Oxidation and Reduction Concepts Chemistry Tutorial

Key Concepts

• Oxidation: an atom or ion loses one or more electrons (e^-):

  	reactant	→	products
  Oxidation reaction:	M	→	M+	+	e-

• Oxidation reaction: electrons (e^-) are a product of the oxidation reaction.
• Oxidation reaction: the atom or ion becomes more positive due to the loss of negatively charged electrons.
• Reduction: an atom or ion gains one or more electrons (e^-):

  	reactants			→	product
  Reduction reaction:	X	+	e-	→	X-

• Reduction: electrons (e^-) are a reactant in the reduction reaction.
• In a reduction reaction, the atom or ion beomes more negative due to the gain of negatively charged electrons.

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Oxidation Reactions

An oxidation reaction has occurred if a species (atom or ion) has lost one or more electrons.

Electrons carry a negative charge, so, when a species (atom or ion) loses an electron, the product must be more positively charged than the reactant.

1. Consider the oxidation reaction in which a neutral, highly reactive, sodium atom (Na) loses an electron to form the more stable, much less reactive, sodium cation (Na⁺):

   	reactant	→	products
   Oxidation reaction:	Na	→	Na+	+	e-

   The sodium atom has been oxidised because it has lost an electron.

   The charge on sodium has been increased from 0 for the neutral atom to 1+ for the sodium ion (Na⁺).

   Sodium has become more positive.

2. Consider the oxidation reaction in which the copper(I) ion, Cu⁺, loses an electron to form the copper(II) ion, Cu²⁺:

   	reactant	→	products
   Oxidation reaction:	Cu+	→	Cu2+	+	e-

   A copper ion with a charge of 1+ loses an electron to produce a copper ion with a charge of 2+

   Cu⁺ has been oxidised because it has lost an electron.

   The charge on copper has been increased from 1+ for the Cu⁺ ion to 2+ for the Cu²⁺ ion.

   Copper has become more positive.

3. Consider the oxidation reaction in which lead(II) ions (Pb²⁺) lose 2 electrons (2e^-) to produce lead(IV) ions (Pb⁴⁺):

   	reactant	→	products
   Oxidation reaction:	Pb2+	→	Pb4+	+	2e-

   A lead ion with a charge of 2+ loses 2 electrons to produce a lead ion with a charge of 4+

   Pb²⁺ has been oxidised because it has lost 2 electrons.

   The charge on lead has been increased from 2+ for the Pb²⁺ ion to 4+ for the Pb⁴⁺ ion.

   Lead has become more positive.

Reduction Reactions

A reduction reaction has occurred if a species (atom or ion) has gained one or more electrons.

Electrons carry a negative charge, so, when a species (atom or ion) gains an electron, the product must be more negatively charged than the reactant.

1. Consider the reduction reaction in which an atom of highly reactive fluorine (F) gains an electron to produce the much more stable and less reactive fluoride ion (F^-):

   	reactants			→	product
   Reduction reaction:	F	+	e-	→	F-

   An electrically neutral¹ fluorine atom, an atom with no overall charge, gains an electron to form the negatively charged fluoride ion (anion).

   The fluorine atom has been reduced because it has gained an electron.

   The charge on fluorine has decreased from 0 for the neutral atom to 1- for the fluoride ion.

   Fluorine has become more negative.

2. Consider the reduction reduction in which an atom of oxygen (O) gains two electrons to form the oxide ion (O^2-):

   	reactants			→	product
   Reduction reaction:	O	+	2e-	→	O2-

   A neutral oxygen atom, an atom with no overall charge, gains 2 electrons to form the negatively charged oxide ion (O^2-).

   The oxygen atom has been reduced because it has gained 2 electrons.

   The charge on oxygen has decreased from 0 for the neutral atom to 2- for the oxide ion (O^2-).

   Oxygen has become more negative.

3. Consider the reduction reaction in which tin(IV) ions, Sn⁴⁺, gain 2 electrons to produce tin(II) ions, Sn²⁺:

   	reactants			→	product
   Reduction reaction:	Sn4+	+	2e-	→	Sn2+

   The Sn⁴⁺ ion gains 2 electrons to form the Sn²⁺ ion.

   Sn⁴⁺ has been reduced because it has gained 2 electrons.

   The charge on tin has decreased from 4+ for Sn⁴⁺ to 2+ for Sn²⁺ .

   Tin has become more negative (or, less positive).

Worked Example: Determining if a Reaction is an Oxidation or Reduction Reaction

Question 1. Consider the reaction given below:
    Fe³⁺_(aq) + 3e^- → Fe_(s)
    Is this an oxidation reaction or a reduction reaction?
    Justify your answer.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   (i) Determine if the reaction is an oxidation or reduction reaction.
   (ii) Justify your answer.

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Balanced chemical equation:

   Fe³⁺_(aq) + 3e^- → Fe_(s)

3. What is the relationship between what you know and what you need to find out?

   Definition of oxidation: species loses electrons (electrons are a product)

   Definition of reduction: species gains electrons (electrons are a reactant)

4. Determine whether the reaction is an oxidation or a reduction of iron

   Electrons are a reactant.

   Fe³⁺ gains 3 electrons to produce Fe_(s)

   Therefore this is a reduction reaction.

5. Is your answer plausible?

   In a reduction reaction, negatively charged electrons are gained by a species, so the species must become more negatively charged (or, less positively charged).
   Fe³⁺ has a charge of 3+
   Fe_(s) has no overall charge, that is, its charge is 0
   The charge on iron has become less positive, +3 is more positive than 0.
   Therefore this is a reduction reaction.

   Since this result agrees with our previous way of deciding the nature of this reaction, we are confident that our answer is plausible.

6. State your solution to the problem "is this an oxidation or reduction reaction":

   (i) This is a reduction reaction

   (ii) because Fe³⁺ has gained 3 electrons to form Fe_(s) and reduction is defined as a gain of electrons.