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
