Transcript Slide 1

IB Topic 9: Oxidation and Reduction
9.1 Introduction to oxidation and reduction
9.1.1 Define oxidation and reduction in terms of electron
loss and gain.
9.1.2 Deduce the oxidation number of an element in a
compound.
9.1.3 State the names of compounds using oxidation
numbers.
9.1.4 Deduce whether an element undergoes oxidation or
reduction in reactions using oxidation numbers.
1
9.1.1 Define oxidation and reduction in terms
of electron loss and gain.
Many important chemical reactions
involve a transfer of electrons
Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
• In this reaction, the Mg will
_________________ electrons to
become Mg 2+
• In this reaction, the 2H+ will
_________________ electrons to
become H2
2
9.1.1 Define oxidation and reduction in terms
of electron loss and gain.
Oxidation: a loss of electrons
Reduction: a gain of electrons
LEO goes GER
Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
Which substance is being oxidized?
_________
Which substance is being reduced?
_________
3
You can’t have one… without the other!
• Reduction (gaining electrons) can’t happen
without an oxidation to provide the electrons.
• You can’t have 2 oxidations or 2 reductions in the
same equation. Reduction has to occur at the
cost of oxidation
LEO the lion says GER!
o l x
s e i
e c d
t a
r t
o i
n o
s n
GER!
a l e
i e d
n c u
t c
r t
o i
n o
s n
9.1.2 Deduce the oxidation number of an
element in a compound.
All elements in a compound (even covalent
compounds) can be assigned oxidation
numbers. It is a convenient form of
“bookkeeping” and will assist in balancing
complex equations.
9.1.2 Deduce the oxidation number of an
element in a compound.
Rules for assigning oxidation numbers (o.n)
1. For any atom in its elemental form (not bonded to anything else), the
o.n. is 0.
Mg, S, H2, Cl2, P4 all are 0
2. For any monatomic ion, the o.n. equals the charge on the ion.
Mg2+ is +2; Cl- is -1
3. The o.n. of oxygen is usually -2, except peroxides like Na2O2 and
H2O2, where it is -1
4. The o.n. of hydrogen is +1 when bonded to a nonmetal and -1 when
bonded to a metal.
5. The o.n of F- is -1. The o.n. of the other halogens is -1 except when
combined with oxygen.
6. The sum of the o.n. of all atoms in a neutral compound is zero.
7. The sum of the o.n. in a polyatomic ion equals the charge on the ion.
6
9.1.2 Deduce the oxidation number of an
element in a compound.
Determine the oxidation number (state) of each element in
the following compounds.
•
•
•
•
H2S
S8
SCl2
Na2SO3
•
•
SO42H2O2
H is +1 (rule 4) ; S is -2 (rule 2)
S is 0 (rule 1)
Cl is -1 (rule 5); S is +2 (rule 6)
Na is +1 (rule 2); O is -2 (rule 3);
S is +4 (rule 6)
O is -2 (rule 3); S is +6 (rule 7)
H is +1 (rule 4); O is -1 (rule 3-peroxide)
7
9.1.2 Deduce the oxidation number of an
element in a compound.
Determine the oxidation number (state) of each element in
the following compounds.
•
•
•
•
•
•
•
•
•
P2O5
NaH
Cr2O7-2
SnBr4
HClO4
NO2N2
Ca(NO3)2
BaO2
P is _____; O is _____
Na is _____; H is _____
Cr is_____; O is _____
Sn is_____; Br is_____
H is _____; Cl is _____; O is _____
N is _____; O is _____
N is _____
Ca is _____; N is _____; O is _____
Ba is _____; O is _____
8
9.1.2 Deduce the oxidation number of an
element in a compound.
Determine the oxidation number (state) of each element in
the following compounds.
•
•
•
•
•
•
•
•
•
P2O5
NaH
Cr2O7-2
SnBr4
HClO4
NO2N2
Ca(NO3)2
BaO2
P is +5; O is -2
Na is +1; H is -1
Cr is +6; O is -2
Sn is +4; Br is -1
H is +1; Cl is +7; O is -2
N is +3; O is -2
N is 0
Ca is +2; N is +5; O is -2
Ba is +4; O is -2
9
9.1.3 State the names of compounds using
oxidation numbers.
Many metals can have more than one oxidation number.
• Exceptions are alkali metals (all +1), alkaline earth metals (all +2),
zinc (+2), aluminum (+3) & silver (+1).
These are identified using Roman numerals to denote the
charge.
• Iron(II) is Fe2+ (ferrous)
• Copper(I) is Cu+1
Iron(III) is Fe3+ (ferric)
Copper(II) is Cu+2
Name the following compounds using oxidation numbers.
SnCl4
_________________________________
Cr(NO3)3
_________________________________
KOH
_________________________________
PbSO4
_________________________________
CuBr
_________________________________
10
9.1.3 State the names of compounds using
oxidation numbers.
Binary covalent compounds can also be named using
Roman numerals but most use prefixes.
CO
CO2
PCl3
Carbon (II) oxide or carbon monoxide
Carbon (IV) oxide or carbon dioxide
Phosphorus (III) chloride or phosphorus trichloride
Name the following compounds using oxidation numbers.
What are their common names?
SO3
PCl5
N2O
_________________________________________
_________________________________________
_________________________________________
11
9.1.4 Deduce whether an element undergoes
oxidation or reduction in reactions using
oxidation numbers.
Deduce what is being oxidized and what is being reduced
in:
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
• Assign oxidation numbers to both reactants and
products:
Mg + 2HCl  MgCl2 + H2
Mg:0; H:+1; Cl:-1  Mg:+2; Cl:-1; H:0
• Which species increased in o.n.? That is the one being
oxidized (losing electrons)
Mg is going from 0 to +2 so is oxidized
• Which species decreased in o.n.? That is the one being
reduced (gaining electrons)
H is going from +1 to 0 so is reduced
12
9.1.4 Deduce whether an element undergoes
oxidation or reduction in reactions using
oxidation numbers.
Deduce what is being oxidized and what is being reduced
in:
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
• Assign oxidation numbers to both reactants and
products:
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
Cu:0; Ag:+1; N:+5; O:-2  Cu:+2; N:+5; O:-2; Ag:0
• Which species increased in o.n.? That is the one being
oxidized (losing electrons)
Cu is going from 0 to +2 so is oxidized
• Which species decreased in o.n.? That is the one being
reduced (gaining electrons)
Ag is going from +1 to 0 so is reduced
13
9.1.4 Deduce whether an element undergoes
oxidation or reduction in reactions using
oxidation numbers.
Deduce what is being oxidized and what is being reduced
in:
I2O5(s) + 5CO(g)  I2(s) + 5CO2(g)
• Assign oxidation numbers to both reactants and
products:
I2O5(s) + 5CO(g)  I2(s) + 5CO2(g)
I:__; O:__; C:__; O:__ I:__; C:__; O:__
• Which species increased in o.n.? That is the one being
oxidized (losing electrons)
•
Which species decreased in o.n.? That is the one being
reduced (gaining electrons)
14
9.1.4 Deduce whether an element undergoes
oxidation or reduction in reactions using
oxidation numbers.
Deduce what is being oxidized and what is being reduced
in:
2Hg2+(aq) + N2H4(aq)  2Hg(l)  N2(g) + 4H+(aq)
Deduce what is being oxidized and what is being reduced
in:
Cl2(g) + H2O(l)  HCl(aq) + HClO(aq)
15
IB Topic 9: Oxidation and Reduction
9.2 Redox equations
9.2.1 Deduce simple oxidation and reduction halfequations given the species involved in a redox reaction.
9.2.2 Deduce redox equations using half-reactions.
9.2.3 Define the terms oxidizing agent and reducing
agent.
9.2.4 Identify the oxidizing and reducing agents in redox
equations.
16
9.2.1 Deduce simple oxidation and reduction
half-equations given the species involved in a
redox reaction.
Consider the reaction when copper metal is placed in a
solution of silver ions.
Half-reactions
Copper metal loses 2 electrons: Cu  Cu2+ + 2eSilver ion gains 1 electron: Ag+ + e-  Ag
17
9.2.1 Deduce simple oxidation and reduction
half-equations given the species involved in a
redox reaction.
The # of e- ____ has to equal the # of e- _________
The silver half-reaction has to be multiplied by 2
Cu  Cu2+ + 2e2Ag+ + 2e-  2Ag
Overall:
18
9.2.1 Deduce simple oxidation and reduction
half-equations given the species involved in a
redox reaction.
Consider the reaction when iodide ions are added to
chlorine water.
Which is more reactive, chlorine or iodine?__________
Since they are nonmetals will the more reactive species
gain or lose electrons? _________
19
9.2.1 Deduce simple oxidation and reduction
half-equations given the species involved in a
redox reaction.
Consider the reaction when iodide ions are added to
chlorine water.
Half-reactions:
2I-  I2 + 2eCl2 + 2e-  2Cl-
20
9.2.1 Deduce simple oxidation and reduction
half-equations given the species involved in a
redox reaction.
Since # electrons lost = # electrons gained, the overall
reaction is:
2I- + Cl2  I2 + 2Cl-
21
Balancing Half Equations
• Balanced half-reactions can be
added together to get overall,
balanced equation.
Zn(s) ---> Zn2+(aq) + 2eCu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Balancing Equations
for Redox Reactions
Some redox reactions have equations that must
be balanced by special techniques.
Mn = +7
Fe = +2
Mn
+ = +2 Fe = +3
MnO4- + 5 Fe2+ + 8 H
---> Mn2+ + 5 Fe3+ + 4 H2O
Balancing Equations
Consider the
reduction of Ag+
ions with copper
metal.
Cu + Ag+
--give--> Cu2+ + Ag
Balancing Equations
Step 1:
Divide the reaction into half-reactions,
one for oxidation and the other for reduction.
Ox
Cu ---> Cu2+
Red
Ag+ ---> Ag
Step 2:
Balance each element for mass.
Already done in this case.
Step 3:
Balance each half-reaction for charge
by adding electrons.
Ox
Cu ---> Cu2+ + 2eRed
Ag+ + e- ---> Ag
Balancing Equations
Step 4:
Multiply each half-reaction by a factor
so that the reducing agent supplies as many
electrons as the oxidizing agent requires.
Reducing agent
Cu ---> Cu2+ + 2eOxidizing agent
2 Ag+ + 2 e- ---> 2 Ag
Step 5:
Add half-reactions to give the overall
equation.
Cu + 2 Ag+ ---> Cu2+ + 2Ag
The equation is now balanced for both charge and
mass.
Balancing Equations
Balance the following in acid solution—
VO2+ + Zn ---> VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn ---> Zn2+
Red
VO2+ ---> VO2+
Step 2:
Balance each half-reaction for mass.
Ox
Zn ---> Zn2+
Red
2 H+ + VO2+ ---> VO2+ + H2O
Add H2O on O-deficient side and add H+
on other side for H-balance.
Balancing Equations
Step 3:
Ox
Red
Balance half-reactions for charge.
Zn ---> Zn2+ + 2ee- + 2 H+ + VO2+ ---> VO2+ + H2O
Step 4:
Multiply by an appropriate factor.
Ox
Zn ---> Zn2+ + 2eRed 2e- + 4 H+ + 2 VO2+
---> 2 VO2+ + 2 H2O
Step 5:
Add balanced half-reactions
Zn + 4 H+ + 2 VO2+
---> Zn2+ + 2 VO2+ + 2 H2O
Tips on Balancing Equations
• Never add O2, O atoms, or
O2- to balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end to
make sure mass and charge
are balanced.
• PRACTICE!
9.2.2 Deduce redox equations using halfreactions.
Balancing more complex red-ox reactions using halfreactions
1) Write the unbalanced equation in ionic form.
2) Write separate half-reactions for the oxidation and
reduction processes, adding the appropriate electrons.
3) Balance the half-reactions. You may add H2O and H+ to
balance oxygen and hydrogen as needed.
4) Multiply each half-reaction by an appropriate number to
make the number of electrons equal in both.
5) Add the half-reactions to show an overall equation.
6) Add the spectator ions and balance the equation.
30
9.2.2 Deduce redox equations using halfreactions.
S(s) + HNO3(aq)  SO2(g) + NO(g) + H2O(l)
1) Write the unbalanced equation in ionic form.
2) Write separate half-reactions for the oxidation and
reduction processes, adding the appropriate electrons.
31
9.2.2 Deduce redox equations using halfreactions.
S(s) + HNO3(aq)  SO2(g) + NO(g) + H2O(l)
3) Balance the half-reactions. You may add H2O and H+ to
balance oxygen and hydrogen as needed.
4) Multiply each half-reaction by an appropriate number to
make the number of electrons equal in both.
32
9.2.2 Deduce redox equations using halfreactions.
S(s) + HNO3(aq)  SO2(g) + NO(g) + H2O(l)
5)
Add the half-reactions to show an overall equation.
6)
Add the spectator ions and balance the equation.
33
9.2.2 Deduce redox equations using halfreactions.
Cr2O72-(aq) + Cl-(aq) + H+(aq)  Cr3+(aq) + Cl2(g) + H2O(l)
1) Write the unbalanced equation in ionic form.
2) Write separate half-reactions for the oxidation and
reduction processes, adding the appropriate electrons.
3) Balance the half-reactions. You may add H2O and H+ to
balance oxygen and hydrogen as needed.
4) Multiply each half-reaction by an appropriate number to
make the number of electrons equal in both.
5) Add the half-reactions to show an overall equation.
6) Add the spectator ions and balance the equation.
34
9.2.2 Deduce redox equations using halfreactions.
Cu(s) + NO3-(aq) + H+(aq) Cu2+(aq) + NO2(g) + H2O(l)
1) Write the unbalanced equation in ionic form.
2) Write separate half-reactions for the oxidation and
reduction processes, adding the appropriate electrons.
3) Balance the half-reactions. You may add H2O and H+ to
balance oxygen and hydrogen as needed.
4) Multiply each half-reaction by an appropriate number to
make the number of electrons equal in both.
5) Add the half-reactions to show an overall equation.
6) Add the spectator ions and balance the equation.
35
9.2.3 Define the terms oxidizing agent and
reducing agent.
Oxidizing Agent (oxidant):
• Substances that are able to
oxidize other substances
• They are themselves reduced
• Substances that readily gain
electrons
Reducing Agent (reductant):
• Substances that are able to
reduce other substances
• They are themselves oxidized
• Substances that readily lose
electrons
Oxidizing
Agents
Reducing
Agents
Oxygen, O2
Carbon, C
Halogens
F2,Cl2,Br2,I2
Carbon
monoxide, CO
Manganate(VII)
ions (H+/MnO4-)
Alkali Metals
Li, Na, K
Dichromate(VI)
Hydrogen
ions (H+/Cr2O72-) sulfide, H2S
Nitric acid, HNO3 Sulfur dioxide,
SO2 (SO32-(aq))
36
9.2.4 Identify the oxidizing and reducing
agents in redox equations.
Consider the following reactions:
1) S(s) + HNO3(aq)  SO2(g) + NO(g) + H2O(l)
2) Cr2O72-(aq) + Cl-(aq) + H+(aq)  Cr3+(aq) + Cl2(g) + H2O(l)
3) Cu(s) + NO3-(aq) + H+(aq) Cu2+(aq) + NO2(g) +
H2O(l)
For each reaction, state the oxidizing agent and the
reducing agent
37
9.2.4 Identify the oxidizing and reducing
agents in redox equations.
Consider the following reaction:
Cd(s) + NiO2(s) + H2O(l)  Cd2+(aq) + Ni2+(aq) + OH-(aq)
1.
What is the oxidizing agent?
2.
What is the reducing agent?
3.
Balance the equation
38
IB Topic 9: Oxidation and Reduction
9.3 Reactivity
9.3.1 Deduce a reactivity series based on the chemical
behavior of a group of oxidizing and reducing agents.
9.3.2 Deduce the feasibility of a redox reaction from a
given reactivity series.
39
Let’s Go Back to
Electronegativity
• Define electronegativity
• What element theoretically has the
greatest electronegativity? Why?
• What element theoretically has the lowest
electronegativity? Why?
• The more willing a substance is to give up
electrons, the more reactive it is.
• This brings us to reactivity series…
Reactivity Series
•
•
•
•
•
•
•
•
•
•
•
K
Na
Li
Ca
Mg
Al
Zn
Fe
Pb
Cu
Ag
Reactivity Series
• The more willing a substance is to give up
electrons, the more reactive it is and the higher
in the reactivity series it is.
• A substance that gives up electrons is going
through _____________.
• So the higher in a reactivity series a substance
is, the better of a(n) ____________ agent it is.
Metals higher in the series can displace metal ions
lower in the series from a solution.
– Example: zinc can react with iron ions to form
zinc ions and precipitate (create) iron metal.
TABLE OF STANDARD
REDUCTION POTENTIALS
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
To determine an oxidation
from a reduction table, just
take the opposite sign of the
reduction!
reducing ability
of element
9.3.1 Deduce a reactivity series based on the
chemical behavior of a group of oxidizing and
reducing agents.
45
9.3.1 Deduce a reactivity series based on the
chemical behavior of a group of oxidizing and
reducing agents.
Activity Series of Metals
•
•
•
•
•
•
Metals at the top of the chart are easily oxidized.
Metals at the top of the chart are most reactive.
Metals above hydrogen react with acids*.
Metals will react with ions of metals below them.
Metals at the top of the chart are good reducing agents.
Ions of metals at the bottom of the chart are good
oxidizing agents.
*Copper does react with nitric acid but the nitrogen in HNO3
is reduced, not the H as in other acid-base reactions.
Balance the following and identify the o.a. & r.a.
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
46
9.3.1 Deduce a reactivity series based on the
chemical behavior of a group of oxidizing and
reducing agents.
Activity Series of Halogens
•
•
•
•
F2 + 2e-  2FCl2 + 2e-  2ClBr2 + 2e-  2BrI2 + 2e-  2I-
•
•
•
•
Halogens
Halogens
Halogens
Halogens
agents.
Decreasing
Reactivity
at the top of the chart are easily reduced.
at the top of the chart are most reactive.
will react with ions of halogens below them.
at the top of the chart are good oxidizing
47
9.3.2 Deduce the feasibility of a redox reaction
from a given reactivity series.
Will an aqueous solution of iron(II) chloride oxidize
magnesium metal? If so, write the redox equation.
•
Because Mg is above Fe2+ in the activity series, we
predict the reaction will occur:
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
Will bromine water, Br2(aq) displace Cl- in a solution of
NaCl? If so, write the redox reaction.
•
The reaction will not occur since Cl is above Br in the
activity series.
48
9.3.2 Deduce the feasibility of a redox reaction
from a given reactivity series.
•
•
•
•
Which of the following metals will be oxidized by Pb(NO3)2; Zn, Cu,
Fe? Write any redox reaction that occurs.
Predict whether a reaction occurs when the following reagents are
mixed: Cl2(aq) and KI(aq); Br2(aq) and LiCl(aq). Write any redox
reaction that occurs.
Write balanced chemical equations for the following reactions. If no
reaction occurs, simply write NR
– Zinc metal is added to a solution of silver nitrate
– Iron metal is added to a solution of aluminum sulfate
– Hydrochloric acid is added to cobalt metal
– Hydrogen gas is bubbled through a solution of iron(II) chloride
– Fluorine gas is bubbled through a solution of sodium iodide
Balance the following redox reaction. Identify the oxid & red
agents:
Zn + H+ + NO3-  Zn2+ + N2O + H2O
49
IB Topic 9: Oxidation and Reduction
9.4 Voltaic cells
9.4.1 Explain how a redox reaction is used to produce
electricity in a voltaic cell.
9.4.2 State that oxidation occurs at the negative electrode
(anode) and reduction occurs at the positive electrode
(cathode).
50
9.4.1 Explain how a redox reaction is used to
produce electricity in a voltaic cell.
• Voltaic cell: Two different half-cells
connected together to enable electron
transfer during the redox reaction to
produce energy in the form of electricity.
The electrons are produced at the half-cell
that is most easily oxidized.
9.4.1 Explain how a redox reaction is used to
produce electricity in a voltaic cell.
A strip of zinc is placed in a copper
solution.
• Write the oxidation reaction
• Write the reduction reaction
• Write the overall reaction
• Describe two observable changes
52
9.4.1 Explain how a redox reaction is used to
produce electricity in a voltaic cell.
This reaction can be used to perform
electrical work by using a voltaic
(galvanic) cell.
The transfer of electrons takes place
through an external pathway.
Metal strips are placed in solutions of
their ions. The metal strips are
connected by a wire for flow of
electrons.
The solutions are connected by a salt
bridge or separated by a porous
glass barrier. This maintains
electrical neutrality.
Electrons flow from the anode to the
cathode.
53
9.4.2 State that oxidation occurs at the
negative electrode (anode) and reduction occurs
at the positive electrode (cathode).
At the Zn electrode (anode):
• Oxidation occurs
• Negative electrode
• Electrons are produced and flow
through the external circuit
toward the cathode
• Zn2+ ions produced and migrate
away from the electrode
• Negative ions (anions) from the
salt bridge migrate into the
solution to balance the increase in
positive charges.
54
9.4.2 State that oxidation occurs at the
negative electrode (anode) and reduction occurs
at the positive electrode (cathode).
At the Cu electrode (cathode):
• Reduction occurs
• Positive electrode
• Electrons come from the anode
and move into the electrode
• Cu2+ ions migrate to the electrode
and gain electrons producing Cu
• Positive ions (cations) from the
salt bridge migrate into the
solution to balance the decrease
in positive charges.
55
Zn/Cu Voltaic Cell
+
Anode,
negative,
source of
electrons
Cathode,
positive,
sink for
electrons
Zn(s) ---> Zn2+(aq) + 2eEo = +0.76 V
Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
--------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Eo = +1.10 V
9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
• Salt bridge: creates a circuit by allowing ions to flow
through to balance the ionic charges of the solutions…
– Electrons DO NOT flow through the salt bridge, though!!!
57
9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
58
Voltaic Cells in Action
• http://www.youtube.com/watch?v=A0VUso
eT9aM
9.4.2 State that oxidation occurs at the
negative electrode (anode) and reduction
occurs at the positive electrode (cathode).
A voltaic cell similar to that shown in the previous slide is constructed.
One electrode compartment consists of a cadmium strip placed in
a solution of Cd(NO3)2 and the other has a nickel strip placed in a
solution of NiSO4. Cadmium is a more reactive metal than nickel.
a) Write the half-reactions that occur in the two electrode
compartments. Write the overall reaction.
b) Which electrode is the anode and which is the cathode?
c)
Indicate the signs of the electrodes.
d) Which way do electrons flow?
e) In which directions do the cations and anions migrate through the
solution?
60
9.4.2 State that oxidation occurs at the
negative electrode (anode) and reduction
occurs at the positive electrode (cathode).
A voltaic cell similar to that shown in slide 38 is constructed. One
electrode compartment consists of a silver strip placed in a
solution of AgNO3 and the other has a nickel strip placed in a
solution of NiSO4. Nickel is a more reactive metal than silver.
a) Write the half-reactions that occur in the two electrode
compartments. Write the overall reaction.
b) Which electrode is the anode and which is the cathode?
c)
Indicate the signs of the electrodes.
d) Which way do electrons flow?
e) In which directions do the cations and anions migrate through the
solution?
61
IB Topic 9: Oxidation and Reduction
9.5 Electrolytic cells
9.5.1 Describe, using a diagram, the essential components
of an electrolytic cell.
9.5.2 State that oxidation occurs at the positive electrode
(anode) and reduction occurs at the negative electrode
(cathode).
9.5.3 Describe how current is conducted in an electrolytic
cell.
9.5.4 Deduce the products of the electrolysis of a molten
salt.
62
9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
• Electrolytic cell: Used to make nonspontaneous redox reactions occur by
providing energy in the form of electricity
from an external source.
9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
•
•
•
•
In an electrolytic cell, electricity is
supplied from an external source
and is used to make a nonspontaneous reaction take place.
The substance that conducts
electricity in the cell is an
electrolyte (substance containing
ions).
Electrolytes do not conduct when
solid because ions are not free to
move and they have no
delocalized electrons.
Electrolytes do conduct when
molten or dissolved in water
because the ions are free to move
toward opposite charged
electrodes
64
9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
• The electrolyte conducts
electricity by the movement of
ions within it.
• Chemical reactions occur at
each electrode so that the
electrolyte is decomposed in
the process.
2NaCl(l)  2Na(l) + Cl2(g)
65
9.5.2 State that oxidation occurs at the positive
electrode (anode) and reduction occurs at the
negative electrode (cathode).
• Oxidation occurs at the
positive electrode (anode)
because negative ions (anions)
are attracted to it,
2Cl-(l)  Cl2(g) + 2e-
• Reduction occurs at the
negative electrode (cathode)
because positive ions (cations)
are attracted to it.
Na+ + e-  Na(l)
66
9.5.3 Describe how current is conducted in an
electrolytic cell.
1)
2)
3)
4)
5)
The electrochemical cell is a
voltaic cell producing electricity
from a chemical reaction. The
anode produces electrons.
Electrons flow from the anode of
the voltaic cell to the cathode of
the electrolytic cell.
Positive ions flow toward the
cathode and gain electrons
(become reduced).
Negative ions flow toward the
anode and lose electrons
(become oxidized).
Electrons flow from the anode of
the electrolytic cell to the
cathode of the voltaic cell
67
Electrolytic Cells in Action
• http://www.youtube.com/watch?v=lVK8Rx
kmOec&NR=1
Comparing Voltaic & Electrolytic
Cells
Create a chart, differentiating between:
• Charge of anode and cathode
• Site of oxidation and reduction
• Direction of electron flow
• Electrode that gains mass
• Electrode that loses mass
• Spontaneous or not
9.5.4 Deduce the products of the electrolysis
of a molten salt.
• Electrolysis: Passage of electric current
through an electrolyte
70
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.
19.8
9.5.4 Deduce the products of the electrolysis of a
molten salt.
• Because the salt has been
heated until it melts, the Na+
ions flow toward the negative
electrode and the Cl- ions flow
toward the positive electrode
Electrolysis of NaCl:
• Cathode (-): Na+ + e- → Na
• Anode (+): 2 Cl- → Cl2 + 2 e-
9.5.4 Deduce the products of the electrolysis
of a molten salt.
Sketch a cell for the electrolysis of molten MgBr2.
• Indicate the directions in which ions and electrons
move.
• Label the anode and cathode, indicating the charge
and the type of reaction occurring.
• Give the electrode reactions.
73
Electrolysis in Action
• http://www.youtube.com/watch?v=i9xS9tKMpc
Terms to Know
•
•
•
•
•
•
•
Oxidation
Reduction
Oxidation number
Oxidizing agent
Reducing agent
Reactivity series
Anode
•
•
•
•
•
•
•
Cathode
Electrolysis
Electrolyte
Half cell
Salt bridge
Voltaic cell
Electrolytic cell