Oxidation and Reduction IB Core Objective 911 Define oxidation and reduction in terms of electron loss and gain Define Give the precise meaning of a word phrase or physical quantity ID: 258098
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Slide1
Topic 9
Oxidation and ReductionSlide2
IB Core Objective
9.1.1 Define
oxidation
and
reduction
in terms of electron loss and gain.
Define: Give the precise meaning of a word, phrase or physical quantity.Slide3
9.1.1 Define
oxidation
and
reduction
in terms of electron loss and gain.
Oxidation: The loss of electrons
Fe
2+
(
aq
)
→
Fe
3+
(
aq
) + e
-
Reduction: The gain of electrons
2H
+
(
aq
) + 2e
-
→
H
2
(g)Slide4
9.1.1 Define oxidation
and
reduction
in terms of electron loss and gain.
Helpful Mnemonic
This is Leo the Lion
LEO
goes
GER
L
oss of
E
lectrons is
O
xidation
G
ain of
E
lectrons is
R
eductionSlide5
9.1.1 Define oxidation
and
reduction
in terms of electron loss and gain.
Or another if you prefer…
OIL RIG
O
xidation
Is L
oss of electrons.Reduction Is Gain of electrons.Slide6
IB Core Objective
9.1.2 Deduce the oxidation number of an element in a compound.
Deduce: Reach a conclusion from the information given.Slide7
9.1.2 Deduce the oxidation number of an element in a compound.
In order to keep track of what loses electrons and what gains them, we assign
oxidation numbers
.Slide8
9.1.2 Deduce the oxidation number of an element in a compound.
A species is
oxidized
when it loses electrons.
Here, zinc loses two electrons to go from neutral zinc metal to the Zn
2+
ion.Slide9
9.1.2 Deduce the oxidation number of an element in a compound.
A species is
reduced
when it gains electrons.
Here, each of the H
+
gains an electron and they combine to form H
2
.Slide10
9.1.2 Deduce the oxidation number of an element in a compound.
It may be easier to find what is being reduced and oxidized by splitting the equation into “half equations”.
For example, with Zn(s) + 2H
+
(
aq
)
→ Zn
2+(aq) + H2
(g)It can be split up as: Zn(s) → Zn2+(aq) + 2e- and 2H+(aq) + 2e-
→
H
2
(g)Slide11
9.1.2 Deduce the oxidation number of an element in a compound.
It is not always easy to split equations into half equations.
Consider the following reaction:
Can you tell which is being oxidized?
If not, then we need to use
oxidation numbers
.
N
2(g) + 3H2(g) 2NH3(g)Slide12
9.1.2 Deduce the oxidation number of an element in a compound.
Oxidation Number
The charge that an atom would have if all covalent bonds were broken so that the more electronegative element kept all the electrons.Slide13
9.1.2 Deduce the oxidation number of an element in a compound.
Oxidation Number Rules
Elements in elemental state = 0
F = -1 (always)
O = -2 (except in H
2
O
2
where its +1)H = +1 (except in hydrides H-)
Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1)The sum of the oxidation numbers in a neutral compound is 0.The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.Slide14
9.1.2 Deduce the oxidation number of an element in a compound.
Find the oxidation number for the following:
Nitrogen in N
2
=
Carbon in CH
4
=
Sulfur in H2
SO4 =Phosphorous in PCl4+ =Iodine in IO4- =Answers: 0, -4, +6, +5, +7
Elements in elemental state = 0
F = -1 (always)
O = -2 (except in H
2
O
2
where its +1)
H = +1 (except in hydrides H
-
)
Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1)Slide15
IB Core Objective
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
Deduce: Reach a conclusion from the information given. (
Obj
3)Slide16
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
Let’s go back to the equation:
What is the oxidation number for nitrogen on both sides?
Has it been oxidized or reduced?
Answer: Oxidation number goes from 0 to -3. It has gained electrons, therefore it has been reduced.
N
2(g)
+ 3H
2(g)
2NH3(g)Slide17
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
Consider the reaction between MnO
4
−
and C
2
O
4
2−
:MnO4−(aq) + C2O42−(aq) Mn2+(
aq
)
+ CO
2
(
aq
)Slide18
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
MnO
4
−
+ C
2
O
4
2-
Mn2+
+ CO
2
First, assign
oxidation numbers.
+7
+3
+4
+2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
Next, find out if carbon and manganese are being oxidized or reduced.Slide19
IB Core Objective
9.1.3 State the names of compounds using oxidation numbers.
State:
Give a specific name, value or other brief answer without explanation or calculation. (
Obj
1)Slide20
9.1.3 State the names of compounds using oxidation numbers.
For elements that have a variable oxidation number, the oxidation state is signified by Roman numerals.
Example: Fe
+3
would be written as Iron(III)
How would you write the following?
FeCl
2
FeCl3 MnO
4- Cr2O3Answers: iron(II) chloride, iron(III) chloride, permanganate (VII), chromium(III) oxideChallenge: How would you write the formula for ammonium dichromate?Answer: (NH4)2Cr2
O
7Slide21
Ammonium dichromate volcano
(NH
4
)
2Cr
2
O
7 --> Cr
2O3 + 4 H2O + N2
Is chromium oxidized or reduced in this reaction?Is nitrogen oxidized or reduced in this reaction?Answer: Chromium is reduced from +6 to +3 Nitrogen is oxidized from +3 to 0Slide22
IB Core Objective
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Deduce:
Reach a conclusion from the information given.Slide23
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Let’s look at an equation that we worked with before….
What is wrong with this equation?
Answer: It is not balanced!
We have worked with half equations before (zinc and hydrogen). Now we’ll dig deeper.
MnO
4
−
+ C
2
O
4
2-
Mn
2+
+ CO
2Slide24
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
General rules for balancing half equations
1) Balance atoms being oxidized or reduced
2) Add H
2
0 to balance Oxygen atoms
3) Add H
+(aq) to balance Hydrogen atoms4) Add e- to balance chargeSlide25
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Oxidation Half-Reaction
C
2
O
4
2
− CO2To balance the carbon, we add a coefficient of 2:
C
2
O
4
2
−
2
CO
2Slide26
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Oxidation Half-Reaction
C
2
O
4
2
− 2 CO2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.C2O
4
2
−
2 CO
2
+ 2 e
−Slide27
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Reduction Half-Reaction
MnO
4
−
Mn
2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.MnO4−
Mn
2+
+ 4 H
2
OSlide28
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Reduction Half-Reaction
MnO
4
−
Mn
2+ + 4 H2O To balance the hydrogen, we add 8 H+ to the left side.8 H
+
+
MnO
4
−
Mn
2+
+ 4 H
2
OSlide29
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
Reduction Half-Reaction
8
H
+
+ MnO
4
− Mn2+ + 4 H2O To balance the charge, we add 5 e− to the left side.
5 e
−
+
8 H
+
+ MnO
4
−
Mn
2+
+ 4 H
2
OSlide30
IB Core Objective
9.2.2 Deduce
redox
equations using half-equations.
Deduce:
Reach a conclusion from the information given.Slide31
9.2.2 Deduce
redox
equations using half-equations.
Combining the Half-Reactions
Now
we evaluate the two half-reactions together:
C
2
O
42− 2 CO2 + 2 e−5 e− + 8 H+ + MnO4−
Mn
2+
+ 4 H
2
O
To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.Slide32
9.2.2 Deduce
redox
equations using half-equations.
Combining the Half-Reactions
5
C
2
O
4
2− 10 CO2 + 10 e−10 e− + 16 H+ + 2 MnO4− 2 Mn
2+
+ 8 H
2
O
When we add these together, we get:
10 e
−
+ 16 H
+
+ 2 MnO
4
−
+ 5 C
2
O
4
2
−
2 Mn
2+
+ 8 H
2
O + 10 CO
2
+10 e
−Slide33
9.2.2 Deduce
redox
equations using half-equations.
Combining the Half-Reactions
10
e
−
+ 16 H
+
+ 2 MnO4− + 5 C2O42−
2 Mn
2+
+ 8 H
2
O + 10 CO
2
+10 e
−
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16 H
+
+ 2 MnO
4
−
+ 5 C
2
O
4
2
−
2 Mn
2+
+ 8 H
2
O + 10 CO
2Slide34
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a
redox
reaction.
9.2.2 Deduce
redox
equations using half-equations.
Practice
Given two half-equations:
Cr2O
72-(aq) → Cr3+(aq)Fe2+ → Fe3+
Deduce the half-equations for each, then deduce the
redox
equation.Slide35
Answer
Cr
2
O
72-
(
aq
) + 14H+
(aq) + 6Fe2+(aq
) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)Slide36
IB Core Objective
9.2.3 Define the terms oxidizing agent and reducing agent.
Define:
Give the precise meaning of a word, phrase or physical quantity. (
Obj
1)Slide37
9.2.3 Define the terms oxidizing agent and reducing agent.
Oxidizing agent: Substance that is reduced and causes the oxidation of another substance in a
redox
reaction.
Reducing agent: Substance that is oxidized and causes the reduction of another substance in a
redox
reaction.
I am oxidizing agent man.
I am here to take your electrons.Slide38
IB Core Objective
9.2.4 Identify the oxidizing and reducing agents in
redox
equations.
Identify: Find an answer from a given number of possibilities. (
Obj
2)Slide39
9.2.4 Identify the oxidizing and reducing agents in redox
equations.
Identify the oxidizing and reducing agents in the following equations:
Sn
2+
(
aq
)
+ 2Fe3+(aq
) → Sn4+(aq) Fe2+(aq)Mg(s) + 2HCl(aq) → MgCl
2(
aq
)
+ H
2(g)Slide40
9.2.4 Identify the oxidizing and reducing agents in
redox
equations.
Deduce the following half equations, deduce the
redox
equation, and identify the oxidizing agent and the reducing agent.
MnO
4
-
(aq) → Mn-2(aq)
SO
2(
aq
)
→
SO
4
2-
(
aq
)Slide41
IB Core Objective
9.3.1 Deduce a reactivity series based upon the chemical
behaviour
of a group of oxidizing and reducing agents.
Deduce:
Reach a conclusion from the information given.Slide42
9.3.1 Deduce a reactivity series based upon the chemical
behaviour
of a group of oxidizing and reducing agents.
Recall in acids and bases that a strong acid had a weak conjugate base.
Same in
redox
reactions. The conjugate of a powerful oxidizing agent is a weak reducing agent.
F
2 + 2e- ↔ 2F
-Strong oxidizing agentWeak reducing agentSlide43
9.3.1 Deduce a reactivity series based upon the chemical
behaviour
of a group of oxidizing and reducing agents.
Mr. F can really attract the electrons (more electronegative).
When Mr. F has the electrons, he doesn’t want to let them go.
So although he is a good oxidizing agent, he is a poor reducing agent. (He doesn’t like to reduce the number of his electrons!)Slide44
9.3.1 Deduce a reactivity series based upon the chemical behaviour
of a group of oxidizing and reducing agents.
Think back to Topic 3 on Periodicity.
What are the trends in
electronegativity
?Slide45
9.3.1 Deduce a reactivity series based upon the chemical
behaviour
of a group of oxidizing and reducing agents.
Compare
What exception do you see?
Hydrogen (Lithium is another
e
xception)Slide46
IB Core Objective
9.3.2 Deduce the feasibility of a
redox
reaction from a given reactivity series.
Deduce: Reach a conclusion from the information given.Slide47
9.3.2 Deduce the feasibility of a
redox
reaction from a given reactivity series.
Cl
2(
aq
)
+ 2I
-(aq) →
I2(aq) + 2Cl-(aq)Feasible? A: YesI2(aq) + 2Cl-(aq)
→
Cl
2(
aq
)
+ 2I
-
(
aq
)
Feasible?
A: No
Chlorine attracts electrons more strongly than iodine, so chlorine is a better oxidizing agent.Slide48
9.3.2 Deduce the feasibility of a redox
reaction from a given reactivity series.
Zn
(s)
+ Cu
2+
(
aq
) → Cu(s) + Zn
2+(aq)Feasible?A: YesCu(s) + Zn2+(aq) → Zn(s) + Cu2+(
aq
)
Feasible?
A: No
These examples are all
displacement reactions
, because they involve a more reactive metal or non-metal displacing the reactive one from its salt.Slide49
IB Core Objective
9.4.1 Explain how a
redox
reaction is used to produce electricity in a Voltaic cell.
Explain: Give a detailed account of causes, reasons or mechanisms.Slide50
9.4.1 Explain how a redox
reaction is used to produce electricity in a Voltaic cell.
A
Voltaic cell
is a device for converting
chemical energy
into
electrical energy
using a redox reaction. Slide51
9.4.1 Explain how a redox
reaction is used to produce electricity in a Voltaic cell.
Anode(-): Oxidation, forms a negative charge
Cathode(+): Reduction, forms a positive charge
2
+
2
+
e
-
e
-
e
-
e
-
e
-
e
-
2
+
2
+
e
-
e
-Slide52
9.4.1 Explain how a
redox
reaction is used to produce electricity in a Voltaic cell.
Lets harness some Energy!!
2
+
2
+
2
+
2
+
2
+
Zn(s)
Cu(s)
2
+
2
+
e
-
e
-
2
+
2
+
2
+
e
-
e
-
Problem, the highly negative charge on electrode causes (+) ions to be attracted back
Solution
Balance (-) charge by replacing it with some more negative ionsSlide53
9.4.1 Explain how a
redox
reaction is used to produce electricity in a Voltaic cell.
Lets harness some Energy!!
2
+
2
+
2
+
2
+
2
+
Zn(s)
Cu(s)
2
+
2
+
e
-
e
-
2
+
2
+
2
+
http://www.dynamicscience.com.au/tester/solutions/chemistry/redox/galvan5.swf
+
-
+
-
+
-
+
-
+
-
e
-
e
-
2
+Slide54
IB Core Objective
9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).
State: Give a specific name, value or other brief answer without explanation or calculation.Slide55
9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).
A typical cell looks like this.
The oxidation occurs at the
anode
.
The reduction occurs at the
cathode
.
Which of the metals is being reduced?
So which is the cathode?Slide56Slide57
9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).
Lead and zinc are set up in a voltaic cell.
Which one would be oxidized? Which one is being reduced?
A: Zinc is being oxidized. Lead is being reduced.
Which one would be the cathode and which would be the anode?
Zinc would be the anode, lead is the cathode.Slide58
IB Core Objective
9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.
Describe: Give a detailed account.Slide59
9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.
Homework:
Draw a diagram of an electrolytic cell.
Provide a brief description what is happening at each step, including the components, where oxidation and reduction is occurring, how current is conducted, and the products of a molten salt.
If you do this effectively, you will have down objectives 9.5.1 – 9.5.4Slide60
9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.
Need to have a liquid containing ions, which is called an
electrolyte
.Slide61
IB Core Objective
9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode).
State: Give a specific name, value or other brief answer without explanation or calculation.Slide62
9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode).
The anode attracts anions.
When the anions reach they anode, they lose electrons.
So are they oxidized or reduced?
A: oxidized
When
cations
reach the cathode they gain electrons and they are reduced.Slide63
IB Core Objective
9.5.3 Describe how current is conducted in an electrolytic cell.
Describe: Give a detailed account.Slide64
9.5.3 Describe how current is conducted in an electrolytic cell.
Electricity is supplied from an external source and used to create a non-spontaneous reaction.
Electrolyte solution can conduct electricity because the ions move towards oppositely charged electrodes.Slide65
IB Core Objective
9.5.4 Deduce the products of the electrolysis of a molten salt.
Deduce: Reach a conclusion from the information given.Slide66
9.5.4 Deduce the products of the electrolysis of a molten salt.
Sodium chloride
Negative chloride ions are attracted to the positive ions. There they lose electrons and are oxidized to chlorine gas:
2Cl
-
(l)
→
Cl
2(g) + 2e-
Positive sodium ions are attracted to the negative cathode. They gain electrons and are reduced to sodium metal:Na+(l) + e- → Na(l)Slide67
9.5.4 Deduce the products of the electrolysis of a molten salt.
Question
For every 2 mol of electrons that flow through the circuit, how many mol of chlorine gas and sodium metal will be produced?
A: 1 mol of chlorine gas and 2 mol of sodium.Slide68
IB HL Objective
19.1.1 Describe the standard hydrogen electrode.
Describe:
Give a detailed account. (
Obj
2)Slide69
19.1.1 Describe the standard hydrogen electrode.
Electrode:
An electrical conductor through which electric current leaves or enters
Anode:
Negative electrode where oxidation takes place.
Cathode
:
Positive
electrode where reduction takes place.Slide70
19.1.1 Describe the standard hydrogen electrode
.
The potential of any two electrodes can be compared using this apparatus
You will learn more about this voltaic cell later (SL topic).Slide71
19.1.1 Describe the standard hydrogen electrode.
Electrode potentials from the voltaic cell are measured relative to the
standard hydrogen electrode
(SHE).Slide72
19.1.1 Describe the standard hydrogen electrode.
The reference half-reaction is the reduction of H
+
(
aq
) to H
2
(g): 2H
+(aq) + 2e- ↔ H2(g)Slide73
IB HL Objective
19.1.2 Define the term
standard electrode potential (E
ѳ
).
Define:
Give the precise meaning of a word, phrase or physical quantity. (
Obj
1)Slide74
Electromotive Force (emf)
Water only spontaneously flows one way in a waterfall
.
Likewise, electrons only spontaneously flow one way in a
redox
reaction—from higher to lower potential energy.Slide75
Electromotive Force (
emf
)
The
potential difference between the anode and cathode in a cell is called the
electromotive force (
emf
)
.
It is also called the cell potential, and is designated Ecell.19.1.2 Define the term standard electrode potential (Eѳ).Slide76
19.1.2 Define the term standard electrode potential (E
ѳ
).
The standard hydrogen electrode (SHE) is defined as having a potential of zero.
Standard electrode potentials also refer to the conditions. In the SHE, the platinum electrode is surrounded by H
2
gas at 1
atm
(1.01 x 105 Pa), electrode is immersed in strong acid at 1.00 mol dm-3, and is kept at 298 K.
in a standard hydrogen electrode is equal to 0 V. is the standard reduction potential.Slide77
19.1.2 Define the term
standard electrode potential (E
ѳ
).
The electrode potential at standard conditions can be found through this
equation:
E
cell
=
E
red
(cathode)
−
E
red
(anode)
There is also another way….If the half-equation is being oxidized instead of reversed, just flip the values .
Example: K ↔ K
+
+e
-
E
ѳ
= +2.92 V.
Then add the two values from the two half-reactions together!Slide78
IB HL Objective
19.1.3 Calculate cell potentials using standard electrode potentials.
Calculate:
Find a numerical answer showing the relevant stages in the working (unless instructed not to do so). (
Obj
2)Slide79
19.1.3 Calculate cell potentials using standard electrode potentials.
Reduction
potentials for many electrodes have been measured and tabulated.
Standard Electrode PotentialsSlide80
19.1.3 Calculate cell potentials using standard electrode potentials.
For the oxidation in this cell,
For the reduction,
E
red
=
−
0.76 V
E
red
= +0.34 V
Slide81
19.1.3 Calculate cell potentials using standard electrode potentials.
E
cell
=
E
red
(cathode)
−
E
red
(anode)
= +0.34 V
−
(−0.76 V)
= +1.10 VSlide82
IB HL Objective
19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values.
Predict:
Give an expected result. (
Obj
3)Slide83
19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values.
Standard electrode potentials allow predictions to be made about which reactions could theoretically occur.
If the cell potential is negative, a spontaneous reaction cannot occur.
If the cell potential is positive, then the reaction could occur spontaneously. Slide84
19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values.
Can copper metal reduce hydrogen ions to hydrogen gas?
Cu ↔ Cu
2+
+ 2e
-
E
ѳ
= -0.34 V2H+ + 2e- ↔ H2
Eѳ = 0.00 VEѳcell = -0.34 VAnswer= NoSlide85
IB HL Objective
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
Predict:
Give an expected result.
Explain:
Give a detailed account of causes, reasons or mechanisms. Slide86
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
Question
How would you know if a solution has electrolytes in it?
A: It conducts electricity.
Electrolytes are easy to determine the products, since there is one positive ion and one negative ion.
So sodium chloride may dissociate in solution. Anything else?Slide87
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
Water is a poor conductor of electricity, but it still dissociates into ions:
H
2
O
(l)
↔ H
+
(aq) + OH-(
aq)In order to electrolyze water, need to add something more to conduct the current that easily produces ions, but won’t be oxidized/reduced. So a small amount of sulfuric acid is added.Slide88
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
If water is electrolyzed, what would the products be? Half-equations for each ion?
A: for the hydrogen ions:
2H
+
(
aq
) + 2e
- → H2(g)
for the hydroxide ions:4OH-(aq) → O2(g) + 2H2O(l) + 4e-Slide89
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
You will also need to know aqueous sodium chloride for this objective.
We will be going over this more with the SL students, since they will need to know this as well.Slide90
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
What if we were to electrolyze copper(II) sulfate solution? What are equations for ions being formed in solution?
A: CuSO
4
(
aq
)
→
Cu2+(aq) + SO42-(
aq)H2O(l) ↔ H+(aq) + OH-(aq)Inert (non-reactive) platinum or graphite electrodes can be used.Slide91
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
Based on the dissociation equations, which would be attracted to the positive electrode? Which would be attracted to the negative electrode? What are the half-reactions for these? Be sure to take into account
E
ѳ
potentials.
(-) electrode: Cu
2+
(
aq) + 2e- →
Cu(s)(+) electrode: 4OH-(aq) → 2H2O(l) + O2(g) + 4e-Because copper is below hydrogen in the reactivity series, it will gain electrons instead of the hydrogen. Because the sulfate would have a more positive value, it would keep its electrons over the hydroxide.Slide92
19.2.1 Predict and explain the products of electrolysis of aqueous solutions.
What if a copper electrode is used? What would the half equations be at each electrode?
(-) electrode: Cu2+(
aq
) + 2e-
→
Cu(s)
(+) electrode: Cu(s)
→ Cu2+(aq) + 2e-Slide93
IB HL Objective
19.2.2 Determine the relative amounts of the products formed during electrolysis.
Determine: Find the only possible answer.Slide94
19.2.2 Determine the relative amounts of the products formed during electrolysis.
In the electrolysis of water, what would the mol ratio be for products?
For oxygen gas:
4OH
-
(
aq
)
→ O2(g) + 2H
2O(l) + 4e-We need four mol of electrons to produce one mol of oxygen.For hydrogen gas:2H+(aq) + 2e- → H2(g)Four mol of electrons would produce two mol of hydrogen gas. Therefore the ratio would be 2 mol H
2
: 1 mol O
2Slide95
19.2.2 Determine the relative amounts of the products formed during electrolysis.
What factors will influence the amount of products produced and the rate?
The magnitude of current (increasing the flow of electrons). The stronger the current, the faster the reaction will take place.
Time: More time that current is allowed to pass, more products will be formed.
Charge on the ions: Look at the half-equations, and you can determine the mol ratios.Slide96
IB HL Objective
19.2.3 Describe the use of electrolysis in electroplating.
Describe: Give a detailed account.Slide97
19.2.3 Describe the use of electrolysis in electroplating.
Electroplating: Cathode becomes coated in a layer of metal (negative electrode).
Cathode is where reduction takes place.Slide98
19.2.3 Describe the use of electrolysis in electroplating.
Electroplating can be used to purify substances.
For example, copper is used for electrical wiring, and it needs to be pure otherwise the resistance increases.
So the positive electrode is impure copper and the negative electrode becomes pure copper.