topic 14 Flashcards
what does oxidation result in, in terms of oxidation no.?
oxidation results in the oxidation number becoming more positive
what does reduction result in, in terms of oxidation number
reduction results in oxidation number becoming more negative
which blocks of the periodic table tend to undergo oxidation
s-block, d-block
which block on the periodic table tends to undergo reduction
p-block elements (further to the right of the periodic table) tend to undergo reduction
what is a redox reaction
- occur in electrochemical cells
- involves both reduction and oxidation
- one species oxidised, one species reduced
what is a reducing agent
loses e- as it itself is oxidised and vice versa for oxidising agents
what is an electrochemical cell
an electrochemical cell is a device capable of generating a potential difference from redox reactions
is created by joining 2 different half cells together
why do electrochemical cells use redox reaction
- use redox reactions since the electron transfer between products creates a flow of electrons
- this flow of charged particles is an electrical current that flows between electrodes in a cell
what is a potential differece
produced between the two electrodes which can be measured using a voltmeter
what are electrochemical cells made up of
two solutions, metal electrodes and a salt bridge
what is a salt bridge
a tube of unreactive ions that can move between the solutions to carry the flow of charge, whilst not interfering with the reaction
draw and label an example of an electrochemical cell set up
- the position of the lamp is where the voltmeter can be placed to measure the potential difference between the two half cells
- this is called EMF or Ecell
what happens when you connect two half cells
- one side undergoes reduction and the other undergoes oxidation → redox
- electrons will flow from a more reactive metal to a less reactive metal
what observation can you make for the electrode that is made of the more reactive metal
the electrode wil become thinner as more of its positive metal ions will be produced which produces electrons to be transferred to the less reactive metal
this half cell would be undergoing oxidation
what observation can you make for the electrode that is made of the less reactive metal
this half cell would accept electrons from the more reactive metals half cell meaning that its undergoing reduction
the more reactive metal electrode will become thicker as its metal ions will receive electrons to produce its solid metal which will build up at the bottom of the electrode
what is the function of the salt bridge
- allows ions to flow through which balances the charges in the cell
- it completes the circuit
how do you set up an electrochemical cell
- obtain the metals under investigation and clean them with sandpaper → remove impurities
- metal used as electrode
- wash the surface of the metal with propanone as some metals have grease on the surface → wear gloves moving forward to prevent contamination
- place the metal into a solution containing the ion of the same metal
- e.g Cu electrode would go into a beaker containing CuSO4 solution
- this will produce the copper ions needed
- if youre using an oxidising agent that contains oxygen you need to add an acid too → needs to be acidified
- make the salt bridge from filter paper soaked in saturated KNO3 or KCL → should be linking the 2 beakers and each end submerged in each beaker but not touching the electrode
- connect the electrodes with wires, crocodile clips and a voltmeter→ a reading will appear will appear if set up correctly
- measures potential difference between two electrodes
what is a half cell
one half of one electrochemical cell
how can a half cell be made
can be made up of a metal dipped in its ions or a platinum electrode with 2 aqueous ions
how does an electrochemical cell work
electricity goes up the electrode, into the voltmeter and then down into the next beaker
what is electrode potential
- each half cell has an electrode potential value which is measured in volts
- tells us how easily the half cell gives up electrons (oxidised)
how do you get the overall cell equation
- always show electrons on left hand side of the equation
-reduction in forwards direction so if its an oxidation reaction you flip the equation to electrons are on the left- combine the two equations to get the overall cell equation
what is standard electrode potential and its symbol
the voltage produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions
E°
what is the standard hydrogen electrode used for and why is it used
is used as a reference on all half-cell potentials as it has a standard electrode potential of zero
what does it mean if the standard electrode potential is positive
substances are more easily reduced and will gain electrons
what does it mean if the standard electrode potential is negative
- substances are more easily oxidised and will lose electrons
- to become more stable
what acronym can be used to remember which half cell is undergoing oxidation or reduction
NO PRoblem
the most negative half cell will undergo oxidation
the most positive half cell will undergo reduction
what is the standard hydrogen electrode
used as a reference to measure standard electrode potentials
it has a cell potential of 0.00V measured under standard conditions
what are the standard conditions for the standard hydrogen electrode
- solutions of 1.0 moldm-3 conc.
- to get 1 moldm-3 of H+ ions you need 1moldm-3 of HCL or 0.5 moldm-3 H2SO4
- a temperature of 298K
- 100kPa pressure
what is the standard hydrogen electrode made up of
- the cell consists of a hydrochloric acid solution, hydrogen gas and platinum electrodes
- platinum electrodes are chosen as they are
- metallic → will conduct electricity
- inert → wont interfere with the reaction
- platinum electrodes are chosen as they are
draw the set up for the standard hydrogen electrode on its own
draw the set up for measuring electrode potentials of half cells against the SHE which is used as a reference
what is the electrochemical series
the electrochemical series is a list of half cell reactions and their standard electrode potentials
what is a strong reducing and oxidising agent
strong reducing agent → easily loses e-
strong oxidising agent → gain electrons more easily
which species would be the most powerful reducing agent
most negative standard electrode potential value
which species would be the most powerful oxidising agent
most positive standard electrode potential value
how do you calculate standard cell potential using standard electrode potential
f there are two positives or two negatives it is the most negative that is oxidised
how can electrode potential values change
if the conditions deviate away from the standard conditions
equilibrium position changes depending on reaction conditions
what conditions a half cells electrode potential affected by
- temperature
- concentration
- pressure
why are standard conditions used to measure electrode potentials
ensures that we get the same values which allows us to make comparisons
how are cell represented (cell notation)
the most negative half cell potential goes to the left of the double line
- single solid lines show a physical state change
- double line represents the salt bridge between the beakers
how would you represent a cell with 2 aqueous ions using cell notation using the example of magnesium and iron and its ions
- follows the same rules for cell notation
- but separate the two ions with a comma and NOT a line
- this is because they are in the same physical state
also include the platinum electrode when using 2 non metals as a half cell
- this is because they are in the same physical state
how can you predict whether a reaction is feasible using the electrochemical series
1) identify which is being oxidised
2) take the oxidised equation and reverse it so that electrons are on the rhs → write the 2 equations next to each other
number of electrons in each equation has to be the same in order to combine them so multiply if needed
3) combine the 2 equations to obtain the feasible reaction
4) compare this equation to the reaction stated in the equation
5) can confirm by calculating standard cell potential. all feasible reactions will have a positive standard cell potential value (Eocell)
how can you predict whether a disproportionation reaction is likely to proceed
same steps for predicting feasibility
the ion that youre predicting to disproportionate should be on the left hand side and be shown to be disproportionated
why may a reaction that is calculated to be feasible using standard electrode potentials not go ahead
non standard conditions -> e.g increasing conc. of a species on the right hand side will cause for the equilibrium to shift to the left
- kinetics may not be favourable (kinetic inhibition)
- the rate of reaction is so slow that it appears that there is no reaction
- if the reaction has a high activation energy due to the reactants being kinetically stable
what is the relationship between cell potential and total entropy change of a reaction
the larger the cell potential, the larger the total entropy change during the cell reaction
what is the relationship between the equilibrium constant as cell potential
what are batteries used for
to store energy
how can you establish the overall reaction in a lithium ion battery
need to know the half equations at each electrode
Li has the most negative E° value → oxidation occurs here
the overall equation on discharge (equations above combined):
Li + CoO2 -> Li+[CoO2]-
work out E°
how do rechargeable batteries work
- rechargeable batteries work by plugging them in to supply a current
- this current forces electrons to flow in the opposite way
- forces equilibrium to shift the other way
- reverse the overall discharge equation to show a battery recharging
how is electricity generated
electricity is generated by a continuous supply of chemicals rather than a ‘ready store’ like in batteries
explain this diagram of an alkaline hydrogen-oxygen fuel cell
- hydrogen feed (numbers correspond to those on the diagram)
hydrogen is fed in here
it reacts with OH- ions in solution in 9 (on the diagram)
2H+(g) + 4OH-(aq) → 4H2O(l) + 4e- - flow of electrons
electrons produced in reaction 1 travel through a platinum electrode → good conductor of e- but inert - component
the flow of electrons is used to power something e.g a car - oxygen feed
oxygen is fed here. it reacts with the water and the 4e- made from step one to make OH- ions
O2(g) + 2H2O(l) + 4e- → 4OH-(aq) - negative electrode
electrons flow to the negative electrode which is made from platinum (just like the anode) - electrolyte
the electrolyte is made from KOH solution → carried the OH- ion from the cathode to the anode
positive electrode
electrons flow from the positive electrode
- water eliminated
the product of the reaction in step 1 is released into the surroundings - movement of OH- ions
OH- ions produced from reaction 4 are carried towards the anode via the electrolyte
what is an electrolyte
electrolytes are substances that have a natural positive or negative charge when dissolved in water
what are ion exchange membranes
- these line the platinum electrodes and allow OH- ions to pass through but not hydrogen and oxygen gas
- this is important because you want to utilise the electrons that are generated
how does the hydrogen-oxygen fuel cell work in acidic conditions
- this reaction can also work in acidic conditions
- instead of OH- ions you have H+ ions moving across a polymer electrolyte membrane
what other types of fuel cells are there
some fuel cells use hydrogen rich molecules such as methanol and ethanol
how do newer fuel cells that use hydrogen rich molecules work
some newer fuel cells now use the fuel directly without the need to convert hydrogen first. this is how it works:
- the alcohol is oxidised at the anode with water present
CH3OH + H2O → CO2 + 6e- + 6H+ - the H+ ions pass through the electrolyte and are oxidised to water
6H+ + 6e- + 1.5O2 → 3H2O
how do you find the conc of MnO4- by titrating against a reducing agent like Fe2+
- reducing agent Fe2+ in a conical flask → has a known volume and unknown concentration
a. add excess dilute sulfuric acid into this as well → to ensure that you have sufficient H+ ions to allow the reduction of the oxidising agent - oxidising agent in burette with a known conc e.g manganate (VII) ions
- add MnO4- ions in the burette to the conical flask until you see the faint colour of MnO4- appear → end point
- add drop by drop near end pt
- sharp colour change of MnO4- → purple. theyll immediately react with the reducing agent until the reducing agent is used up
- read how much oxidising agent has been added. read from the bottom of the meniscus and at eye level + record results that are concordant
how do you find the conc of an oxidising agent
- to find the conc. of oxidising agent, reverse the method used to find the conc of reducing agent (oxidising agent in conical flask)
how do you fnd the conc of a reagent using a titration
example using KMnO4 and iron II sulphate
write out equation and balance it
calculate number of moles of MnO4- ions
use equation to find out molar ratio in order to work out the moles of Fe2+
calculate conc. →moles/vol
how can you use redox titrations to find the percentage of iron in iron tablets
mass of iron/ mass of tablet x 100 = % mass of iron in tablet
what are the 3 steps involved in finding out the conc of IO3- in an iodine-sodium thiosulphate titration
- use oxidising agent (KIO3) to oxidise iodide ions to iodine
- carry out a titration to work out moles of iodine produced in step 1
- use moles of iodine in step 2 to work out conc. of IO3-
what is step one of finding out the conc of IO3- in an iodine-sodium thiosulphate titration
measure out volume e.g 25cm3 of KIO3 which will produce the IO3- ion
add excess acidified potassium iodide solution (KI) to the KIO3 solution
- the I- ions are oxidised to I2.
- the more conc. the oxidising agent, the more I- ions oxidised
what is step 2 of finding out the conc of IO3- in an iodine-sodium thiosulphate titration
- add solution from step 1 into conical flask
- add sodium thiosulphate into the conical flask and look out for a pale yellow colour
- since the colour change is difficult to see - add 2cm3 of starch
- turns blue/black in the presence of iodine
- keep adding until blue colour disappears → means that all of the iodine has reacted
- so the volume of sodium thiosulphate added can be used to work out the number of moles of iodine
what is step 3 of finding out the conc of IO3- in an iodine-sodium thiosulphate titration
what errors can occur in a redox titration
- ensure that the starch indicator is added when most of the iodine has reacted otherwise the blue colour would take a long time to disappear
- make starch solution only when youre ready to use it
- the copper (I) iodide precipitate makes it difficult to see the colour of the solution
- keep solution as cool as possible → iodine produced in the reaction can evaporate readily at room temp
- lead to false titre
what should you remember when drawing an electrochemical cell
when drawing : the half cell that’s being reduced is on the left and the half cell being oxidised should be on the right
recall the equation involving faradays constant to find ΔG
ΔG = -nFE⦵
n= moles of electrons (number in front of e-)
F- faradays constant
E⦵ - of the cell
how can the equation to calculate gibbs free energy be used to find out whether a reaction is feasible
/ how can it be used to find out whether the cell potential will be positive or negative
if E⦵ is positive it makes the whole equation positive, meaning that ΔG is negative
making the reaction feasible
if ΔG is zero, cell potential will be 0
if ΔG is negative, cell potential is positive
recall the reactivity series
What happens if half cells are set up the wrong way around where the reduced half cell is on the left rather than the oxidised?
The reaction wont be feasible if the half cells are set up the wrong way around
How are half cells set up
Oxidised half cell on the left
Reduced half cell on the right
How do you calculate emf when given cell set up
follow the way that the cell is set up with the values
So put the values in the same order that the cells are in
what are the advantages of using an ethanol fuel cell rather than a hydrogen fuel cell
easier to transport
renewable
easier to store
less flammable
carbon neutral
what is a disadantag of using a hydrogen fuel cell compared with direct combustion of hydrogen
high cost
short life span
cost of catalyst
what is the main advantage of using a fuel cell over direct combustion of hydrogen
fuel cell is more efficient
less heat loss
releasing energy in a more controlled manner
what is the main advantage of hydrogen as a fuel compared to fossil fuels
water is the only product
what is a disadvantage of using a hydrogen- oxygen fuel cell compared with a rechargeable battery when providing electrical energy for a motor vehicle
hydrogen and oxygen have to be stores under pressure
leakage of hydrogen
hard to transport
hydrogen is flammable
requires continual replenishment of H2 and O2
why are platinum electrodes made by coating porous material with platinum rather than using platinum rods
to increase the surfae area
why is the Ecell for a hydrogen-oxygen fuel cell operating in acidic conditions identical to that of an alkaline fuel cell
reaction is the same
Which electrode does oxidation occur at usually
anode
Which electrode does reduction occur at
Cathode
When calculating Ecell of the reaction, if you’re given two electrode potentials that are in the forward direction (reduction), what should you do to the electrode potential of the species that’s being oxidised
For the electrode potential of the species being oxidised put a minus infront of it because electrode potentials are always given with reduction in the forward direction
So by putting a minus infront of it, it would be reversed so that it would give the electrode potential of the oxidation reaction
which side is the standard hydrogen electrode always drawn at
left hand side
