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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

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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)

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35
Q

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

A

Al3+ = oxidising agent = reduced

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36
Q

identify the oxidising agent in this cell

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

A

MnO2 = reduced

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37
Q

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

A

the surface area of the platimum electrode

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38
Q

state the meaning of the term electrochemical series

A

a list of electrode potential values in numerical order

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39
Q

which is the weakest reducing agent in the table above

A

[Co(H2O06]2+ = most positive number

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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

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41
Q

platinum electrode =

A

Reaction = 2H+ (aq) +2e- –> H2(g)

as hydrogen gas is used

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42
Q

platinum electrode + MnO4-

A

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

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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

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44
Q

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

A

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

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45
Q

purpose of the salt bridge =

A

maintain charge balance

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46
Q

name the substance used as electrode B in the diagram above

A

platinum

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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

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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

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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

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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

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51
Q

oxidisng agent =

A

electron acceptor

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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+

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53
Q

is Br2 and oxidising agent

A

yes not Br-

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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

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55
Q

cell diagram

A

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

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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

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57
Q

hydrogen electrode

A

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

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

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58
Q

disproportionation

A

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

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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)

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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

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61
Q

further down table =

A

more likely to reduce

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62
Q

reduced vs oxidising agent

A

Oxidised (reducing agent) Reduced (oxidising agent)

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63
Q

Zn and Cu cell equation

A

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

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64
Q

anode vs cathode

A

Anode = negtive electrode (right side)

Cathode = positive electrode (left side)

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65
Q

Ni2 and H+

A

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

Negative –> positive

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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

81
Q

electrodes

A

-iron electrode is dipped into a solution of Fe2+ ions

-if 2 metal ions are present, use a platinum electrode

82
Q

iron reversible equation

A

Fe2+ (aq) + 2e- ⇌ Fe (s)

83
Q

movement of electrons in electrochemical cells

A

Electrochemical cells = 2 half cells joined together

Electrons flow from more reactive to less reactive metal

84
Q

anode and cathode

A

Arrow from electrode to ion = anode

Arrow from ion to electrode = cathode

Anode = oxidiation = loss of electrons

Cathode = reduction = electrode recieves ions

85
Q

electrode potential

A

Electrode potential is measured in volts

In electrochemical cells we always write equations in the reduced form but if the ion is oxidised it can later be flipped before being turned to ionic equation

86
Q

ionic equation of Zn and Cu

A

1) Zn (s) ⇌ Zn2+ (aq) + 2e-

2) Cu2+ (aq) + 2e- ⇌ Cu (s)

3) Zn(s) + Cu2+ (aq) ⇌ Zn2+ (aq) + Cu(s)

87
Q

concentration of H+ ions

A

1 moldm-3 of H+ ions in solution

For diprotic acids e.g H2SO4 = 0.5 moldm-3 of H+ ions in solution

88
Q

more positive vs more negative value

A

More positive value = stronger oxidising agent

More negative value = stronger reducing agent

89
Q

cell notation

A

-reduced form l oxidised form ll oxidised form l reduced form

Oxidation state of Zn = O

Oxidation state of Zn2+ = +2

2 aqueous ions = present with comma e.g Fe3+, Fe2+ l Pt (s)

90
Q

predicting reaction feasibility

A

-identify which element is oxidised

-write oxidised and reduction equations

-combine the 2 equations to write the ionic equation

-e.g Mg (s) + Cu2+ (aq) ⇌ Mg2+ (aq) + Cu (s)

-all feasible reactions will have a positive value

91
Q

lithium ion cell

A

Lithium ion cell:

-Electrode A = LiCoO2

-Electrode B = graphite

-Electrolyte = lithium salt dissolved in an organic solvent

Li ⇌ Li+ + e-

Li+ + CoO2 + e- ⇌ Li+[CoO2]-

Overall = Li + CoO2 ⇌ Li+ [CoO2]

92
Q

electrolyte

A

Electrolyte = acts as a conductive pathway for ions to move from one electrode to another

93
Q

fuel cell

A

1) Hydrogen feed = reacts with OH- ions in solution (2H2 (g) + 4OH- (aq) –> 4H2O + 4e-)

2) Flow of electrons = electrons prduced in reaction 1 travel

3) Component = flow of electrons is used to power something

4) Oxygen feed = oxygen reacts with water and 4 electrons made from step 1 to make OH- ions (O2 (g) + 2H2O (l) + 4e- –> 4OH- (aq))

5) negative electrode (cathode) = electrons flow to negative electrode which is made from platinum

6) Electrolyte is made from KOH solution. Carries OH- ions from cathode to the anode

7) positive electrode (anode) = electrons flow from the positive electrode which is made from platinum

8) Water is emitted

9) OH- ions produced from reaction 4 are carried towards the anode via the electrolyte

94
Q

true or false - anode = negative

A

false = anode is positive

95
Q

advantages and disadvantages of fuel cells

A

Advantages of fuel cells:

-more efficient than internal combustion engines

-more energy is converted t kinetic energy

-don’t need to be recharged just ready supply of oxygen and hydrogen

-no CO2 emissions

Disadvantages of fuel cells:

-hydrogen is highly flammable and must be stored and transported correctly

-expensive to transport and store hydrogen

-energy is required to make hydrogen and oxygen. Uses fossil fuels

96
Q

when to use platinum strip

A

when neither side of the electrons half equation has a solid metal to conduct the electrons

97
Q

EMF when Mg = 2.87 and standard hydrogen electrode is present

A

2.87

bc H2 = 0.00v

98
Q

in non standard conditions if equilbirium shifts to the left meaning

A

more electrons are produced
more negative

99
Q

why is zinc a stronger reducing agent than copper

A

turns into its aqueous ions faster

100
Q

single vs double displacement =

A

single = always redox
double = never redox

101
Q

A student wants to use an aqueous calcium sulphate solution to dissolve a zinc wire. They predict this will produce a zinc sulphate solution and solid calcium.

can this occur

A

No
Zinc is a weaker reducing agent
Ca is a weaker oxidisng agent

102
Q

salt bridge requirements

A

-should not react with ions in solution
-as it may alter the concentrations + electromotive force

103
Q

Daniell cell

A

zinc + copper

104
Q

Double A batteries are designed so that the casing is also the negative electrode.

Suggest one advantage and one disadvantage to this design

A

-advantage = AA batteries can be lighted

-disadvantage = zinc casting becomes thinner = more zinc is oxidised to zinc ions = leakage

105
Q

importance of ions moving=

A

balance charge

106
Q

dry cell =

A

no free liquid

107
Q

structure of AA battery

A

zinc case (negative electrode)
positive electrode in centre
paste instead of free liquid
porous separator

108
Q

why is carbon rod used

A

conducts electrcitiy but does not react with other elements

109
Q

oxidiation state for the reduction of Mn

A

manganese is reduced from an oxidation state of +4 to +3

110
Q

porous separator

A

Component
B
is the porous separator - it is used to keep the reducing agent and the oxidising agent coming into contact. This stops them from directly reacting, which forces electrons to travel through the wire.

The separator is also porous enough to allow for the transfer of ions, this balances the charge and prevents charge build-up around the electrode

111
Q

alkaline battery =

A

brass pin
electrolyte = KOH
zinc = inside

112
Q

forming oxides reactions

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.

113
Q

why are metal oxidies ionic

A

. They are ionic
because of the large electronegativity difference
between metal and O
-The increased charge on the cation makes the ionic
forces stronger (bigger lattice enthalpies of
dissociation) going from Na to Al so leading to
increasing melting points

114
Q

why dont Al2O3 and SiO2 dissolve

A

Al2O3 and SiO2 do not dissolve in water
because of the high strength of the Al2O3
ionic lattice and the SiO2 macromolecular
structure, so they give a neutral pH 7

115
Q

why are ionic oxides basic

A

The ionic oxides are basic because the oxide ions accept
protons to become hydroxide ions in this reaction (acting as a
Bronsted-Lowry base)
MgO (s) + H2O (l)  Mg(OH)2
(s) pH 9
Mg(OH)2
is only slightly soluble in water as its lattice is stronger
so fewer free OHions are produced and so lower pH.

116
Q

Oxides + acids = salts

A

Na2O (s) + 2 HCl (aq)  2NaCl (aq) + H2O (l)
Na2O (s) + H2SO4
(aq)  Na2SO4
(aq) + H2O (l)
MgO (s) + 2 HCl (aq)  MgCl2
(aq) + H2O (l)
Or ionic equations
Na2O (s) + 2H+
(aq)  2Na+
(aq) + H2O (l)
MgO (s) + 2 H+
(aq)  Mg2+ (aq) + H2O (l)

117
Q

aluminium oxide acting as an acid

A

Aluminum oxide acting as a acid
Al2O3( s)+ 2NaOH (aq) + 3H2O (l) –> 2NaAl(OH)4 (aq)

Al2O3 (s)+ 2OH- (aq) + 3H2O (l) –> 2Al(OH)4- (aq)

118
Q

aluminium oxide acting as a base

A

Aluminum oxide acting as a base
Al2O3(s)+ 3H2SO4 (aq) –> Al2(SO4)3(aq) + 3H2O (l)

Al2O3 + 6HCl –> 2AlCl3 + 3H2O

Or ionic: Al2O3 + 6H+ –> 2Al3+ + 3H2O

119
Q

reaction of sodium metal in water

A

-Sodium metal reacts rapidly with water to form a colourless solution of sodium hydroxide (NaOH) and hydrogen gas (H2). The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.

120
Q

several oxidation states

A

Chlorine and sulfur both have more than one type of oxide since they have different oxidiation states

Amphoteric oxides like Al2O3 have several oxidiation states

121
Q

ions

A

Phosphate ion = PO4^3-

Anion of H2SO4 = SO4^2-

122
Q

summary of period 3 reactions

A

Na2O = ionic = dissolves in water = basic (pH 14)

MgO = stronger ionic = slightly soluble = basic (pH 10)

Al2O3 = ionic = insoluble due to high lattice enthalphy = amphoteric

SiO2 = insoluble = giant covalent = acidic

P4O10 = simple covalent = violent reaction in water = acidic (P4O10 + 2OH- –> 4PO4^3- + 6H2O)

SO2 = gas = dissolves = acidic

SO3 = liquid = reacts violently

123
Q

equation for sulfur oxide + water

A

SO2 + H2O –> H2SO3

124
Q

explain, using an equation why silicone oxide is classified as an acid oxide

A

SiO2 + 2NaOH –> Na2SiO3 + H2O

-SiO2 = donate oxide ions + forms salt + water with a base

125
Q

equation for P4O10 + MgO

A

P4O10 + 3MgO –> 3Mg(PO4)2

126
Q

equation for phosphorous with an excess of oxygen

A

P4 + 5O2 –> P4O10

127
Q

describe a test you could carry out to distinguish between sodium oxide and the product of the reaction in part A

A

-add water to each and measure pH with pH meter
-Na = pH 12
-P = pH 1-2

128
Q

state the type of crystal structure of silicone dioxide and sulfur trioxide

A

SiO2 = giant covalent
Sulfur = simple molecular

129
Q

sulfur trioxide + KOH

A

SO£ + KOH –> KSO4 + H2O

130
Q

how covalent character of aluminum oxide arises

A

The aluminium cation is very small.

This means that the aluminium cation is closer to the oxide ion.

The cation is also highly charged enough to distort the electron cloud.

Therefore, the electron cloud appears more covalent.

131
Q

insoluble

A

SiO2

Al2O3

132
Q

formation of hydroxide ion from Na2O

A

Sodium oxide dissociates in water into the
Na+ ion and the
O2- ion. The O2- ion attracts protons very strongly, resulting in the formation of the hydroxide ion.

133
Q

SO3

A

pyramidal
107

134
Q

pH decreases

A

across period

135
Q

silicone dioxide reacts with

A

strong bases

136
Q

Na2SiO3

A
137
Q

sodium hydroxide + phosporic acid

A

3NaOH (aq) + H3PO4 (aq) → Na3PO4 (aq) + 3H2O (l)

138
Q

why is the electrode value not the same as standard electron

A

not in standard conditions

139
Q

what change needs to be done in the diagram to allow the cell to go into completion

A

remove the voltmeter

140
Q

you are given a daniel cell with a known mass of zinc electrode. (0.5) how would you check this

A

-allow zinc to discharge until 0.5
-confirm by colorimic measurement
-weight Zn before and after experiment

141
Q

suggest why the sampling technique has no effect on the concentration

A

remaining concentration of solution remains unchanged

142
Q

why is it important to use excess potassium iodie

A

so all the chlorine reacts

143
Q

give the ionic equation for the reaction between I2 and S2O3^2-

1/2I2 –> I- (reduced)

1/2S4O6 –> S2O3^2- (oxidised)

use the values provided

A

1/2I2 + S2O3^2- –> 1/2S4O6 + I-

144
Q

true or false - when explaining choice of acid from the table write the redox equation

A

true

145
Q

heterogenous =

A

catalyst is in a different phase to the reactants

146
Q

Explain the meaning of the term autocatalyst.
Explain, using equations where appropriate, why the reaction is slow at first and
then goes quickly.

A

autocatalyst = products of the reaction catalyst the reaction

slow as negative ions repel + activation energy is high

attraction between oppositely charged ions / negativereactant ion(s) and positive catalyst / Mn2+ / Mn3

Mn2+ + MnO4– + 8 H+ → 5 Mn3+ + 4 H2O

2 Mn3+ + C2O42– → 2 Mn2+ + 2 CO2

147
Q

suggest the function of the porous separator in the above diagram

A

allows ions to move faster

148
Q

Suggest why the EMF values of the acidic and alkaline hydrogen–oxygen fuel cells
are the same.

A

same overall reaction

149
Q

identify an ionic compound that could be used as a salt bridge

A

potassium nitrate

150
Q

State how, if at all, the EMF of this cell will change if the surface area of the platinum
electrode is increased.

A

no change

151
Q

The voltmeter V shown in the diagram above is replaced by a bulb.
Give an equation for the overall reaction that occurs when the cell is operating.

A

Mg + 2HCl –> MgCl2 + H2

152
Q

Use data from the table to explain why fluorine reacts with water.

A

flourine oxidises water
more positive value = better oxidising agent

2F2 + 2H2O → 4F– + 4H+ + O2

153
Q

Suggest one reason why the cell cannot be electrically recharged.

A

reactions are not reversible

154
Q

A lead–acid cell can be recharged.
Write an equation for the overall reaction that occurs when the cell is being
recharged.

A

2PbSO4 + 2H2O → Pb +PbO2 + 2HSO4– + 2H+lead species correct on correct sides of equation 1equation balanced and includes H2O,HSO4– and H+ (or H2SO4)

155
Q

give one reason why the e.m.f of the lead acid cell changes after several hours

A

reagents/ions are used up

156
Q

explain why voltage remains constant in a fuel cell

A

reagents supplied continiously
concentrations of reagents remains constant

157
Q

Write an equation for the spontaneous cell reaction.

Pt|Fe2+(aq), Fe3+(aq)||Tl3+(aq),Tl+
(aq)|Pt

A

Tl3+ + 2 Fe2+ → 2Fe3+ + Ti+

158
Q

H2O + SO2 + HgO → H2SO4 + Hg

2H2O + SO2 + Cl2 → H2SO4 + 2HCl

A

In Reaction 1, identify the substance that donates oxygen and therefore is the
oxidising agent. = HgO
Hg2+ +2e- –> Hg

Write a half-equation for the oxidation process occurring in reaction 2.

2H2O + SO2 → H2SO4 + 2e–

159
Q

Calculate the e.m.f. of this cell and state, with an explanation, how this e.m.f.
will change if the concentration of Fe3+(aq) ions is increased.

A

emf increases
more Fe3+ ions to accept electrons
becomes more positive

160
Q

Give one economic and one
environmental disadvantage of this method of producing hydrogen.

A

economic = expensive costs o high temperature

environmental = releases CO2

161
Q

Explain how the salt bridge D provides an electrical connection between the two
electrodes.

A

it has mobile ions

162
Q

In the external circuit of this cell, the electrons flow through the ammeter from right
to left.
Suggest why the electrons move in this direction.

A

The Cu2+ ions / CuSO4 in the left-hand electrode more concentrated
So the reaction of Cu2+ with 2e− will occur (in preference at) left-hand electrode/ Cu → Cu2+ + electrons at right-hand electrode

163
Q

Explain why the current in the external circuit of this cell falls to zero after the cell
has operated for some time.

A

eventually copper ions in each electrode will be at the same concentration

164
Q

Suggest why the recharging of a lithium cell may lead to release of carbon
dioxide into the atmosphere.

A

Electricity for recharging the cell may come from power stations burning(fossil) fuel

Aqa a level chemistry (19 decks)

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