Many
chemical reactions are electrical in nature and involve a competition for
electrons.
2Mg(s) + O2(g) →
2MgO(s)
There
is a competition for electrons between the magnesium and the oxygen atoms. Since oxygen has a higher electronegativity,
it gains two electrons from the magnesium.
- magnesium – loses 2 electrons (and undergoes oxidation)
- oxygen – gains 2 electrons (and undergoes reduction)
This
is the basis of reduction-oxidation reactions (or redox reactions,
for short).
Oxidation Numbers
How
can you tell a redox reaction has occurred?
Look for a change in
the oxidation number of an element during the course of a reaction.
Oxidation
numbers are used by chemists to keep track of the positive or negative
character of atoms or ions in redox reactions.
Then, it is possible to determine which species has been oxidized (lost electrons) and
which has been reduced (gained electrons).
An oxidation number is the real or
apparent charge an atom or ion has when all bonds are assumed to be ionic. You are already familiar with oxidation
numbers – up to this point, we have called them valences or charges.
Rules for Determining Oxidation Numbers
- In free elements, each atom has an oxidation number of zero.
ex. Cu0, Zn0, Fe0,
H20
- For ions consisting of a single atom, the oxidation number is equal to the charge on the ion.
ex. Cl- (-1), Na+ (+1), Pb2+
(+2), Fe3+ (+3)
- The oxidation number of H is +1 and of O is -2 in most compounds (except in hydrides where the oxidation number of H is -1 and peroxides where the oxidation number of O is -1).
- The algebraic sum of the oxidation numbers in a neutral compound must be zero; in a polyatomic ion, the sum must be equal to the ion charge.
ex. HSO3- ON(H) + ON(S) + 3ON(O) = -1
+1 + x + 3(-2)
= -1
x = +4
TryIts!.
Find the oxidation number of the underlined element in each case: HNO3, MnO4-, S2O32-
Answers are found at the end of the lesson.
Oxidation- Reduction Reactions
The Loss of Electrons occurs during an Oxidation reaction.
The Gain of Electrons occurs during a Reduction reaction.
To remember this,
think of a lion 🦁: LEO says GER!
ex. For the following reaction, label the oxidation and
reduction reactions, show the changes in oxidation numbers, label the oxidizing
and reducing agents.
2K(s) + MgCl2(aq) → Mg(s)
+ 2KCl(aq)
First
begin by writing out the total ionic equation (remember that all
aqueous substances are broken apart into their ions. All solid, liquid
or gas substances are left well enough alone. Second, cross out any
spectator ions.
2K(s) + Mg2+(aq) + 2Cl-(aq)
→ Mg(s) + 2K+(aq) + 2Cl-(aq)
Third, Write out what is left as the net ionic equation:
2K(s) + Mg2+(aq) → Mg(s)
+ 2K+(aq)
Notice that:
- potassium - ox # goes from 0 to +1, ∴ 1 electron lost ↝ oxidation
- magnesium - ox # goes from +2 to 0, ∴ 2 electrons gained ↝ reduction
- chlorine begins and ends with an ox # of -1, so it is neither oxidized nor reduced; it is simply a spectator ion
Every redox reaction can be broken into 2 half-reactions:
2K0(s) →
2K+(aq) + 2e-
oxidation half reaction
Mg2+(aq)
+ 2e- →
Mg0 (s) reduction half reaction
Notice that each half-reaction is balanced in terms of elements (as usual), but also in terms of charge.
↦ The oxidation half-reaction needed two negative electrons placed on the product side so that the total charge on each side is zero.
↦ The reduction half-reaction needed two negative electrons placed on the reactant side so that the total charge on each side is zero.
Please note that it's not about making the charge zero on each side of the half-reaction, but rather just making sure the charge is equal on each side.
↦ The oxidation half-reaction needed two negative electrons placed on the product side so that the total charge on each side is zero.
↦ The reduction half-reaction needed two negative electrons placed on the reactant side so that the total charge on each side is zero.
Please note that it's not about making the charge zero on each side of the half-reaction, but rather just making sure the charge is equal on each side.
Oxidizing & Reducing Agents
A
substance that reduces another is called
a reducing agent; it gets oxidized
in the process. A substance that oxidizes another is called an oxidizing agent; it gets reduced in the process.
In
the above example,
- K(s) is the reducing agent (because it causes Mg2+(aq) to undergo reduction and it gets oxidized in the process)
- Mg2+(aq) is the oxidizing agent (because it causes K(s) to undergo oxidation and it gets reduced in the process).
TryIts! Answers:
HNO3 +1 + x +3(-2) = 0 MnO4- x + 4(-2) = -1 S2O32- 2x + 3(-2) = -2
x = +5 x
= +7 x = +2
Homework
Oxidation Number Practice - try a few
Redox Reaction Practice - try at least one
Homework and Answer Keys for the remainder of the unit:
This is not on the data table and you may need it:
Cr2+(aq) + 2e → Cr(s) Eored = -0.91 V
Au3+(aq) + 3e → Au(s) Eored = +1.50 V
SO42-(aq) + H2O(l) + 2e → SO32-(aq) + 2OH-(aq) Eored = -0.93 V
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#6-10
#11-12
#13-17
#18-19
#20-23
#24-29
#30-32
#33-35
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