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Unit 4: Electrochemistry
Oxidation-Reduction Reactions
An oxidation-reduction reaction involves two processes:
Oxidation: a species
electrons and is said to be oxidized.
Reduction: a species
electrons and is said to be reduced.
Electrons are transferred from the species that is
to the species that is
.
The
is the species that causes oxidation, so it is the species that is reduced.
The
is the species that causes reduction, so it is the species that is oxidized.
A
shows the number of electrons a species loses via oxidation or gains via reduction.
Ex: Consider the reaction of Cu(s) and Ag+(aq).
is oxidized because it loses 2 electrons.
is reduced because it gains 1 electron.
Electrons are transferred from
to
The oxidizing agent is
because it is reduced.
The reducing agent is
because it is oxidized.
.
The oxidation half reaction is:
The reduction half reaction is:
Determining which species is oxidized and which species is reduced is easier with the net ionic equation.
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Ex 1: Zn(s) + CuSO4(aq)
a) Determine the balanced chemical equation, total ionic equation, and net ionic equation for this reaction.
b) Which species is oxidized?
c) Which species is the oxidizing agent?
Which species is reduced?
Which species is the reducing agent?
d) What is the oxidation half reaction?
What is the reduction half reaction?
Ex 2: Na(s) + KI(aq)
a) Determine the balanced chemical equation, total ionic equation, and net ionic equation for this reaction.
b) Which species is oxidized?
c) Which species is the oxidizing agent?
Which species is reduced?
Which species is the reducing agent?
d) What is the oxidation half reaction?
What is the reduction half reaction?
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Ex 3: Br2(l) + NaCl(aq).
a) Determine the balanced chemical equation, total ionic equation, and net ionic equation for this reaction.
b) Which species is oxidized?
c) Which species is the oxidizing agent?
Which species is reduced?
Which species is the reducing agent?
d) What is the oxidation half reaction?
What is the reduction half reaction?
Ex 4: Fe(s) + Pb(NO3)2(aq).
a) Determine the balanced chemical equation, total ionic equation, and net ionic equation for this reaction.
b) Which species is oxidized?
c) Which species is the oxidizing agent?
Which species is reduced?
Which species is the reducing agent?
d) What is the oxidation half reaction?
What is the reduction half reaction?
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Oxidation Numbers
Chemists use oxidation numbers to determine if an element is oxidized or reduced during a chemical reaction.
Oxidation numbers are assigned according to the following rules:
Examples
1. The oxidation number of an element is 0.
The oxidation number of Na in Na(s) is 0.
2. The oxidation number of an element in a monatomic
ion is equal to the charge of its ion.
The oxidation number of Cu in Cu2+ is 2.
3. The oxidation number of an element in a binary ionic
compound is the charge of its ion.
The oxidation number of Na in NaCl is 1.
The oxidation number of Cl in NaCl is -1.
4. The oxidation number of hydrogen in its compounds
is 1, except in metal hydrides where it is -1.
The oxidation number of H in H2S is 1.
The oxidation number of H in NaH is -1.
5. The oxidation number of oxygen in its compounds
is -2, except in peroxides where it is -1.
The oxidation number of O in Li2O is -2.
The oxidation number of O in H2O2 is -1.
6. In covalent compounds that do not contain H or O,
the more electronegative element is assigned an
oxidation number that equals the negative charge it
normally has in its ionic compounds.
The oxidation number of Cl in PCl3 is −1.
7. The sum of the oxidation numbers of all the elements
in a compound is 0.
The oxidation number of C in CF4 is 4.
The oxidation number of F in CF4 is -1.
4 + 4(-1) = 0
8. The sum of the oxidation numbers of all the elements
in a polyatomic ion equals the charge on the ion.
The oxidation number of N in NO2− is 3.
The oxidation number of O in NO2− is -2.
3 + 2(-2) = -1
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Ex 1: Determine the oxidation number of N in each species.
a) N2
b) NH3
c) NCl3
d) AlN
e) NaNO2
f) NO3−
Ex 2: Determine the oxidation number of S in each species.
a) S8
b) SO3
c) SF6
d) ZnS
e) MgSO4
f) SO32−
Ex 3: Determine the oxidation number of Cr in each species.
a) Cr3+
b) K2CrO4
c) K2Cr2O7
Ex 4: Determine the oxidation number of V in each species.
a) V2+
b) VO2+
c) VO2+
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Identifying Redox Reactions
An increase in oxidation number AND a decrease in oxidation number occur during a redox reaction.
An element is oxidized if it’s oxidation number
.
An element is reduced if it’s oxidation number
.
Ex: Consider the reaction of NiO(s) and C(s).
C is
because its oxidation number increases from 0 to 2.
Ni is
because its oxidation number decreases from 2 to 0.
Ex: Which of the following reactions are redox reactions?
a) H2(g) + Cl2(g) 2 HCl(g)
b) HgO(s) Hg(l) + O2(g)
c) Cl2(g) + 2 HBr(aq)  2 HCl(aq) + Br2(l)
d) ZnS(s) + HCl(aq) ZnCl2(aq) + H2S(g)
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Balancing Redox Reactions
This equation is not balanced because the number of O atoms and the charge on each side is not equal.
MnO4−(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(aq)
We will use the half reaction method to balance reactions like this:
1. Separate the equation into an oxidation half reaction and a reduction half reaction.
2. Balance atoms other than O or H.
3. Balance O atoms by adding H2O(l)
4. Balance H atoms by adding H+(aq).
5. Balance the charge by adding e−.
6. Equalize the number of electrons in the half reactions.
7. Combine the half reactions. Cancel terms common to the left and right sides of the equation.
Ex: Balance the following redox reactions:
a) Cu(s) + Mg2+(aq) Cu+(aq) + Mg(s)
b) Ni2+(aq) + Al(s) Al3+(aq) + Ni(s)
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c) Ag+(aq) + HCHO(aq)  Ag(s) + HCOOH(aq)
d) HNO3(aq) + Cu(s) NO2(g) + Cu2+(aq)
e) ClO4−(aq) + NO2(g)  Cl−(aq) + NO3−(aq)
f) Cr2O72−(aq) + H2O2(aq) Cr3+(aq) + O2(g)