Unit 27 - Chemical Periodicity and Its Applications
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(D3) Unit 27 - Chemical Periodicity and Its Applications
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Unit 27 - Chemical Periodicity and Its Applications
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PEARSON (PEARSON)
Hi, this document covers D3 of unit 27.6 of the 2010 specification of the applied science course. It has been marked as achieved by my teacher. Please use this document for guidance purposes only due to plagiarism.
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Unit 27 - Chemical Periodicity and Its Applications
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27.6
D3
In this titration the volume of chlorine within was bleach was identified. A standard solution of potassium
dichromate was mixed with water. This will then allow it to standardise with Sodium Thiosulphate. Then
potassium iodide is mixed with sulphuric acid into the standard potassium dichromate. starch is used as well.
To determinate the availability of chlorine within the bleach created, bleach is used and then distilled water is
used to fill up the flask. potassium iodine and concentrated ethanoic acid is used. This type of a titration is a
redox titration in particular a back titration. redox titration is when both a oxidising and reducing agents are
used but an indicator is not used like normal acid-base titration this is because there needs to be a colour
change with the reactant which in this titration is potassium iodide. This is a brown solution meaning the
colour change should be colourless this is because within potassium iodide has I- ions allowing it to get
oxidised, when this happens it becomes I2 making it a brown solution. When starch is added, which is an
indicator for I2 allowing it to become colour when there is I2. Back titration is when a titration is done
backwards meaning a standard reagent is used instead of an original sample; this reagent is known and will
be titrated. This titration is a back titration, hence why it was used, as the reagent is sodium thiosulphate as
its used to find the amount of chlorine in sodium hypochlorite. the analyte is bleach.
Half equation:
Cr2O72- + 6I- + 14H+ = 3I2 + 2Cr3+ + 7H2O
Reduc on: Cr072- + 6e- + 14H- ! 2Cr3+ + 7H2O
Oxida on: 2I- ! I2 + 2e-
In these half equations the iodide ions are being oxidised this is because it has a oxidising state of -1 where then
the iodide ion loses one electron making the final oxidising state 0 making it a clear oxidising reaction as it is
losing electrons, making Iodide the reducing agent. Dichromate is being reduced as it is an oxidising agent
because it is gaining electron as the oxidising state goes from +6 electrons to +3, gaining 6 electrons as shown in
the equation.
2S2O32- + I2 = 2I- + S4O62-
Reduc on: 2e- + I2 ! 2I-
Oxida on: 2S2O32- ! S4O62- + 2e-
In these half equations the iodide is being reduced as the oxidising state goes from 0 and then loses -2 electrons
and so it is used as oxidising agent. Sodium thiosulphate is then gaining these two electrons from the iodide has
they are both reacting together, making the oxidising state for sodium thiosulphate +2. By sodium thiosulphate
oxidising it can be used as the reducing agent.
OCl- + 2I- + 2H+ = I2 + Cl- + H2O
Reduc on: OCl- + 2e- + 2H+ ! Cl- + H2O
Oxida on: 2I- ! I2 + 2e-
titi
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