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.

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)  Mg 2+ (aq) + H
2
(g)
• In this reaction, the Mg will
_________________ electrons to become Mg 2+
• In this reaction, the 2H+ will
_________________ electrons to become H
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)  Mg 2+ (aq) + H
2
(g)
Which substance is being oxidized?
_________
Which substance is being reduced?
_________

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
GER
o s e i t o n i x d a r o n s t c l e i a n r o n s t c l e i t o n e d u c
GER!

9.1.2 Deduce the oxidation number of an element in a compound.

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, H
2
, Cl
2
, P
4 all are 0
2. For any monatomic ion, the o.n. equals the charge on the ion.
Mg 2+ is +2; Cl is -1 and 3. The o.n. of oxygen is usually -2, except peroxides like Na
H
2
O
2
, where it is -1
2
4. The o.n. of hydrogen is +1 when bonded to a nonmetal and -1 when bonded to a metal.
O
2
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.

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.
• H
2
S
• S
8
• SCl
2
• Na
2
SO
3
• SO
4
• H
2
O
2-
2
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)

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.
• P
2
O
5
• NaH
• Cr
2
O
• SnBr
7
-2
• HClO
4
4
• NO
• N
2
-
2
• Ca(NO
3
)
2
• BaO
2
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 _____

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.
• P
2
O
5
• NaH
• Cr
2
O
• SnBr
7
-2
• HClO
4
4
• NO
• N
2
-
2
• Ca(NO
3
)
2
• BaO
2
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

9.1.3 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 Fe 2+ (ferrous)
• Copper(I) is Cu +1
Iron(III) is Fe 3+ (ferric)
Copper(II) is Cu +2
Name the following compounds using oxidation numbers.
SnCl
4
Cr(NO
3
KOH
PbSO
CuBr
4
)
3
_________________________________
_________________________________
_________________________________
_________________________________
_________________________________

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 Carbon (II) oxide or carbon monoxide
CO
PCl
2
3
Carbon (IV) oxide or carbon dioxide
Phosphorus (III) chloride or phosphorus trichloride
Name the following compounds using oxidation numbers.
What are their common names?
SO
3
PCl
N
2
5
_________________________________________
_________________________________________
O _________________________________________

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)  MgCl
2
(aq) + H
2
(g)
• Assign oxidation numbers to both reactants and products:
Mg + 2HCl  MgCl
2
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)
+ H
2
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

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) + 2AgNO
3
(aq)  Cu(NO
• Assign oxidation numbers to both reactants and products:
Cu + 2AgNO
3
3
)
 Cu(NO
3
)
2
2
(aq) + 2Ag(s)
+ 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

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:
• Assign oxidation numbers to both reactants and products:
I
I
2
2
O
O
5
5
(s) + 5CO(g)  I
(s) + 5CO(g)  I
2
2
(s) + 5CO
(s) + 5CO
2
(g)
2
(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)

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:
2Hg 2+ (aq) + N
2
H
4
(aq)  2Hg(l)  N
2
(g) + 4H + (aq)
Deduce what is being oxidized and what is being reduced in:
Cl
2
(g) + H
2
O(l)  HCl(aq) + HClO(aq)

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.

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  Cu 2+ + 2e -
Silver ion gains 1 electron: Ag + + e  Ag

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  Cu 2+ + 2e -
2Ag + + 2e  2Ag
Overall:

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? _________

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  I
Cl
2
2
+ 2e -
+ 2e  2Cl -

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 + Cl
2
 I
2
+ 2Cl -

Zn(s) ---> Zn 2+ (aq) + 2e-
Cu 2+ (aq) + 2e- ---> Cu(s)
--------------------------------------------
Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s)

Some redox reactions have equations that must be balanced by special techniques.
Mn = +7 Fe = +2
MnO
4
+ 5 Fe 2+ + 8 H +
Fe = +3
---> Mn 2+ + 5 Fe 3+ + 4 H
2
O

Balancing Equations
Consider the reduction of Ag + ions with copper metal.
Cu + Ag + --give--> Cu 2+ + Ag

Balancing Equations
Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction.
Ox
Red
Cu ---> Cu 2+
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
Red
Cu ---> Cu 2+ + 2e-
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 ---> Cu 2+ + 2e-
Oxidizing agent 2 Ag + + 2 e- ---> 2 Ag
Step 5: Add half-reactions to give the overall equation.
Cu + 2 Ag + ---> Cu 2+ + 2Ag
The equation is now balanced for both charge and mass.

Balancing Equations
Balance the following in acid solution —
VO
2
+ + Zn ---> VO 2+ + Zn
Step 1: Write the half-reactions
2+
Ox
Ox
Zn ---> Zn 2+
Red VO
2
+ ---> VO 2+
Step 2: Balance each half-reaction for mass.
Zn ---> Zn 2+
Red
2 H + + VO
2
+ ---> VO 2+ + H
2
O
Add H
2
O on O-deficient side and add H + on other side for H-balance.

Balancing Equations
Step 3: Balance half-reactions for charge.
Ox
Red
Zn ---> Zn 2+ + 2ee+ 2 H + + VO
2
+ ---> VO 2+ + H
2
O
Step 4: Multiply by an appropriate factor.
Ox Zn ---> Zn 2+ + 2e-
Red 2 e- + 4 H + + 2 VO
2
+
---> 2 VO 2+ + 2 H
2
O
Step 5: Add balanced half-reactions
Zn + 4 H + + 2 VO
2
+
---> Zn 2+ + 2 VO 2+ + 2 H
2
O

• Never add O
2
, O atoms, or
O 2to balance oxygen.
• Never add H
2 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 H
2
O 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.

9.2.2 Deduce redox equations using halfreactions.
S(s) + HNO
3
(aq)  SO
2
(g) + NO(g) + H
2
O(l)
1) Write the unbalanced equation in ionic form.
2) Write separate half-reactions for the oxidation and reduction processes, adding the appropriate electrons.

9.2.2 Deduce redox equations using halfreactions.
S(s) + HNO
3
(aq)  SO
2
(g) + NO(g) + H
2
O(l)
3) Balance the half-reactions. You may add H
2 balance oxygen and hydrogen as needed.
O and H + to
4) Multiply each half-reaction by an appropriate number to make the number of electrons equal in both.

9.2.2 Deduce redox equations using halfreactions.
S(s) + HNO
3
(aq)  SO
2
(g) + NO(g) + H
2
O(l)
5) Add the half-reactions to show an overall equation.
6) Add the spectator ions and balance the equation.

9.2.2 Deduce redox equations using halfreactions.
Cr
2
O
7
2(aq) + Cl (aq) + H + (aq)  Cr 3+ (aq) + Cl
2
(g) + H
2
O (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 H
2 balance oxygen and hydrogen as needed.
O and H + to
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.

9.2.2 Deduce redox equations using halfreactions.
Cu(s) + NO
3
(aq) + H + (aq)  Cu 2+ (aq) + NO
2
(g) + H
2
O(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 H
2 balance oxygen and hydrogen as needed.
O and H + to
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.

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
Oxygen, O
2
Halogens
F
2
,Cl
2
,Br
2
,I
2
Manganate(VII) ions (H + /MnO
4
)
Dichromate(VI) ions (H+/Cr
2
O
7
2)
Nitric acid, HNO
3
Reducing
Agents
Carbon, C
Carbon monoxide, CO
Alkali Metals
Li, Na, K
Hydrogen sulfide, H
2
S
Sulfur dioxide,
SO
2
(SO
3
2(aq))

9.2.4 Identify the oxidizing and reducing agents in redox equations.
Consider the following reactions:
1) S(s) + HNO
3
(aq)  SO
2
(g) + NO(g) + H
2
O(l)
2) Cr
2
O
7
2(aq) + Cl (aq) + H + (aq)  Cr 3+ (aq) + Cl
2
(g) + H
2
O (l)
3) Cu(s) + NO
3
H
2
O(l)
(aq) + H + (aq)  Cu 2+ (aq) + NO
2
(g) +
For each reaction, state the oxidizing agent and the reducing agent

9.2.4 Identify the oxidizing and reducing agents in redox equations.
Consider the following reaction:
Cd (s) + NiO
2
(s) + H
2
O (l)  Cd 2+ (aq) + Ni 2+ (aq) + OH (aq)
1. What is the oxidizing agent?
2. What is the reducing agent?
3. Balance the equation

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.

• 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…

• K
• Na
• Li
• Ca
• Mg
• Al
• Zn
• Fe
• Pb
• Cu
• Ag

• 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.

oxidizing ability of ion Eo (V)
Cu2+ + 2eCu +0.34
2 H+ + 2eH
2
0.00
Zn2+ + 2e-
To determine an oxidation from a reduction table, just take the opposite sign of the reduction!
Zn -0.76
reducing ability of element

9.3.1 Deduce a reactivity series based on the chemical behavior of a group of oxidizing and reducing agents.

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 HNO is reduced, not the H as in other acid-base reactions.
Balance the following and identify the o.a. & r.a.
3
Cu + HNO
3
 Cu(NO
3
)
2
+ NO
2
+ H
2
O

9.3.1 Deduce a reactivity series based on the chemical behavior of a group of oxidizing and reducing agents.
Activity Series of Halogens
• F
2
• Cl
2
• Br
2
• I
2
+ 2e  2F -
+ 2e  2Cl -
+ 2e  2Br -
+ 2e  2I -
Decreasing
Reactivity
• Halogens at the top of the chart are easily reduced.
• Halogens at the top of the chart are most reactive.
• Halogens will react with ions of halogens below them.
• Halogens at the top of the chart are good oxidizing agents.

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 Fe 2+ in the activity series, we predict the reaction will occur:
Mg(s) + Fe 2+ (aq)  Mg 2+ (aq) + Fe(s)
Will bromine water, Br
2
(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.

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(NO
3
)
2
; Zn, Cu,
Fe? Write any redox reaction that occurs.
• Predict whether a reaction occurs when the following reagents are
(aq) and LiCl(aq). Write any redox mixed: Cl
2
(aq) and KI(aq); Br
2 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 + + NO
3
 Zn 2+ + N
2
O + H
2
O

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).

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

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.

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
• Zn 2+ 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.

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
• Cu 2+ 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.

+
Anode, negative, source of electrons
Zn(s) ---> Zn 2+ (aq) + 2e-
Cu 2+ (aq) + 2e- ---> Cu(s)
E o = +0.76 V
E o = +0.34 V
---------------------------------------------------------------
Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s)
E o = +1.10 V
Cathode, positive, sink for electrons

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!!!

9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.

•

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(NO
3 solution of NiSO
4
)
2 and the other has a nickel strip placed in a
. 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?

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 AgNO solution of NiSO
4
3 and the other has a nickel strip placed in a
. 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?

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.

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

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) + Cl
2
(g)

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)  Cl
2
(g) + 2e -
• Reduction occurs at the negative electrode (cathode) because positive ions (cations) are attracted to it.
Na + + e  Na(l)

9.5.3 Describe how current is conducted in an electrolytic cell.
1) The electrochemical cell is a voltaic cell producing electricity from a chemical reaction. The anode produces electrons.
2) Electrons flow from the anode of the voltaic cell to the cathode of the electrolytic cell.
3) Positive ions flow toward the cathode and gain electrons
(become reduced).
4) Negative ions flow toward the anode and lose electrons
(become oxidized).
5) Electrons flow from the anode of the electrolytic cell to the cathode of the voltaic cell

•

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

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 → Cl
2
+ 2 e-

9.5.4 Deduce the products of the electrolysis of a molten salt.
Sketch a cell for the electrolysis of molten MgBr
2
.
• 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.

•

• Oxidation
• Reduction
• Oxidation number
• Oxidizing agent
• Reducing agent
• Reactivity series
• Anode
• Cathode
• Electrolysis
• Electrolyte
• Half cell
• Salt bridge
• Voltaic cell
• Electrolytic cell