standard electrode potentials refers to the conditions of 298K, 100kpa, and 1.00moldm-3 solution of ions

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27
Q

position of EQ

A

-if a rod metal is dipped into a solution of its own ion an equilibium is set up e.g Zn (s) –> Zn2+ + 2e-

-the position of equilibrium determines the electrical potential between the metal strip and the solution of the metal

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28
Q

calculating electrode potential

A

-We cannot directly measure electrode potential. We can instead connect together two different electrodes and measure the potential difference with a voltmeter

Calculate electrode potential:

-EmF = E(right) - E(left) –> more positive electrode is represented on the right

e.g Li+ (aq) + e- –> Li (s) = -3.05volts

2H+ (aq) + 2e- –> H2 (g) = 0.00 volts

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29
Q

steps to drawing electrochemical cells

A

Start from salt bridge ||

Determine sides of electrode

(-) electrode on the left, (+) electrode on the right

Write highest oxidation state species next to the salt bridge

Draw phase boundary | between species of different phase

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30
Q

flow of electrons

A

Electrons flow from negative electrode to the most positive electrode

Anti-clockwise rule to predict the direction of the reaction = Oxidation (negative electrode) goes on the top

Reduction (positive electrode) goes at the bottom

Draw anti clockwise arrow from right side of negative electrode

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31
Q

positive vs negative

A

More positive electrode = reduction (forwards reaction)

More negative electrode = oxidation (backwards reaction)

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32
Q

voltage

A

When connected together the zinc half-cell has more of a tendency to oxidise to the Zn2+ ion and release electrons than the copper half-cell. (Zn –> Zn2+ + 2e-) More electrons will therefore build up on the zinc electrode than the copper electrode. A potential difference is created between the two electrodes. The zinc strip is the negative terminal and the copper strip is the positive terminal. This potential difference is measured with a high resistance voltmeter, and is given the symbol E. The E for the above cell is E= +1.1V.

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33
Q

salt bridge

A

-used to connect up the circuit. Free moving ions conduct the charge

-made from filter paper soaked in potassium nitrate

-The salt should be unreactive with the electrodes and electrode solutions. E.g. potassium chloride would not be suitable for copper systems because chloride ions can form complexes with copper ions. A wire is not used because the metal wire would set up its own electrode system with the solutions

34
Q

current

A

If the voltmeter is removed and replaced with a bulb or if the circuit is short circuited, a current flows. The reactions will then occur separately at each electrode. The voltage will fall to zero as the reactants are used up. The most positive electrode will always undergo reduction. Cu2+ (aq) + 2e-  Cu(s) (positive as electrons are used up) The most negative electrode will always undergo oxidation. Zn(s)  Zn2+ (aq) + 2e- (negative as electrons are given off)

35
Q

which species is more likely to be reduce, Al3+ or Mg2+

A

Al3+ = oxidising agent = reduced

36
Q

identify the oxidising agent in this cell

2MnO2 (s) + 2NH4+ (aq) + 2e- –> Mn2O3 (s) + 2NH3 (Aq) + H2O (l)

A

MnO2 = reduced

37
Q

which change to a hydrogen electrode has no effect on the electrode potential

A

the surface area of the platimum electrode

38
Q

state the meaning of the term electrochemical series

A

a list of electrode potential values in numerical order

39
Q

which is the weakest reducing agent in the table above

A

[Co(H2O06]2+ = most positive number

40
Q

explain why an aqueuous electrolyte is not used for a lithum cell

A

lithium would react with the water
the electorde potential is more negative than water as it is a better reducing agent

41
Q

platinum electrode =

A

Reaction = 2H+ (aq) +2e- –> H2(g)

as hydrogen gas is used

42
Q

platinum electrode + MnO4-

A

reaction = MnO4- (aq) + 8H+ + 5e- –> Mn2+ + 4H2O (l)

43
Q

state 2 conditions needed for the following half cell to have an electrode potential of zero volts

A

-temperature at 298K
-pressure at 100Kpa

44
Q

give the conditions under which the electrode potential of Zn2+/Zn electrode is -0.76V

A

100kpa, 298K, 1moldm-3 of Zn2+

45
Q

purpose of the salt bridge =

A

complete the cell

46
Q

name the substance used as electrode B in the diagram above

A

platinum

47
Q

explain how the salt bridge connects the circuit

A

ions in the ionic substance in the salt bridge move through the salt bridge to maintain charge balance

48
Q

state why the left-hand electrode does not have an electrode potential of +0.34V

A

the concentration of the CuSO4 isnt at 1moldm-3 which is the standard conditions its at 0.15moldm-3

49
Q

The following cell has an EMF of +0.46V

Cu / Cu2+//Ag+/Ag

which statement is correct about the operation of the cell

A

metallic copper is oxidised by Ag+ ions

50
Q

use the data from the table to explain why gold jewellery is unreactive in moist air

A

the EMF of the reaction of Au+ and oxygen is -0.45 so is not spontaneous

51
Q

oxidisng agent =

A

electron acceptor

52
Q

use the data from the table to explain in terms of redox what happens when a soluble gold compound contaning Au+ ions is added to water

A

Au+ ions oxidise water

Observation = effervescence of O2 gas or gold precipitate

Equation = 2Au+ + H2O –> 2Au + 2H+

53
Q

is Br2 and oxidising agent

A

yes not Br-

54
Q

use you data booklet to calculate the E cell for Al/Al3+ and Mn/Mn2+

A

Al3+ + 3e- –> Al (more positive)
Mn2+ + 2e- –> Mn (more negative)

3Mn + 2Al3+ –> 3Mn2+ + 2Al

55
Q

cell diagram

A

Zn(s) | Zn2+ (aq) | | Cu2+ (aq) | Cu (s) E= +1.1V

56
Q

for Fe2+ (aq) –> Fe3+ (aq) + e- there is no
solid conducting surface, a Pt electrode must be
used.
The cell diagram is drawn as:

| Fe3+ (aq), Fe2+ (aq) |Pt

A

if a system includes an electrode that is not a metal then a a platinum electorde must be used

57
Q

hydrogen electrode

A

The hydrogen electrode equilibrium is:
H2(g) –> 2H+ (aq) + 2e

Pt |H2 (g) | H+ (aq)

58
Q

disproportionation

A

-Disproportionation reaction = reduction and oxidation occur in the same reaction from the same element

59
Q

half equations

A

Fe2+ –> Fe3+ + e-

MnO4- –> Mn2+ + 4H2O

(if both elements appear to be oxidised look at oxidation number e.g MnO4- = +7 and Mn2+ = +2 so is reduced)

60
Q

salt bridge

A

Salt bridge = prevent build up of a charge, without it there would be no potential difference of movement. No salt bridge = 2 half cells would be charged and charge flow would stop so potential difference is zero

61
Q

further down table =

A

more likely to reduce

62
Q

reduced vs oxidising agent

A

Oxidised (reducing agent) Reduced (oxidising agent)

63
Q

Zn and Cu cell equation

A

Zn (s) / Zn2+ (aq) / Cu2+ (aq) / Cu (s)

64
Q

anode vs cathode

A

Anode = negtive electrode (right side)

Cathode = positive electrode (left side)

65
Q

Ni2 and H+

A

Ni (s) / Ni2+ (aq) / H+ (aq) / H2(g) / Pt (s)

Negative –> positive

66
Q

calculating electrode potential

A

Left = more negative

Right = more positive

Emf = Eright – Eleft (flow of electrons from 1 half cell to another)

Positive – negative

67
Q

forwards vs backwards reaction

A

Forward reaction = more positive = reduction

Backwards reaction = more negative = oxidiation

68
Q

metals

A

When metals react they loose electrons to form positive ions

If a rod of metal is dipped into a solution of its own ions then an equilbirum is set up = Zn (s) –> Zn2+ (aq) + 2e-

69
Q

dynamic equilbirium

A

Half cell and the strip of metal is an electrode

Dynamic equilbirim will be established when the rate at which ions are leaving the surface is exactly equal to the rate at which they are joining it again

70
Q

direction of arrows

A

Arrow away from the metal/ion = oxidiation

Arrow towards the metal/ion = reduction

71
Q

small electrode potentials

A

Li, K, Ca etc = small electron potentials = good reducing agent so will get oxidised

Down the reactivity series = more likely to be reduced (positive number)

72
Q

larger electrode potentials

A

The metal forms at the half cell

If the species has a large electrode potential the species will be a good oxidising agent

Calcium is a weaker reducing agent than lithium

73
Q

what is a half cell

A

Half cell = solid lead electrode combined with ts solution with its own

74
Q

more positive value =

A

oxidising agent

75
Q

what affects value of electrode potential

A

-temperature

-pressure

-solution concentration

–> reference for this standard hydrogen electrode

76
Q

standard hydrogen electrode

A

-H2(g) at 100kpa

-298K

-electrode potential = 0.00volts

Concentration = 1moldm-3

77
Q

negative reading on voltmeter

A

If the reading on voltmeter is negative then we can conclude that the species on the right looses an electron and is therefore being oxidised. This means that it is acting as a reducing agent.

78
Q

define standard electrode potential

A

the potential difference between a standard hydrogen electrode and the half cell with an ion concentration at 1moldm-3, gas at 100kpa and 298K

79
Q

anode

A

Anode = oxidation (metal looses electrons so is oxidised into metal ions)

80
Q

cathode

A

Cathode = reduction (metal ions gain electrons so is reduced to metal atoms)

More negative electrode (anode) is drawn on the left

reduction - oxidation

Aqa a level chemistry (18 decks)

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