Transcript Chemistry Tutorial

The Wright Stuff
Chemistry Tutorial
REDOX REACTIONS
by
Dr John G Wright
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Press PgDn or left click for the next slide
Redox Reactions
The tutorials are divided into five main sections, each
covering a separate topic concerning redox reactions, plus
a problems section.
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Oxidation and Reduction
Oxidation Numbers (this set of 18 slides)
Disproportionation
Balancing Half Equations
Electrochemical Cells
Problems
The individual topics can be viewed as separate slide shows.
If a slide show is missing from the web page, it means it is
being updated and improved. (Or perhaps I just haven’t
finished writing it yet!)
Oxidation and Reduction
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It is assumed that you have read the first tutorial on Oxidation
and Reduction and are familiar with the meaning of these
terms, especially the electron-based definitions.
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Students on introductory level courses may only need to read
the previous section on Oxidation and Reduction, but those on
advanced courses, such as A-level and above, will need to
study both tutorials, with the emphasis being placed on the
modern electron-based definition of oxidation and reduction.
Oxidation Numbers
This section is about calculating the oxidation state or
oxidation number of an element in a compound. It uses a set
of rules which require little or no previous knowledge of the
chemical being examined. (The oxidation number is similar to
the valency of the element but has a + or - sign.)
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Oxidation Numbers
First, the definition.
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The oxidation number of an element is the charge which the
atom would have if the element was acting as an ion in the
species being studied.
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This can seem a bit strange at times, especially when you know
that the compound you are examining is covalent. But it
enables us to easily work out whether an element is oxidised or
reduced in a reaction. We just compare the oxidation number
before and after the reaction and work out whether it has
gained or lost electrons.
There are a series of simple rules we use to calculate the
oxidation number of an element.
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Oxidation Numbers
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The rules are given in order of importance. If you reach a rule
which gives you data to work from and a later rule appears to
contradict the earlier result, just ignore it, you use the earlier
result. I.e. later rules do not overrule earlier decisions that you
have already made.
Oxidation Numbers
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The oxidation number of an uncombined element is zero.
I.e. the element in the element itself is zero.
The algebraic sum of the oxidation numbers of all the atoms in
a compound equals zero.
The algebraic sum of the oxidation numbers of all the atoms in
an ion equals the charge on the ion.
Algebraic sum means take account of the sign of the charge.
Learn these rules.
Oxidation Numbers
But where do you start? We find that some atoms always have
the same oxidation number in their compounds or ions. As
stated earlier, the first rules overrule the later rules.
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Fluorine is always ALWAYS -1 in its compounds.
Group I metals are always ALWAYS +1 in compounds.
Group II metals are always +2 in their compounds.
Oxygen is almost always -2 in compounds, except in peroxides
(H2O2, ROOR, etc.) where it is -1. (Or when overruled by one
of the above rules.)
Halides are often -1, but other numbers are often possible.
Hydrogen can be +1 or -1. (If it is the first element given in a
formula, it is usually +1, if given second or third, it may be -1.
If in doubt, the most electronegative element is +ve and the
other element will be -ve.)
Oxidation Numbers
Let’s look at these rules in action in a simple problem.
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Calculate the oxidation number of Mn in KMnO4.
K and O are in our table of known oxidation numbers, and
KMnO4 is a compound.
K is +1 (Group I metal) and oxygen is almost always -2.
K + Mn + (4 x O) = 0 (that’s a zero)
So +1 + Mn + (4 x -2) = 0
i.e. +1 + Mn - 8 = 0
i.e. Mn - 7 = 0
so Mn = +7 (We say +7 and not 7, because it is a charge.)
Easy? It’s just simple arithmatic, not chemistry.
Oxidation Numbers
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Here’s another example, this time involving an ion.
Calculate the oxidation number of chlorine in the ion ClO3Oxygen comes before chlorine in our known table, so it’s rule
takes preference, and it has a charge of -2.
Cl + (3 x O) = -1 ( the -1 is the charge on this ion)
Cl + (3 x -2) = -1
Cl - 6 = -1
Cl = +5
(Again, it’s +5 and not 5 as it’s the charge on the “ion”.)
Oxidation Numbers
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Calculate the oxidation number of phosphorus in POCl2F
Oxygen is almost always -2, chlorine is usually -1 and fluorine
is always ALWAYS -1. As P is the unknown, we can reason
that although Cl can have other values, then -1, the commonest
value, is the one to use (otherwise the problem is unsolvable).
P + O + (2 x Cl) + F = 0
P - 2 - 2 -1 = 0
P-5=0
P = +5
But phosphorus can have other values for it’s oxidation
number, depending on the compound under investigation.
Oxidation numbers
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What is the oxidation number for P in NaH2PO3?
Na = +1, H = +1 usually, and O = -2 almost always.
Na + 2H + P + 3O = 0
+1+2+P-6=0
P - 3 = 0, so P = +3
What is charge on the Cr in K2Cr2O7 ?
K and O are in the table of fixed values.
2K + 2Cr + 7O = 0
+ 2 + 2Cr - 14 = 0
2Cr - 12 = 0, so 2Cr = +12, and Cr = +6
Oxidation Numbers
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To avoid any confusion when an element can have several
oxidation numbers, the oxidation number is usually mentioned
in the compound’s name, written as Roman numerals in
brackets after the element to which it refers.
In names like “elementate(X)”, the number refers to “element”
and not the associated oxygens.
So if we look at the examples we’ve just done, we get the
following names:KMnO4
potassium manganate(VII)
NaClO3
sodium chlorate(V)
POCl2F
phosphorus(V) oxydichlorofluoride
NaH2PO3 sodium dihydrogenphosphate(III)
K2Cr2O7 potassium dichromate(VI)
Oxidation numbers
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Check for yourself that the numbers in the following
compounds’ names are the same as the oxidation number of the
element with which they are associated
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LiNO3
NaNO2
POCl3
CrCl3
TiO2
lithium nitrate(V),
PtCl4 platinum(IV) chloride
sodium nitrate(III), NaBrO4 sodium bromate(VII)
phosphorus(V) oxychloride
chromium(III) chloride, TiO titanium(II) oxide
titanium(IV) oxide
Did you get them all right?
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Oxidation Numbers
An important use of oxidation numbers is in the recognition of
oxidation and reduction. They let us quickly see when an
element has gained or lost electrons, and hence been oxidised
or reduced. If the oxidation number becomes more positive,
oxidation has occurred, becoming more negative means that
reduction has occurred.
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Consider the equation below.
2FeCl2 + Cl2  2FeCl3
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The oxidation number of the iron changes thus:
Fe2+  Fe3+ ( + e- of course)
sometimes written as Fe(II)  Fe(III) + ei.e. Fe2+ has been oxidised. This example is easy to spot as an
example of oxidation, but sometimes it’s a bit harder.
Oxidation Numbers
Consider a reaction where the following change occurs (the
other reagents are not important at this stage).
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I-  IO3Is this oxidation or reduction?
Calculating the oxidation numbers reveals all.
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In I-, the iodine is -1, while in IO3- it is +5. (Don’t take my
word for it, calculate the oxidation numbers yourself.) So the
iodide ion has lost electrons, and been oxidised to +5.
I-  IO3- + 6ewhich could also be written as
I(I)  I(V) + 6e-
Oxidation Numbers
Here are a few more examples showing the use of oxidation
numbers to discover whether oxidation or reduction occurs.
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3MnO2 + KClO3 + 6KOH  3K2MnO4 + KCl + 3H2O
Mn = +4
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Mn = +6
Cl = +5
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Cl = -1
i.e Mn has lost electrons and been oxidised, while
Cl has gained electrons and been reduced.
4FeCO3 + O2  2Fe2O3 + 4CO2
Fe = +2
 Fe = +3
O=0 
O = -2
C = +4
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C = +4
i.e. Fe has lost electrons and been oxidised, while the
oxygen has gained electrons and been reduced. The carbon
hasn’t changed. Check these oxidation numbers for yourself.
The End
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I hope you have enjoyed this tutorial. It is the second in a
series of tutorials on oxidation and reduction. The tutorials
become progressively more advance, but unless it says so on
the frist couple of pages, are all intended for A-level students.
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I also hope you have increased your understanding of
chemistry, learned something useful about chemistry and that it
will increase your marks in examinations.
Bye for now, Dr John G Wright,
The Wright Stuff
www.sky-web.net
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