Redox (II): Oxidation and Reduction

Edexcel A level Chemistry (2017)

Topic 14: Redox (II): Oxidation and Reduction

Here are two learning objectives:

14/1: To understand the terms ‘oxidation’ and ‘reduction’ in terms of electron transfer, applied to s-, p- and d-block elements.

14/2: To understand the terms ‘oxidation’ and ‘reduction’ in terms of changes in oxidation number, applied to s-, p- and d-block elements.

Oxidation and Reduction

Elements have a tendency either to gain or lose electrons.

Metals tend to lose their outer shell electrons

Non-metals tend to gain electrons to complete their outer electrons shells.

Oxidation means the loss of electrons and the formation of positive ions or cations.

Reduction means the gain of electrons and the formation of negative ions or anions.

For example: Sodium (Group 1) tends to lose its outer shell electron leaving a full outer shell.

Na    ⟶    Na⁺     +     e–

Chlorine (Group 7(17)) tends to gain one electron to complete its outer shell.

Cl      +     e–      ⟶    Cl^–    

Ions formed tend to have full outer shells of electrons.

Sodium burns in chlorine gas producing white sodium chloride.

The process involves the transfer of an electron between a chlorine atom and a sodium atom.

Ionic sodium chloride is the product of the reaction:

2Na(s)      +     Cl₂(g)     ⟶      2Na⁺Cl^–(s)

There is an electrostatic attraction between the positive sodium cation and the negative chloride anion.

Since a new bond is formed between ions of sodium and chlorine the process evolves heat and is exothermic.

This is a redox reaction between an s block metal and a p block non-metal.

Redox is an abbreviation of reduction and oxidation.

It is a neat way of analyzing many reactions.

e.g. the displacement reaction between zinc and copper(II) sulfate solution:

Zinc    +     Copper sulfate    ⟶    Zinc sulfate     +   copper

Zn(s)      +     CuSO₄(aq)    ⟶    ZnSO₄(aq)         +      Cu(s)     

                   Blue solution        colourless solution   brown solid

Ionically:

Zn(s)      +     Cu²⁺(aq)    ⟶    Zn²⁺(aq)     +   Cu(s)     

           

We see that zinc loses two electrons (oxidized) in the reaction and donates them to the copper ion (reduced) to form copper.

Zn(s)    ⟶    Zn²⁺(aq)     +   2e–    oxidation half-equation

Cu²⁺(aq)   +   2e–     ⟶     Cu(s)    reduction half-equation 

The half-equations compete our analysis of the redox reaction.

Redox and oxidation number

Oxidation number is artificial.

Oxidation number imposes a set of rules on redox reactions.

You need to know the rules before you can use oxidation number to analyse a redox reaction.

Oxidation number rules depend on the electronegativity of the elements.

What are the oxidation number rules?

Oxidation number rules:

Rule 1: The oxidation number of an uncombined element is zero.

e.g.  Cl in Cl₂ is 0, P in P₄ is 0 etc…….

Rule 2: The oxidation number of an ion is the charge on that ion.

e.g. Cl^–  is –1, O^2– is –2, Na⁺ is +1 etc………

Rule 3: The oxidation numbers of the elements in a compound always add up to zero.

e.g. in H₂O: O is -2 and H is +1 so that –2 +(+1+1) = 0

Rule 4: The oxidation numbers of a polyatomic ion always add up to the charge on that ion.

e.g. sulfate ion SO₄^2–   

O is –2 and S is +6 so that  +6 + (–2–2–2–2) =  –2 

Rule 5: There are some fixed oxidation numbers:

Oxygen combined with another element is always –2 unless in a peroxide (its –1) or with fluorine (its +1)

Fluorine combined with another element is always –1

Group 1 metals are always +1

Group 2 metals are always +2

Aluminum is always +3

In a binary compound, the more electronegative element always takes the negative oxidation number.

Oxidation means an increase in oxidation number.

Reduction means a decrease in oxidation number.

Let’s analyse a metal displacement reaction again this time using oxidation number.

Here is the displacement reaction between zinc and copper(II) sulfate solution:

Zinc    +     Copper sulfate    ⟶    Zinc sulfate     +   copper

Zn(s)      +     CuSO₄(aq)    ⟶    ZnSO₄(aq)          +      Cu(s)     

                   blue solution        colourless solution   brown solid

Ionically:

Zn(s)      +     Cu²⁺(aq)    ⟶    Zn²⁺(aq)     +   Cu(s)     

          

Add in the oxidation numbers gives this:

Zn(s)      +     Cu²⁺(aq)    ⟶    Zn²⁺(aq)     +   Cu(s)     

0                     +2                         +2                       0

The oxidation number of zinc increases by +2

The oxidation number of copper decreases by +2

Copper is reduced, zinc is oxidized and by the same oxidation number value 2.