Topic 1.5: Introduction to oxidation and reduction

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1.5a Define and use the concept of oxidation number
The oxidation number of an element in a chemical species is a measure of the extent of the oxidation of that element. The more positive the oxidation number the greater the extent of oxidation.  Oxidation number is also a measure of the extent of reduction of an element. Using oxidation numbers it is possible to decide whether oxidation or reduction (redox) has occurred. An atom is said to be oxidised when its oxidation number increases and reduced when its oxidation number decreases.

The oxidation state can be defined as the number of electrons that need to be added or removed from either a positive/negative ion in order to get a neutral atom.
E.g. Fe²⁺ needs to gain two electrons for it to become neutral iron atom therefore its oxidation state is +2.
Task 1.5a  Explain what is happening if the oxidation number of 
sulfur changes from +4 in SO₂ to +6 in SO₄^2-.
Chlorine changes from -1 in chloride to 0 in chlorine,
iron changes from +3 in Fe³⁺ to +2 in Fe²⁺,
aluminium changes from + 3 in Al₂O₃ to zero in aluminium,
carbon changes from +2 in carbon monoxide to +4 in carbon dioxide.

1.5b The strength of oxidising and reducing agents
Oxidising agents causes oxidation.  Their strength varies.  Some oxidising agents with their relative strengths are:
F₂ > MnO₄^- > Cl₂ > Cr₂O₇^2- > Br₂ > I₂
Reducing agents cause reduction.  They too have variable strengths.  Some reducing agents and their strengths are:
Na > Zn > H₂
Task 1.5b Use the terms oxidation, reduction, oxidising agent, oxidant, oxidised, reducing agent reductant, reduced and oxidation number in explaining what is happening in the following reaction:
Cu²⁺(aq) + Zn(s) ---> Cu(s) + Zn²⁺(aq)

1.5c Oxidation, reduction and electron transfer
OILRIG

• Oxidation is the process where electrons are lost.
• Reduction is the process where electrons are gained.
• An oxidising agent is an electron acceptor.
• A reducing agent is an electron donor.

Task 1.5c Explain how metals and halogens might act as oxidants or reductants by gaining or losing electrons.

1.5d Using ionic half equations
The reaction of acidified potassium manganate (VII) with iron (II) sulphate solution:

Reduction half equation: MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- ------> Mn²⁺_(aq) + 4H₂O_(l)

Oxidation half equation: Fe²⁺_(aq)  -----> Fe³⁺_(aq) + e^-

When combining these equations, the number of electrons used must equal the number  formed, therefore the oxidation equation above must be multiplied by 5 
MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- ------> Mn²⁺_(aq) + 4H₂O_(l)
5Fe²⁺_(aq)  -----> 5Fe³⁺_(aq) + 5e^-
and then the equations must be added together.
MnO₄^-_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) ------> Mn²⁺_(aq) + 4H₂O_(l)   +  5Fe³⁺_(aq)
In this equation the manganate (VII) is the oxidising agent and the iron (II) is the reducing agent.  The element manganese is reduced and the element iron is oxidised.
Half equations can be balanced as follows:
e.g. balance the equation: NO₃^- --> NH₄⁺
a) balance oxygen atoms with water
b) balance hydrogen atoms with hydrogen ions
c) balance the charges
a) gives  NO₃^- --> NH₄⁺ + 3H₂O
b) gives  NO₃^- + 10H⁺ --> NH₄⁺ + 3H₂O  
c) gives 8e^- + NO₃^- + 10H⁺ --> NH₄⁺ + 3H₂O  
Task 1.5d.1 constructing half equations
Task 1.5d.2 combining half equations

1.5e Deducing oxidation number
Oxidation number can be calculated by applying a set of rules:

• The oxidation state of all uncombined elements is 0.
• The sum of the oxidation states of each element in an ion is equal to the charge on the ion.
• In neutral molecules, the algebraic sum of the oxidation numbers is 0.
• The oxidation state of hydrogen is +1 (except when combined with group I and II metals as hydrides).
• The oxidation state of oxygen is -2 (except in peroxides, OF₂ and oxygen gas).

Task 1.5e Work out the oxidation number of the following:
(a) N in NO, NO₂, NO₃^-
(b) Mn in MnSO₄, Mn₂O₃, MnO₄^-
(c) As in AsO₂^-, As₂O₃, AsO₄^-
(d) Cr in CrO₄^-, Cr₂O₇^2-, CrO₃
(e) I in I^-, IO₃^-, I₂

1.5f Oxidation, reduction and changes in oxidation number
MnO₄^-_(aq) + 8H⁺_(aq) + 5 Fe²⁺_(aq) ------> Mn²⁺_(aq) + 4H₂O_(l)   +  5 Fe³⁺_(aq)
In the equation above it can be seen that manganese is reduced because its oxidation number is decreased.
in  MnO₄^- ox no of Mn + ox no of four oxygen atoms = -1
                ox no of Mn  + 4*(-2) =  -1
                ox no of Mn = -1 - 4*(-2) = +7
in Mn²⁺ ox no of manganese = charge on ion = +2
Therefore manganese has been reduced from ox state +7 to Ox state +2.
In the equation is can also be seen that the oxidation no of iron increases from +2 to +3 (the charges on the ions) so iron is oxidised
Explain why some surfaces are super black using redox.  Needs Powerpoint.