Download 3202 Chapter 18 - Eric G. Lambert School

Unit 4:
Electrochemistry
Chemistry 3202
Introduction
Electrochemistry involves the study of
electron movement or electric current in
chemical reactions.
 Electrochemical reactions involve the
transfer of electrons between reactants.

Terms
oxidation
reduction
oxidizing agent
reducing agent
half-reaction
oxidation number
redox reaction
electrochemical cell
electrolytic cell
electrolysis
electroplating
fuel cell
pyrometallurgy
hydrometallurgy
CHAPTER 18
Introduction
eg.
Ag(s) + CuCl(aq) → Cu(s) + AgCl(aq)
Which species loses an electron?
Which species gains an electron?
Which species has no electron
change?
Oxidation and Reduction
OXIDATION occurs when an apple turns
brown or steel rusts upon exposure to air
(reaction with oxygen)
 the Loss of Electrons in a reaction is called
Oxidation (LEO)
 a substance is oxidized when it loses
electrons

Oxidation and Reduction
REDUCTION
 the Gain of Electrons in a reaction is called
Reduction (GER)
 a substance is reduced when it gains
electrons

LEO
GER
Oxidation
Is
Loss
Reduction
Is
Gain
Oxidation and Reduction
eg.
Why is this NOT an example of an
electrochemical reaction?
CaCl2 + 2 AgNO3 → Ca(NO3)2 + 2 AgCl
Oxidation and Reduction

Oxidation-Reduction reactions are usually
referred to as redox reactions.
Oxidation and Reduction
eg. Cu(s) metal placed in AgNO3(aq)
Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq)→ Cu2+(aq) + 2 NO3-(aq)+ 2 Ag(s)
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)
NET IONIC EQUATION:
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
loses 2 electrons – Cu is oxidized
 Ag+ gains 1 electron – Ag+ is reduced
 NO3-(aq) is not changed by this reaction
and is called a spectator ion
 Cu
Oxidation and Reduction
eg. Write the NET IONIC EQUATION
for the reaction between Cl2(g) and
NaI(aq). Identify the species being
oxidized and the species being
reduced.
Cl2(g) + NaI(aq) → NaCl(aq) + I2(s)
Oxidizing and Reducing Agents
 the
substance that is oxidized in a
redox reaction is called the reducing
agent (RA).
 the substance that is reduced in a
redox reaction is called the oxidizing
agent (OA).
Oxidizing and Reducing Agents
eg. Identify the OA and the RA in the
following redox reactions
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
Cl2(g) +
2 I-(aq) → 2 Cl-(aq) + I2(s)
Half Reactions
A
redox reaction can be separated
into two parts called half-reactions.
 The oxidation half-reaction shows the
loss of electrons.
 The reduction half-reaction shows the
gain of electrons.
Half Reactions
eg.
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
oxidation: Cu(s) →Cu2+(aq) + 2e-
reduction: 2 Ag+(aq) + 2e- → 2 Ag(s)
Half Reactions
Write the half-reactions for:
Sn4+(aq) + Pb2+(aq) → Sn2+(aq) + Pb4+(aq)
F2(g) + 2 Br -(aq) → 2 F-(aq) + Br2(s)
Write the overall equation for:
Al3+ + 3e- →Al(s)
Mg(s) → Mg2+(aq) + 2e-
Oxidation and Reduction
 p.
715 #’s 1-4
 p. 716 #’s 5-7
Oxidation Numbers
 The
oxidation number is the actual or
hypothetical charge on an element or
ion using an assigned set of rules.
species
Clion
Mg2+
ion
Mg
atom
C in
CH4
oxidation
#
-1
+2
0
-4
Determining Oxidation Numbers
(See p. 724)
1.
Pure elements → oxidation # = 0
 eg.
1.
Cl2(g), Mg(s), P4(s), Fe(s)
Simple ions → oxidation # = ion charge
Cl- → oxidation # = -1
 Mg 2+ → oxidation # = +2
 eg.
Determining Oxidation Numbers
3.
Hydrogen has an oxidation number of +1
except in metal hydrides.
 eg.
H is +1 in NH3, H2O, and HCl
 H is -1 in NaH and CaH2
4.
Oxygen has an oxidation number of -2
except in peroxides such as H2O2 and in
OF2.
Determining Oxidation Numbers
5.

If H or O are not in a covalent compound,
the atom with the highest
electronegativity has the oxidation
number that matches its charge.
eg. NF3
vs.
NI3
Determining Oxidation Numbers

The sum of all oxidation numbers in a
compound is ZERO.
 eg.
CH4
H →
 C must be ???
Determining Oxidation Numbers

The sum of all oxidation numbers in a
polyatomic ion equals the ion charge.
CO32O →
 C + 3(-2) = -2
 C = +4
 eg.
Page 726 - #’s 9 – 12
Extra Practice (oxidation #’s)
Determine the oxidation number of:
a) S in SO2
d) Cr in Cr2O72b) Cl in HClO4
e) I in MgI2
c) S in SO422. What is the oxidation number of nitrogen in
each of the following?
 N 2O
NO
NO2
NH3
 N 2H 4
NaNO3
N2
NH4Cl
1.
Identifying Redox Reactions
Redox reactions can be identified by
looking at the oxidation numbers of the
species involved.
 Changes in oxidation numbers indicate a
reaction is a redox reaction.

Identifying Redox Reactions

An increase in oxidation number indicates
oxidation
 (ie.

the reducing agent)
A decrease in oxidation number indicates
a reduction
 (ie.
the oxidizing agent)
Identifying Redox Reactions
eg. Use oxidation numbers to show why
this is a redox reaction:
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
-4
0
-2
-1
Identifying Redox Reactions

eg.Using oxidation numbers, explain why
this is NOT a redox reaction:
 CaCO3(s) →
CaO(s) + CO2(g)
Page 728 - #’s 13 – 15
 Handout Electro #1

Extra Practice (identifying redox)
1. Determine whether the following are redox
reactions. For the redox reactions, identify the
OA and the RA.
a) Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
b) Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
c) Br2(l) + 2 NaI(aq) → 2 NaBr(aq) + I2(s)
Extra Practice (identifying redox)
d) 2 NaCl(l) → 2 Na(l) + Cl2(g)
e) HCl(aq) + NaOH(aq) → HOH(l) + NaCl(aq)
f) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
g) 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
Balancing Redox Equations
Redox equations are written by first
balancing the ½ reactions and then
combining the balanced ½ reactions.
 Three ways to balance redox equations:

 reactions
in acidic solution
 reactions in basic solution
 standard reduction potentials table
Reactions in Acidic Solution (p.
732)
1.
2.
3.
4.
5.
Write an unbalanced half reaction
Balance everything but O and H
Balance O by adding water
Balance H by adding H+
Balance the charges with é
Reactions in Acidic Solution (p.
732)
balance each ½ reaction using the 5 steps
 multiply the ½ reactions to cancel
electrons
 combine the ½ reactions
 CHARGE AND # OF ATOMS MUST BE
BALANCED

Reactions in Acidic Solution
eg. Balance each ½ reaction in acidic
solution.
a) MnO4-(aq) → Mn2+(aq)
b) HNO2(aq) → N2O(g)

Page 732 #’s 19, 20
MnO4-(aq) → Mn2+(aq)
MnO4-(aq) → Mn2+(aq)
MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
HNO2(aq) → N2O(g)
2 HNO2(aq) → N2O(g)
2 HNO2(aq) → N2O(g) + 3 H2O(l)
4 H+(aq) + 2 HNO2(aq) → N2O(g) + 3 H2O(l)
4 e- + 4 H+(aq) + 2 HNO2(aq) → N2O(g) + 3 H2O(l)
(Reduction: OA is HNO2(aq) )
Reactions in Acidic Solution

eg.Balance the following redox equation in
acidic solution.

IO3-(aq) + NO2(g) → NO3-(aq) + I2(s)
Reactions in Acidic Solution
Solution:
 write and balance each half reaction
IO3-(aq) → I2(s)
 NO2(g) → NO3-(aq)

Reactions in Acidic Solution
IO3-(aq) → I2(s)
2 IO3-(aq) → I2(s)
2 IO3-(aq) → I2(s)
+ 6 H2O(l)
12 H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l)
10 e- + 12 H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l)
(Reduction: OA is IO3- )
Reactions in Acidic Solution
NO2(g) → NO3-(aq)
NO2(g) → NO3-(aq)
H2O(l) + NO2(g) → NO3-(aq)
H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq)
H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) + 1 e(Oxidation: RA is NO2(g) )
Oxidation:
H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) + 1 eReduction:
10 e- + 12 H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l)
10 H2O(l) + 10 NO2(g) → 10 NO3-(aq) + 20 H+(aq) + 10 eOverall equation:
4 H2O(l) + 10 NO2(g) + 2 IO3-(aq)
→ I2(s) + 10 NO3-(aq) + 8 H+(aq)
Extra Practice (balance in acidic conditions)
CH3OH(l) + MnO4-(aq) → CH2O(l) + Mn2+(aq)
x5
CH3OH(l) → CH2O(l)
CH3OH(l) → CH2O(l) + 2 H+(aq) + 2 e-
x2
MnO4-(aq) → Mn2+(aq)
5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
5 CH3OH(l) → 5 CH2O(l) + 10 H+(aq) + 10 e10 e- + 16 H+(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)
__ CH3OH(l) + __ H+(aq) + __MnO4-(aq) →
__ CH2O(l) + __Mn2+(aq) + __ H2O(l)

p. 739 # 27
Balance the following redox equation in acidic
solution.
N2O(g) + SO42-(aq) → HNO2(aq) + H2SO3(aq)
N2O(g) → HNO2(aq)
3 H2O(l) + N2O(g) → 2 HNO2(aq) + 4 H+(aq) + 4 e-
SO42-(aq) → H2SO3(aq)
SO42-(aq) + 4 H+(aq) + 2 e- → H2SO3(aq) + H2O(l)
Reactions in Basic Solution
1.
2.
3.
4.
5.
6.
7.
Write an unbalanced half reaction
Balance everything but O and H
Balance O and H the same
Adjust for basic conditions with OH (=H)
Balance the charges with é
Cancel H+ and OHBalance charges by adding electrons
Reactions in Basic Solution
balance each ½ reaction using the 7 rules
from p. 733
 multiply the ½ reactions to cancel
electrons
 combine the ½ reactions
 CHARGE AND # OF ATOMS MUST BE
BALANCED

Reactions in Basic Solution
eg. Balance this ½ reaction in basic
solution
ClO-(aq) → Cl-(aq)
Reactions in Basic Solution
ClO- → Cl- + H2O
2 H+ + ClO- → Cl- + H2O
2 OH- + 2 H+ + ClO- → Cl- + H2O + 2 OH2 H2O + ClO- → Cl- + H2O + 2 OH-
1 H2O + ClO- → Cl- + 2 OH2e- + H2O + ClO- → Cl- + 2 OH-
Reactions in Basic Solution
eg. Balance the following redox equation
under basic conditions.
SO42-(aq)+ NO(g) → SO32-(aq) + NO2-(aq)
2 H+ + SO42- → SO32- + H2O
2 OH- + 2 H+ + SO42- → SO32- + H2O + 2 OH-
2 H2O + SO42- → SO32- + H2O + 2 OHH2O + SO42- → SO32- + 2 OH2 e- + H2O + SO42- → SO32- + 2 OH-
H2O + NO → NO2- + 2 H+
2 OH- + H2O + NO → NO2- + 2 H+ + 2 OH-
2 OH- + H2O + NO → NO2- + 2 H2O
2 OH- + NO → NO2- + H2O
2 OH- + NO → NO2- + H2O + 1 e-
Reactions in Basic Solution
2 OH- + NO → NO2- + H2O + 1 eMultiply by 2 to cancel electrons!!
4 OH- + 2 NO → 2 NO2- + 2 H2O + 2 e2 e- + H2O + SO42- → SO32- + 2 OH-
2 OH- + 2 NO + SO42- → SO32- + 2 NO2- + H2O
ClO− + CrO2− → CrO42− + Cl2
CrO2− → CrO42−
ClO− → Cl2
p. 734 #’s 23, 24
 p. 739 #’s 27, 28
 Handout Electro #2
 Handout Electro #3
