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P6 M4 D3 Unit 27 - describe analytical procedures based on redox principles, explain analytical procedures in terms of oxidation states of the substances involved - BTEC Applied Science Extended Diploma£4.49
P6 M4 D3 Unit 27 - describe analytical procedures based on redox principles, explain analytical procedures in terms of oxidation states of the substances involved - BTEC Applied Science Extended Diploma
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P6 – Describe analytcal procedures based on redox reactonss
M4 – Explain analytcal procedures in terms of oxidaton states of the substances involveds
D3 – Review the applicability of an analytcal technique involving oxidaton and reducton to a
range of industriess
Oxidation is seen as the loss of electrons and the reduction is seen as the gain of electrons. An
example of how some molecules can be oxidising agents and some can be reducing agents can be
seen below. The copper oxide and the magnesium oxide are both involved in this reactionn they are
ionic.
In terms of being specifc to this reactionn the magnesium in this would be reducing coppern they’ll
do that by providing them with electrons which would neutralise the charge of the molecules. This
means that the magnesium is a reducing agent. When it comes to coppern it would get rid of
electrons from the magnesium which would result with the magnesium ions being formed. The
copper molecules would become oxidising agents.
In most cases it is possible to write ionic equations for the redox reactions. or examplen when a
magnesium reduces copper oxide to copper an example of the equation can be seen below.
This equation can be divided into two partsn for example in the frst equation you can see that the
magnesium lost two electronsn the copper on the other hand has gained those two electrons. These
two equations can be seen as the half equations. The electrons which are lost through the reactions
are oxidised by the oxidation reaction. Those electrons which are gainedn it happens by reduction
reaction.
There is also a possible reaction between the hydrogen peroxide and also the manganite(VII) ions.
The manganate ions would oxidise hydrogen peroxide into oxygen gas. The manganate(VII) ions
would be the oxidising agents and the hydrogen peroxide would be the reducing agent. The way in
which the reaction takes place is with the potassium manganate(VII) solution which also has
hydrogen peroxide acidifed with the dilute sulphuric acid. When the chemicals reactn the
manganate(VII) ions would be reduced to manganese(II) ions.
In order to balance the oxygen on the right siden it would need to have two hydrogen ions being
added to the right side of the equation. It is then important to also add the electronsn in this casen
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