Chapter 23 - Redox reactions Flashcards
what is reduction (2) and what is oxidation (2)
Reduction is:
- gain of electrons
- decrease in oxidation number
oxidation is:
- loss of electrons
- increase in oxidation number
what is the oxidising agent and what is the reducing agent
the oxidising agent is:
- reduced
- the thing that gains the electrons from the thing it’s oxidising
- the stronger the oxidising agent, the better it is at gaining electrons
the reducing agent is:
- oxidised
- the thing that loses electrons to the thing getting reduced
- the stronger the reducing agent, the better it is at losing electrons
what is the standard method for forming ionic equations and what do ionic equations show
ionic equations only show the ions that react
1) write out the full equation
2) balance
3) cancel any spectator ions
what do half equations show
they show what happens to the electrons specifically
e.g. Cu2+ + 2e- —> Cu
what is the method for constructing a full equation from half equations under acidic conditions
1) write out the half equations
2) add any missing species,
- add H2O to one side of the equation to balance any oxygens
- then add any hydrogen ions required to balance the hydrogens
3) balance any charges by adding a number of electrons
4) scale such that the numbers of electrons produced/used balance
5) write out full equation and cancel any species on both sides
6) check everything balances
what is the method for constructing a full equation from half equations under alkaline conditions
1) write out the half equations
2) add missing species:
- balance oxygens with H2O on one side
- balance hydrogens with H+ on the other as always
3) but because its under alkaline conditions we can’t have H+ ions so we need to add OH- ions ON BOTH SIDES to cancel them
4) form any water molecules from H+ and OH- ions
5) scale accordingly so that the electrons balance
6) add and cancel
7) check everything works
how can we use oxidation numbers to balance equations
1) write out equation
2) identify changes in oxidation number
3) balance such that the sum of the oxidation number changes is 0
4) balance anything else that needs balancing
what is the reduction of manganate equation
MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O
Mn(VII) —> Mn(II)
what is the procedure for a manganate (VII) titration
1) add standard solution of KMnO4 to the burette
2) add other solution to conical flask using pipette and add an excess of sulfuric acid but no indicator
3) as the MnO4- is added, it is reduced and turns colourless, so the end point is when the first permanent pink colour occurs
why do we dissolve iron (II) sulfate in sulphuric acid before titrating with KMnO4
to ensure there are excess H+ ions as KMnO4 reduces using H+
where should we read the miniscus from in manganate titrations
the top instead of the bottom
what can we use our results to the iron (II) sulphate titration with MnO4- to do
- calculate the percentage purity of iron in the tablet
- to calculate the number of moles of water when the salt is hydrated
what is another method we can use to calculate the number of moles of waters of crystallisation instead of the titration
- weigh the hydrated salt
- heat in a crucible
- weigh the anhydrous salt
- do some calculation
what is the most common thing used to reduce the manganate ions
iron (II), it is oxidised to iron (III)
Fe2+ —> Fe3+ + e-
what occurs in an iodine-thiosulfate reaction, give the equations
thiosulfate is oxidised:
2S2O3(2-) —-> S4O6(2-) + 2e-
iodine is reduced:
I2 + 2e- —-> 2I-
give the overall eqaution for the thiosulfate-iodine titration
2S2O3(2-) + I2 —> S4O6(2-) + 2I-
what else can we use the thiosulfate-iodine titration to work out
- the ClO- content in household bleach
- the Cu2+ content in copper (II) compounds
- the Cu content in a copper alloy
what is the procedure for the iodine-thiosulfate titration
1) add a standard solution of Na2S2O3 to the Burette
2) prepare a solution of oxidising agent, add to a conical flask using a pipette, then add an excess of potassium iodide, the oxidising agent e.g. ClO- will react with the iodide ions and form iodine, the solution is orange-brown
3) titrate it against the S2O3(2-), this reacts with the iodine forming iodide and it gradually fades from brown to yellow so no clear end point
4) to solve this, when the end point is close, add starch
- blue-black forms immediately
- as I2 reforms I- it fades to almost colourless
how can we use the iodine - thiosulfate titration to analyse chlorate
ClO- + 2I- + 2H+ —-> Cl- + I2 + H2O
then the reaction as usual where 1mol I2 reacts with 2 mol of S2O3(2-)
so
1mol (ClO-) = 2mol(S2O3(2-))
how can we use the iodine-thiosulfate titration to analyse copper
- dissolve copper salts in water or acid
- react copper alloys with HNO3 then neutralise
2Cu(2+) + 4I- —> 2CuI + I2
then reaction as usual so
2mol(Cu2+) = 1mol(I2) = 2mol(S2O3(2-))
so 1mol(Cu2+) = 1mol(S2O3(2-))
what is a half cell
what is a voltaic cell
- a half cell contains the species present in a half equation
- it converts chemical energy to electrical energy and contains two half cells
how does a metal-metal ion half cell work
- metal rod partially submerged in a solution of its metal ions
- an equilibrium of
metal ion(x+) + ye- metal (x-y)+
how can we represent a phase boundary in a half cell
using a vertical line
e.g.
Zn | Zn2+
when do electrons flow in a voltaic cell
when two half cells of different equilibrium positions (electrode potentials) are connected
what are ion-ion half cells
- made of mixtures of different ions of the same element
- e.g. Fe(3+) + e- Fe(2+)
- it uses an inert platinum electrode in order to transport electrons in and out of the half cell
generally what happens in a full voltaic cell
- the more reactive metal loses electrons and is oxidised at the negative electrode (anode)
- the less reactive metal gains electrons and is reduced at the positive electrode (cathode)
how do we measure the something’s tendency to lose electrons and what kit do we use to do this
- measured using a standard electrode potential (E)
- we use a hydrogen half cell
- this is a tube where hydrogen gas is pumped in, partially submerged in a solution containing H+ ions with an inert platinum electrode to transport electrons
- it is all under standard conditions:
- 298K
- 101Kpa
- [H+] = 1moldm^-3
what do we need to watch out for when making sure solutions are at 1moldm^-3 in standard conditions
- the ionic ratios may not be 1-1
e. g. H2SO4 we only need 1/2 mol of to get 1moldm^-3 of H+ ions in solution
what is the value for a standard hydrogen electrode
0V
define standard electrode potential
“The standard electrode potential is the EMF of a half-cell connected to a standard hydrogen half-cell under standard conditions of 298K, 101Kpa, and 1moldm^-3 solitutions”
how can we measure a standard potential
1) set up a hydrogen half cell
2) set up another half cell
3) connect using wires and a voltmeter in parallel
4) link using a salt bridge
5) measure EMF
what is our salt bridge
usually filter paper soaked in KNO3(aq)
what can we analyse from our standard electrode values
- MPBOA = more positive, better oxidising agent
- a greater E value implies better oxidising agent so more easily reduced so gains electrons more easily so equilibrium sits further to the right
how can we measure cell potentials
- use the same kit as for measuring standard electrode potentials
- prepare 2 standard half cells
- connect and record EMF
what is the E value for a cell
Modulus(DeltaE)
how can we generally make predictions about redox systems based off of E values
- the most negative system has a tendency to be oxidised
- the most positive system has a tendency to be reduced
- therefore if the reactants line up correctly it works
you want it to go anticlockwise when the more negative is on top
if you have a table where the E values increase down the table which systems will react with each other
bottom reacts with both above it, where bottom goes to the right
middle only one above where it goes to the right
top none
what is the limitation based on reaction rate of using E values to predict feasibility
- even though a reaction can be feasible, sometimes its activation energy is too high so it has a very slow rate of reaction
what is the limitation based on concentration when using E values to predict feasibility
- E values only apply to standard conditions (1moldm^-3)
- actual concentrations may vary, this means positions of equilibrium may change according to le chatalier’s and therefore change how they react
what do you need to form cells and what are the types of cell
- to form a cell you need two electrodes with different electrode potentials
- types are primary, secondary and fuel cells
what is the key feature of a primary cell
- non-rechargable
- reactions that take place in the cells are non-reversible
what is the key feature of a secondary cell
- rechargeable
- reactions that take place inside the cell are reversible
what is the most common type of secondary cell, what are the equations and where do they happen
lithium ion
at negative electrode (graphite coated in lithium)
Li —> Li+ + e-
at positive electrode (CoO2)
CoO2 + Li+ + e- —> LiCoO2
what are the risks of lithium ion batteries
- flammable
- toxic materials
- can explode
- short life cycle
- don’t work well at high temperatures
what is a fuel cell
- works very similarly to a normal cell but uses the energy from the reaction of a fuel with oxygen instead
- fuel and O2 flow into the cell, the electrolyte remains in the cell, water flows out
what are some features of a fuel cell
- can run continuously providing the fuel and O2 are constantly provided
- do not need to be recharged
what is the most common sort of fuel cell and what are the two types
hydrogen fuel cell, acidic and alkaline
what are the redox systems for an alkaline hydrogen fuel cell
2H2O + 2e- —– H2 + 2OH- E = -0.83V
1/2O2 + H2O + 2e- —– 2OH- E = + 0.4V
what are the redox systems for a hydrogen fuel cell under acidic conditions
2H+ + 2e- —- H2
1/2 O2 + 2H+ + 2e- —- H2O
what to remember if there’s any mention of copper with an iodide salt like potassium iodide or a white precipitate
2Cu2+ + 4I- —-> 2CuI + I2
which is the positive electrode and which is the negative electrode
- the positive electrode is where electrons are used
- the negative electrode is where electrons are produced
what are two advantages and two disadvantages of using hydrogen fuel cells over fossil fuels
ADV:
- greater efficiency
- non-polluting, only H2O formed
DISADV:
- H2 is difficult to store
- H2 is difficult to initially manufacture
if you are given an overall equilibrium reaction with an E value, how can you tell which species is the better oxidising agent
- if the value of E is positive then you know equilibrium sits to the right, hence the species on the left is the better oxidising agent as it is the one being reduced (more easily)
