Electrochemical cells + period 3 Flashcards
identify which species can be used to reduce VO2- ions to VO2+
reagent: Fe2+
justification: EƟ VO2+(/ VO2+) > EƟ Fe3+(/Fe2+) > EƟ VO2+(/V3+)
ionisation energy across period 3
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.
melting point across period 3
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
reaction of sodium and magnesium in water
-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.
period 3 oxides
Na2O
MgO
Al2O3
SiO2
P4O10
SO2
SO3
Cl2O
electrical conductivity of period 3 elements
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.
sodium + oxygen
Sodium + oxygen –> 4Na + O2 –> 2Na2O (oxidation state of Na = +1)
magnesium + oxygen
Magnesium + oxygen –> 2Mg + O2 –> 2MgO (oxidation state of Mg = +2)
aluminium + oxygen
4Al + 3O2 –> 2Al2O3 (Al = +3 oxidation state)
silicon + oxygen
Si + O2 –> SiO2 (Si = +4 oxidation state)
phosphorus + oxygen
P4 + 5O2 –> P4O10 (P = +5 oxidation state)
sulfur + oxygen
S + O2 –> SO2 (+4 oxidation state)4 oxidation state_
sodium + water
Sodium + water –> 2Na + 2H2O –> 2NaOH + H2
Na2O
-dissolves and reacts to form a solution
-basic oxide
-pH 14
MgO
-ionic
-slightly soluble
-less soluble than Na2O due to higher lattice enthalpy
Al2O3
SO2
insoluble
high lattic energy (Al)
strong covalent bonds (SO2)
chlorine + water
Cl2(g) + H2O(l) ⇌ HCl(aq) + HClO(aq)
Chlorine + water = dissolves in water to give green solution
amphoteric
Amphoteric –> (of a compound, especially a metal oxide or hydroxide) able to react both as a base and as an acid
H3PO4
H3PO4 = tetrahedral shape (double bond for first oxygen) and ionic is –3 charge
element X
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
element Y
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
element Z
Element Z forms an amphoteric oxide that has a high boiling point:
-bonding = ionic
-element = Al2O3
-amphoteric = reacts to acids and bases
burning of period 3 elements
-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.
summary of equations period 3
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)
metal vs non-metal oxides
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
electrochemical cells
-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
position of EQ
-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
calculating electrode potential
-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
steps to drawing electrochemical cells
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
flow of electrons
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
positive vs negative
More positive electrode = reduction (forwards reaction)
More negative electrode = oxidation (backwards reaction)
voltage
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.
salt bridge
-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
current
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)
which species is more likely to be reduce, Al3+ or Mg2+
Al3+ = oxidising agent = reduced
identify the oxidising agent in this cell
2MnO2 (s) + 2NH4+ (aq) + 2e- –> Mn2O3 (s) + 2NH3 (Aq) + H2O (l)
MnO2 = reduced
which change to a hydrogen electrode has no effect on the electrode potential
the surface area of the platimum electrode
state the meaning of the term electrochemical series
a list of electrode potential values in numerical order
which is the weakest reducing agent in the table above
[Co(H2O06]2+ = most positive number
explain why an aqueuous electrolyte is not used for a lithum cell
lithium would react with the water
the electorde potential is more negative than water as it is a better reducing agent
platinum electrode =
Reaction = 2H+ (aq) +2e- –> H2(g)
as hydrogen gas is used
platinum electrode + MnO4-
reaction = MnO4- (aq) + 8H+ + 5e- –> Mn2+ + 4H2O (l)
state 2 conditions needed for the following half cell to have an electrode potential of zero volts
-temperature at 298K
-pressure at 100Kpa
give the conditions under which the electrode potential of Zn2+/Zn electrode is -0.76V
100kpa, 298K, 1moldm-3 of Zn2+
purpose of the salt bridge =
maintain charge balance
name the substance used as electrode B in the diagram above
platinum
explain how the salt bridge connects the circuit
ions in the ionic substance in the salt bridge move through the salt bridge to maintain charge balance
state why the left-hand electrode does not have an electrode potential of +0.34V
the concentration of the CuSO4 isnt at 1moldm-3 which is the standard conditions its at 0.15moldm-3
The following cell has an EMF of +0.46V
Cu / Cu2+//Ag+/Ag
which statement is correct about the operation of the cell
metallic copper is oxidised by Ag+ ions
use the data from the table to explain why gold jewellery is unreactive in moist air
the EMF of the reaction of Au+ and oxygen is -0.45 so is not spontaneous
oxidisng agent =
electron acceptor
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
Au+ ions oxidise water
Observation = effervescence of O2 gas or gold precipitate
Equation = 2Au+ + H2O –> 2Au + 2H+
is Br2 and oxidising agent
yes not Br-
use you data booklet to calculate the E cell for Al/Al3+ and Mn/Mn2+
Al3+ + 3e- –> Al (more positive)
Mn2+ + 2e- –> Mn (more negative)
3Mn + 2Al3+ –> 3Mn2+ + 2Al
cell diagram
Zn(s) | Zn2+ (aq) | | Cu2+ (aq) | Cu (s) E= +1.1V
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
if a system includes an electrode that is not a metal then a a platinum electorde must be used
hydrogen electrode
The hydrogen electrode equilibrium is:
H2(g) –> 2H+ (aq) + 2e
Pt |H2 (g) | H+ (aq)
disproportionation
-Disproportionation reaction = reduction and oxidation occur in the same reaction from the same element
half equations
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)
salt bridge
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
further down table =
more likely to reduce
reduced vs oxidising agent
Oxidised (reducing agent) Reduced (oxidising agent)
Zn and Cu cell equation
Zn (s) / Zn2+ (aq) / Cu2+ (aq) / Cu (s)
anode vs cathode
Anode = negtive electrode (right side)
Cathode = positive electrode (left side)
Ni2 and H+
Ni (s) / Ni2+ (aq) / H+ (aq) / H2(g) / Pt (s)
Negative –> positive
calculating electrode potential
Left = more negative
Right = more positive
Emf = Eright – Eleft (flow of electrons from 1 half cell to another)
Positive – negative
forwards vs backwards reaction
Forward reaction = more positive = reduction
Backwards reaction = more negative = oxidiation
metals
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-
dynamic equilbirium
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
direction of arrows
Arrow away from the metal/ion = oxidiation
Arrow towards the metal/ion = reduction
small electrode potentials
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)
larger electrode potentials
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
what is a half cell
Half cell = solid lead electrode combined with ts solution with its own
more positive value =
oxidising agent
what affects value of electrode potential
-temperature
-pressure
-solution concentration
–> reference for this standard hydrogen electrode
standard hydrogen electrode
-H2(g) at 100kpa
-298K
-electrode potential = 0.00volts
Concentration = 1moldm-3
negative reading on voltmeter
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.
define standard electrode potential
the potential difference between a standard hydrogen electrode and the half cell with an ion concentration at 1moldm-3, gas at 100kpa and 298K
anode
Anode = oxidation (metal looses electrons so is oxidised into metal ions)
cathode
Cathode = reduction (metal ions gain electrons so is reduced to metal atoms)
More negative electrode (anode) is drawn on the left
reduction - oxidation
electrodes
-iron electrode is dipped into a solution of Fe2+ ions
-if 2 metal ions are present, use a platinum electrode
iron reversible equation
Fe2+ (aq) + 2e- ⇌ Fe (s)
movement of electrons in electrochemical cells
Electrochemical cells = 2 half cells joined together
Electrons flow from more reactive to less reactive metal
anode and cathode
Arrow from electrode to ion = anode
Arrow from ion to electrode = cathode
Anode = oxidiation = loss of electrons
Cathode = reduction = electrode recieves ions
electrode potential
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
ionic equation of Zn and Cu
1) Zn (s) ⇌ Zn2+ (aq) + 2e-
2) Cu2+ (aq) + 2e- ⇌ Cu (s)
3) Zn(s) + Cu2+ (aq) ⇌ Zn2+ (aq) + Cu(s)
concentration of H+ ions
1 moldm-3 of H+ ions in solution
For diprotic acids e.g H2SO4 = 0.5 moldm-3 of H+ ions in solution
more positive vs more negative value
More positive value = stronger oxidising agent
More negative value = stronger reducing agent
cell notation
-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)
predicting reaction feasibility
-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
lithium ion cell
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]
electrolyte
Electrolyte = acts as a conductive pathway for ions to move from one electrode to another
fuel cell
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
true or false - anode = negative
false = anode is positive
advantages and disadvantages of fuel cells
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
when to use platinum strip
when neither side of the electrons half equation has a solid metal to conduct the electrons
EMF when Mg = 2.87 and standard hydrogen electrode is present
2.87
bc H2 = 0.00v
in non standard conditions if equilbirium shifts to the left meaning
more electrons are produced
more negative
why is zinc a stronger reducing agent than copper
turns into its aqueous ions faster
single vs double displacement =
single = always redox
double = never redox
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
No
Zinc is a weaker reducing agent
Ca is a weaker oxidisng agent
salt bridge requirements
-should not react with ions in solution
-as it may alter the concentrations + electromotive force
Daniell cell
zinc + copper
Double A batteries are designed so that the casing is also the negative electrode.
Suggest one advantage and one disadvantage to this design
-advantage = AA batteries can be lighted
-disadvantage = zinc casting becomes thinner = more zinc is oxidised to zinc ions = leakage
importance of ions moving=
balance charge
dry cell =
no free liquid
structure of AA battery
zinc case (negative electrode)
positive electrode in centre
paste instead of free liquid
porous separator
why is carbon rod used
conducts electrcitiy but does not react with other elements
oxidiation state for the reduction of Mn
manganese is reduced from an oxidation state of +4 to +3
porous separator
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
alkaline battery =
brass pin
electrolyte = KOH
zinc = inside
forming oxides reactions
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.
why are metal oxidies ionic
. 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
why dont Al2O3 and SiO2 dissolve
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
why are ionic oxides basic
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.
Oxides + acids = salts
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)
aluminium oxide acting as an acid
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)
aluminium oxide acting as a base
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
reaction of sodium metal in water
-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.
several oxidation states
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
ions
Phosphate ion = PO4^3-
Anion of H2SO4 = SO4^2-
summary of period 3 reactions
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
equation for sulfur oxide + water
SO2 + H2O –> H2SO3
explain, using an equation why silicone oxide is classified as an acid oxide
SiO2 + 2NaOH –> Na2SiO3 + H2O
-SiO2 = donate oxide ions + forms salt + water with a base
equation for P4O10 + MgO
P4O10 + 3MgO –> 3Mg(PO4)2
equation for phosphorous with an excess of oxygen
P4 + 5O2 –> P4O10
describe a test you could carry out to distinguish between sodium oxide and the product of the reaction in part A
-add water to each and measure pH with pH meter
-Na = pH 12
-P = pH 1-2
state the type of crystal structure of silicone dioxide and sulfur trioxide
SiO2 = giant covalent
Sulfur = simple molecular
sulfur trioxide + KOH
SO£ + KOH –> KSO4 + H2O
how covalent character of aluminum oxide arises
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.
insoluble
SiO2
Al2O3
formation of hydroxide ion from Na2O
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.
SO3
pyramidal
107
pH decreases
across period
silicone dioxide reacts with
strong bases
Na2SiO3
sodium hydroxide + phosporic acid
3NaOH (aq) + H3PO4 (aq) → Na3PO4 (aq) + 3H2O (l)
why is the electrode value not the same as standard electron
not in standard conditions
what change needs to be done in the diagram to allow the cell to go into completion
remove the voltmeter
you are given a daniel cell with a known mass of zinc electrode. (0.5) how would you check this
-allow zinc to discharge until 0.5
-confirm by colorimic measurement
-weight Zn before and after experiment
suggest why the sampling technique has no effect on the concentration
remaining concentration of solution remains unchanged
why is it important to use excess potassium iodie
so all the chlorine reacts
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
1/2I2 + S2O3^2- –> 1/2S4O6 + I-
true or false - when explaining choice of acid from the table write the redox equation
true
heterogenous =
catalyst is in a different phase to the reactants
Explain the meaning of the term autocatalyst.
Explain, using equations where appropriate, why the reaction is slow at first and
then goes quickly.
autocatalyst = products of the reaction catalyse 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
suggest the function of the porous separator in the above diagram
allows ions to move faster
Suggest why the EMF values of the acidic and alkaline hydrogen–oxygen fuel cells
are the same.
same overall reaction
identify an ionic compound that could be used as a salt bridge
potassium nitrate
State how, if at all, the EMF of this cell will change if the surface area of the platinum
electrode is increased.
no change
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.
Mg + 2HCl –> MgCl2 + H2
Use data from the table to explain why fluorine reacts with water.
flourine oxidises water
more positive value = better oxidising agent
2F2 + 2H2O → 4F– + 4H+ + O2
Suggest one reason why the cell cannot be electrically recharged.
reactions are not reversible
A lead–acid cell can be recharged.
Write an equation for the overall reaction that occurs when the cell is being
recharged.
2PbSO4 + 2H2O → Pb +PbO2 + 2HSO4– + 2H+lead species correct on correct sides of equation 1equation balanced and includes H2O,HSO4– and H+ (or H2SO4)
give one reason why the e.m.f of the lead acid cell changes after several hours
reagents/ions are used up
explain why voltage remains constant in a fuel cell
reagents supplied continiously
concentrations of reagents remains constant
Write an equation for the spontaneous cell reaction.
Pt|Fe2+(aq), Fe3+(aq)||Tl3+(aq),Tl+
(aq)|Pt
Tl3+ + 2 Fe2+ → 2Fe3+ + Ti+
H2O + SO2 + HgO → H2SO4 + Hg
2H2O + SO2 + Cl2 → H2SO4 + 2HCl
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–
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.
emf increases
more Fe3+ ions to accept electrons
becomes more positive
Give one economic and one
environmental disadvantage of this method of producing hydrogen.
economic = expensive costs o high temperature
environmental = releases CO2
Explain how the salt bridge D provides an electrical connection between the two
electrodes.
it has mobile ions
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.
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
Explain why the current in the external circuit of this cell falls to zero after the cell
has operated for some time.
eventually copper ions in each electrode will be at the same concentration
Suggest why the recharging of a lithium cell may lead to release of carbon
dioxide into the atmosphere.
Electricity for recharging the cell may come from power stations burning(fossil) fuel
2 observations for when sodium reacts in oxygen
-yellow flame
-white powder
phoshporus and oxygen =
white fumes
sodium carbonate = standard hydrogen electrode
carbonate ion will react with acid
H+ + CO3^2- –> HCO3-
state why the left hand electrode doesnt have an electrode potential of 0.34V
conc is not 1.00moldm-3
colour of litmus paper with Na2O
blue
range of melting points =
impure solution
SO3 + 2KOH –>
K2SO4 + H2O
why does Na2O form base
O2- ions react with water to form hydroxide ions
equations of amphoteric Al2O3
Al2O3 + 6Hcl –> 2Al3= + 6Cl- + 3H2O
Al2O3 + 2NaOH + 3H2O –> 2Na+ + 2[Al(OH)4]-
commercial cell =
not in standard conditions
substances and conditions for standard hydrogen electrode
1 moldm-3 of H+ ions
298K platinum electrode
100kpa H2 gas
positive electrode at lithium cell
Li+ + CoO2 + e– → Li^+[CoO2]–
negative electrode at lithium cell
Li → Li+ + e–
discharging a lead acid battery
Discharging a lead–acid battery is the reverse of a disproportionation reaction: lead starts in two different oxidation states and ends up in one oxidation state. The technical term for the reverse of disproportionation is comproportionation, but you absolutely do not need to know that for the exam.
graphite as a support medium in lithium ion cells
-conducts electrons
-allows Li to move
non-rechargable
zinc-carbon + NH4Cl electrolyte
Zn —> Zn2+ + 2e-
2MnO2 + 2NH4+ + 2e- –> Mn2O3 + 2NH3 + H2O
not reversible = H2 not produced as it is quickly oxidised to water
rechargable =
lead-acid
H2 –> 2H+ + 2e-
anode
1/2O2 + 2H+ + 2e- –>
H2O cathode
lithium =
lighter than lead + strong reducing agent so can provide a high voltage
equation for lithium ion battery discharging
Li + CoO2 –> LiCoO2
electrolye in hydrogen feed
aqueous NaOH or KOH
equation for reaction at negative electrode of hydrogen feed
H2 + 2OH- –> 2H2O + 2e-
equation for the reaction at the positive electrode of the hydrogen feed
O2 + 2H2O + 4e- –> 4OH-
summary of alkaline fuel cell
1) initially H2 is oxidised to H+ at the negative electrode
2) O2 is reduced to OH- at the positive electrode
3) H2O is produced at the negative electrode
conventional representation for alkaline fuel cell
Pt I H2 I OH- II O2 I OH- I Pt
The hydrogen electrode is made by coating a porous ceramic material with platinum, instead of using a platinum rod. Suggest why this is the case.
-increase S.A of electrode
-less pt used
Platinum coating is more cost effective than using a platinum rod, which is one advantage. The increased surface area increases the rate of H2 splitting at the electrode, which improves the fuel cell’s performance.
overall equation for alkaline fuel cell
2H2 + O2 –> 2H2O
state why the electrode potential for the standard hydrogen electrode is equal to 0.00v
by definition (under standard conditions)
why are pltinium electrodes made by coating it rather than using a platinum rod
increases surface area so reaction is faster
why is the EMF of a hydrogen oxygen fuel cell in acidic conditions the same of that as an alkaline fuel cell
overall reaction is the same
2H2 + O2 –> 2H2O
oxidation state of O2
-4
oxidation state of O2-
-2
half equation for he reaction occuring at negative electrode during lithium ion cell
Li (s) –> Li+ (aq) + e-
2 properties of platinum
-inert (unreactive)
-conducts electricity
why is water not used as a solvent in lithium ion cell
water would react vigirously with lithium
cell notation for lithium ion cell
Li I Li+ II Li, CoO2 I LiCoO2 I Pt
conventional cell representation for ClO3- an Cl- in aqueous
and SO4^2- and SO3^2- in aqueous
Pt (s) I SO3^2- (aq), SO4^2- II ClO3- (aq), Cl- (aq) I Pt (s)
one essential property of the non-reactive porus separator labelled in the diagram
-allows ions to pass
suggest the function of the carbon rod in the cell
allows electrons to flow
Suggest why a cell often leaks after being used for a long time
Zn is used up (has all reacted)
one environmental advantage of this rechargable cell compared with non-rechagable cell
supplies are not depleted
equation for ethanol-oxygen fuel cell
half equation
C2H5OH + 3O2 –> 2CO2 + 3H2O
C2H5OH + 3H2O -> 2CO2 + 12H+ + 12e-
why is ethanol considered to be carbon neutral
CO2 released by combustion is taken up in photosynthesis
why would potassium chloride not be a suitable salt bridge
Cl- would react with Cu2+
in the external circuit of the cell, the electrons flow through the ammeter from right to left. suggest why the electrons move in this direction
conc of Cu2+ on the left is greater so a reduction is preferred on the left
suggest why the recharging of a lithium cell may lead to release of carbon dioxide in the atmosphere
electrictity may come from burning fossil fuels in power stations
half equations for the electrode reactions in the hydrogen-oxygen fuel cell
H2 –> 2H+ + 2e-
O2 + 4H+ + 4e- –> 2H2O
(hydrogen produces electrons)
(oxygen accepts electrons)
suggest the main advantage of using hydrogen in a fuel cell rather than in an internal combustion engine
greater proportion of energy available from hydrogen-oxygen reaction is converted into useful energy
conventional representation for alkaline hydrogen oxygen fuel cell
Pt (s) I H2 (g) I OH- (aq), H2O (l) II O2 (g) I H2O (l), OH- (aq) I Pt (s)
anode
-anode = oxidation (negative electrode) on the left
cathode
-cathode = reduction (positive electrode) on the right
using comma in conventional cell representation
-if elements are in the same state use comma e.g Li+ (aq), CoO2 (aq)
key notes
-ions move from left to right in electrochemical cell
-Salt bridge allows IONS to move
-Oxidation = flip equation shown on EMF table e.g Zn (s) –> Zn2+ (aq) + 2e-
acidic hydrogen fuel cell
-electrons and hydrogen move from left to right
-H2 –> 2H+ + 2e- (on the left)
-O2 + 4H+ + 4e- –> 2H2O (on the right)
equation for hydrogen-oxygen fuel cell
Complete equation = 2H2 + O2 –> 2H2O
alkaline hydrogen fuel cell
-electrons move from left to right
-OH- move from right to left
-H2 + 2OH- –> 2H2O + 2e- (left)
-2H2O + O2 + 4e- –> 4OH- (Right)
why is Cd(OH)2 the weakest oxidising agent
-has the most negative electrode potential
overall reaction in ethanol-oxygen fuel cell
3O2 + C2H5OH –> 2CO2 + 3H2O
explain why rechargable cells are connected to these solar cells
solar cells dont supply electrical all the time
rechargable cells can store electrical energy for use when solar cells arent working
suggest one reason why many waste disposal centres contan a separate section for cells and batteries
prevent pollution of the environment by toxic substances
explain why such a salt bridge would not be suitable for use in this cell
carbonate ion reacts with the acid. The EMF of the cell would be altered as Al3+ concentration changes
redox reactions of aqueous iron with excess of aqueous silver nitrate
silver ions oxidised Fe2+
Fe2+ further oxidised to Fe3+ since its in excess
when describing standard hydrogen electrode make sure to include
platinum electrode
calculating percentage of iron ions…
ensure to write balanced equation first
explain why the current in the external circuit of this cell falls to zero after the cell has operated for some time
eventually copper ions will be at the same concentration
function of platinum electrode
allows the transfer of electrons
electrically recharged =
reactions are reversible
contstant EMF on graph =
fuel cell
conventional represenation for standard electrode potential of iron
Fe I Fe2+ I H+ I H2 I Pt
suggest one reason other than cost why this type of cell is not recharged
reaction cannot be reversed
suggest the major advantage of using the fuel cell
internal combustion engine wastes more heat energy
why can lithium cells be recharged
reversible reaction so reagents are continiusly supplied
