Module 5 - chapter 23 P1 Flashcards
reduction in terms of electrons and oxidation number
reduction is the gain of electrons or decrease oxidation number
oxidation in terms of electrons and oxidation number
oxidation is the loss of electrons or increase oxidation number
oxidising agent
takes electrons from the species being oxidised (the oxidising agent itself gains electrons so contains the species being reduced)
reducing agent
adds electrons to the species being reduced (the reducing agent itself loses electrons so contains the species being oxidised)
key points for balancing redox equations
- balance electrons (multiply equations until same e-)
- balance for charge
- balance for species (incl H+ and OH-)
how-to balancing ionic equations in acidic conditions
1) write out half equations
2) balance half equations for atoms
3) add water on one side to balance for oxygens if necessary. balance the other side for H+ after this.
4) add electrons to balance charges on both sides
5) balance half equations against each other so the same no. e- are required and produced by multiplying/dividing.
6) merge equations and cancel down. NO ELECTRONS SHOULD REMAIN OR YOU’VE BALANCED WRONG
7) check charge balances, if not something’s wrong
how-to balancing ionic equations in alkaline conditions
1) write out half equations
2) balance the half equation containing a hydroxide with OH- instead of H2O and H+
3) balance the other half equation for acidic conditions (using H2O and H+)
4) add electrons to balance charges on both sides of both equations
5) balance half equations against each other so the same no. e- are required and produced by multiplying/dividing.
6) neutralise any H+ with the same number of OH- to make the environment alkaline. add the same number of OH- to both sides so remains balanced
7) cancel out H+ and OH- to form water
8) combine equations and cancel down
9) check charges balance and no electrons remain
how-to balance a redox equation using oxidation numbers
1) write out the unbalanced equation
2) write out all oxidation numbers
3) identify changes in oxidation numbers and write clearly so you can see how many to balance.
e. g. the element, initial oxidation no., final oxidation no., and the difference between them
4) identify how the oxidation numbers need to balance and what needs to multiply to match them.
5) balance only the atoms which change oxidation state by balancing them within their compounds
6) balance any atoms which have changed due to your balancing
what two redox titrations do you need to know?
potassium manganate (VII) under acidic conditions and sodium thiosulfate for determination of iodine
key points for manganate (VII) titrations
- potassium manganate (VII) KMnO4 in burette
- excess of dilute H2SO4 added to conical flask to provide H+ for reduction of MnO4- ions to MN2+
- reaction is self-indicating
- manganate decolourises when added
- as end-point approached, decolourises less rapidly
- end-point = pale permanent pink colour (excess MnO4- ions)
- swirl flask so end-point not judged too soon
- meniscus read from top
why is the meniscus read from the top in manganate titrations?
KMnO4 is a deep purple colour, bottom of meniscus difficult to read through intense colour.
titre has same difference between both readings if bottom of meniscus used for initial and final readings
examples of reducing agents analysed with manganate titrations
iron (II) ions ethanedioic acid (COOH)2
determining percentage purity
1) calculate the titre in dm3 and use to find moles of the ion added from the burette (MnO4-)
2) work out the moles of the unknown that reacted, using the previous n
3) scale up for moles in full 250cm3 (if used 25cm3 sample in conical flask eg)
4) use moles to find mass of pure sample
5) find percentage purity by comparing to experimental mass (impure)
%purity = mass of pure/mass of impure x100
- when calculating Mr to work out mass from moles, include Mr of any ions in compound, e.g. H+ may be included to make it neutral (in acidified solution)
key points about iodine-thiosulfate titrations
- S2O3 2- are oxidised to S4O6 2- and iodine (I2) is reduced to 2I-
- sodium thiosulfate = Na2S2O3 in burette
- excess of potassium iodide added to oxidising agent in conical flask
- iodide oxidised to iodine = yellow brown
- iodine reduced back to I- in titration
- end-point = pale straw colour and starch added to check
uses of iodine-thiosulfate titrations
analysing:
ClO- content of bleach
Cu2+ content in copper (II) compounds
the Cu content in copper alloys
how starch helps determine the end-point in iodine-thiosulfate titrations
starch is added once the end-point is being approached and a pale straw colour is observed. adding starch makes it go blue-black because iodine tests for starch.
titrate until it becomes creamy white - when only iodide ions remain
what type of titration are iodine-thiosulfate titrations?
back titrations - reacted with a known conc of excess reagent (KI) and then titrated against a second reagent to show how much the initial one was in excess, so the unknown concentration can be calculated
how copper alloys are prepared for iodine-thiosulfate titrations
salts are just dissolved in water but insoluble compounds can be reacted with conc acids to form CU2+ ions
copper alloys e.g. brass and bronze dissolved in conc HNO3 then neutralised with sodium carbonate to form ions, neutralises copper nitrate - nitrate ions are oxidising agents so interfere with end-point by forming iodine from iodide
voltaic cell
converts chemical energy into electrical energy
half cell
contains species present in half equation
Where is equilibrium set up in a half cell?
At the phase boundary where the metal is in contact with its ions
In isolated half cell?
No net transfer of electrons either unit or out of the metal
What are ion/ion half cells?
Contain ions if the same element in different oxidation states, e.g. Fe (II) and Fe (III)
Uses inert platinum electrode
Anode?
Negative electrode - most reactive metal which loses electrons and is oxidised
