Summary EDEXCEL A LEVEL CHEMISTRY UNIT 14 REDOX II NOTES
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Unit 14 - Redox II
Instelling
PEARSON (PEARSON)
I created these notes throughout y12&13 and used them for my A level 2022 exams. I got a grade 9 in GCSE and an A* at a level. These notes were made using the CGP revision guide, edexcel textbook and class notes in order to cover everything in the course. I have also added markschemes from past pap...
14.1 REDOX REACTIONS
Monday, 28 March 2022
16:05
Oxidation is the loss of electrons.
Oxidation number increases.
Reduction is the gain of electrons.
Oxidation number is reduced.
Oxidation and reduction occur simultaneously, known as a redox reaction.
Oxidation numbers show how many electrons an atom has gained or lost to form an
ion.
The charge it would have if it's an ion.
Describes its relative state of oxidation or reduction.
S-block metals tend to react by being oxidised, losing electrons to form positive ions.
Only have one oxidation state, corresponding to their respective group.
P-block metals lose electrons, but non-metals gain electrons to form negative ions.
D-block metals form ions with variable oxidation states, but tend to form positive
ions with positive oxidation numbers.
14.2 BALANCING REDOX EQUATIONS
Monday, 28 March 2022
16:54
Balancing half equations:
Calculate oxidation states on each side of the equation.
Balance the elements which has been oxidised or reduced.
Add electrons in order to balance the oxidation numbers.
Balance the oxygen atoms. For each oxygen atom gained/lost, add/remove a
water molecule.
Balance the hydrogens. For every H gained/lost, add/remove a H+ ion.
Check if the charges on both sides are equal.
Combining half equations:
Multiply the half equations to get the same number of electrons in both half
equations.
Combine the 2 half equations together.
Cancel out electrons or anything which can be.
, 14.3 REDOX TITRATIONS
Monday, 28 March 2022
16:56
In a redox titration, an oxidising agent reacts with a reducing agent.
Transition elements are good at changing oxidation numbers, making them useful as
oxidising and reducing agents as they readily donate or receive electrons.
To work out the concentration of a reducing agent, titrate a known volume of it
against an oxidising agent.
This can be used to find out how many manganate (VII) ions are needed to react with
a reducing agent.
Add a known volume and concentration of Fe²⁺ ions into a conical flask.
Add dilute sulphuric acid to provide a lot of H⁺ ions to allow the oxidising
agent, (MnO₄⁻), to be reduced.
The MnO₄⁻ is a purple colour, so the end point is when the mixture in the flask
becomes purple.
As transition metals change oxidation state, they often also change colour, indicating
when a reaction has finished.
End-point of a titration.
When adding an oxidising agent to a reducing agent, the reaction will continue until
all of the reducing agent is used up.
Next drop will turn mixture into the colour of the oxidising agent.
Core Practical 11
The titration of Fe²⁺ with MnO₄⁻ can be used to calculate the % of iron in iron tablets.
Crush Fe tablets using pestle and mortar, add some H₂SO₄ to make a paste.
Add 100cm³ H₂SO₄ (1.5M) into a 100cm³ flask beaker.
Add paste to beaker and stir to dissolve.
Filter into volumetric flask and remove residue, wash with acid the beaker and
filter paper.
Make solution up to 250cm³ and add stopper to shake to uniformly distribute
molecules.
Pipette a 25cm³ aliquot into a conical flask.
Titrate this Fe(II) aliquot solution with KMO₄ (0.005M) until solution turns
from colourless to pink.
5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Repeat titration until concordant results.
Using mean titre and balanced equation, the % of Fe can be calculated:
Moles of MnO₄⁻ using mean titre x5 is moles of Fe(II) in aliquot.
x10 to obtain moles of Fe(II) in 250cm³.
Calculate mass of Fe.
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