Download Topic 9 - uaschemistry

Topic 9
Oxidation and Reduction
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.
9.1.1 Define oxidation and reduction in terms of
electron loss and gain.
Oxidation: The loss of electrons
Fe2+(aq) → Fe3+(aq) + eReduction: The gain of electrons
2H+(aq) + 2e- → H2(g)
9.1.1 Define oxidation and reduction in terms of
electron loss and gain.
Helpful Mnemonic
LEO goes GER
Loss of Electrons is
Oxidation
Gain of Electrons is
Reduction
9.1.1 Define oxidation and reduction in terms of
electron loss and gain.
Or another if you prefer…
OIL RIG
Oxidation Is Loss of
electrons.
Reduction Is Gain of
electrons.
IB Core Objective
• 9.1.2 Deduce the oxidation number of
an element in a compound.
• Deduce: Reach a conclusion from the
information given.
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.
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 Zn2+ ion.
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 H2.
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) → Zn2+(aq) + H2(g)
It can be split up as:
Zn(s) → Zn2+(aq) + 2eand 2H+(aq) + 2e- → H2(g)
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:
N2(g) + 3H2(g)  2NH3(g)
Can you tell which is being oxidized?
If not, then we need to use oxidation numbers.
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.
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 H2O2 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.
9.1.2 Deduce the oxidation number of an
element in a compound.
Find the oxidation number for the
following:
Nitrogen in N2
=
Carbon in CH4
=
Sulfur in H2SO4
=
Phosphorous in PCl4+ =
Iodine in IO4=
Answers: 0, -4, +6, +5, +7
– Elements in elemental
state = 0
– F = -1 (always)
– O = -2 (except in H2O2
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)
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)
9.1.4 Deduce whether an element undergoes oxidation
or reduction in reactions using oxidation numbers.
Let’s go back to the equation:
N2(g) + 3H2(g)  2NH3(g)
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.
9.1.4 Deduce whether an element undergoes oxidation or
reduction in reactions using oxidation numbers.
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
9.1.4 Deduce whether an element undergoes oxidation or
reduction in reactions using oxidation numbers.
+7
+3
+2
+4
MnO4− + C2O42-  Mn2+ + CO2
First, assign oxidation numbers.
Next, find out if carbon and manganese are being
oxidized or reduced.
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
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)
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?
FeCl2
FeCl3
MnO4- Cr2O3
Answers: iron(II) chloride, iron(III) chloride, permanganate (VII),
chromium(III) oxide
Challenge: How would you write the formula for
ammonium dichromate?
Answer: (NH4)2Cr2O7
Ammonium dichromate volcano
• (NH4)2Cr2O7 --> Cr2O3 + 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 0
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.
9.2.1 Deduce simple oxidation and reduction halfequations given the species involved in a redox reaction.
• Let’s look at an equation that we worked with
before….
MnO4− + C2O42-  Mn2+ + CO2
• 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.
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)
– 2)
– 3)
– 4)
Balance atoms being oxidized or reduced
Add H20 to balance Oxygen atoms
Add H+(aq) to balance Hydrogen atoms
Add e- to balance charge
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Oxidation Half-Reaction
C2O42−  CO2
To balance the carbon, we add a coefficient of 2:
C2O42−  2 CO2
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Oxidation Half-Reaction
C2O42−  2 CO2
The oxygen is now balanced as well. To
balance the charge, we must add 2 electrons
to the right side.
C2O42−  2 CO2 + 2 e−
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Reduction Half-Reaction
MnO4−  Mn2+
The manganese is balanced; to balance the
oxygen, we must add 4 waters to the right
side.
MnO4−  Mn2+ + 4 H2O
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Reduction Half-Reaction
MnO4−  Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+ to the
left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Reduction Half-Reaction
8 H+ + MnO4−  Mn2+ + 4 H2O
To balance the charge, we add 5 e− to the left
side.
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
IB Core Objective
• 9.2.2 Deduce redox equations using halfequations.
• Deduce: Reach a conclusion from the
information given.
9.2.2 Deduce redox equations using half-equations.
Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
To attain the same number of electrons on each
side, we will multiply the first reaction by 5
and the second by 2.
9.2.2 Deduce redox equations using half-equations.
Combining the Half-Reactions
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
9.2.2 Deduce redox equations using half-equations.
Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
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:
Cr2O72-(aq) → Cr3+(aq)
Fe2+ → Fe3+
Deduce the half-equations for each, then
deduce the redox equation.
Answer
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) →
2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
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)
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.
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)
9.2.4 Identify the oxidizing and reducing agents
in redox equations.
Identify the oxidizing and reducing agents in
the following equations:
Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) Fe2+(aq)
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
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.
• MnO4-(aq) → Mn-2(aq)
• SO2(aq) → SO42-(aq)
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.
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.
Strong oxidizing
agent
F2 + 2e- ↔ 2F-
Weak reducing
agent
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!)
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?
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
exception)
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.
9.3.2 Deduce the feasibility of a redox reaction from a
given reactivity series.
Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq)
Feasible?
A: Yes
I2(aq) + 2Cl-(aq) → Cl2(aq) + 2I-(aq)
Feasible?
A: No
Chlorine attracts electrons more strongly than
iodine, so chlorine is a better oxidizing agent.
9.3.2 Deduce the feasibility of a redox reaction from a
given reactivity series.
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
Feasible?
A: Yes
Cu(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.
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.
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.
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
e
e
-
ee e2e+
2+
e
e
9.4.1 Explain how a redox reaction is used to produce electricity in a
Voltaic cell.
• Lets harness some Energy!!
Problem, the highly negative charge on
electrode causes (+) ions to be attracted
back
Zn(s)
- e
e
e
e
2+
2+
2+
2+
2+
Solution
Balance (-) charge
by replacing it
with some more
negative ions
Cu(s)
9.4.1 Explain how a redox reaction is used to produce electricity in a
Voltaic cell.
• Lets harness some Energy!!
+
+
-
-
+
-
Zn(s)
Cu(s)
+
+
-
-
e
e
2+
e
2+
2+
e
2+
2+
http://www.dynamicsci
ence.com.au/tester/solu
tions/chemistry/redox/g
alvan5.swf
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.
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?
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.
IB Core Objective
• 9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
• Describe: Give a detailed account.
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 – 9.5.4
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.
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.
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.
IB Core Objective
• 9.5.3 Describe how current is conducted
in an electrolytic cell.
• Describe: Give a detailed account.
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.
IB Core Objective
• 9.5.4 Deduce the products of the electrolysis
of a molten salt.
• Deduce: Reach a conclusion from the
information given.
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) → Cl2(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)
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.
IB HL Objective
• 19.1.1 Describe the standard
hydrogen electrode.
• Describe: Give a detailed account.
(Obj 2)
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.
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).
19.1.1 Describe the standard hydrogen electrode.
• Electrode potentials from the voltaic cell are
measured relative to the standard hydrogen
electrode (SHE).
19.1.1 Describe the standard hydrogen electrode.
• The reference half-reaction is the reduction of
H+(aq) to H2(g): 2H+(aq) + 2e- ↔ H2(g)
IB HL Objective
• 19.1.2 Define the term standard
electrode potential (Eѳ).
• Define: Give the precise meaning of
a word, phrase or physical quantity.
(Obj 1)
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.
19.1.2 Define the term standard electrode
potential (Eѳ).
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).
• 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 H2 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.
•
0 V.
•
in a standard hydrogen electrode is equal to
is the standard reduction potential.
19.1.2 Define the term standard electrode potential
(Eѳ).
The electrode potential at standard
conditions can be found through this
equation:
 = Ered
 (cathode) − Ered
 (anode)
Ecell
There is also another way….If the half-equation is being oxidized
instead of reversed, just flip the values .
Example: K ↔ K+ +e- Eѳ = +2.92 V.
Then add the two values from the two half-reactions together!
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)
19.1.3 Calculate cell potentials using standard electrode
potentials.
Standard Electrode Potentials
Reduction
potentials for
many electrodes
have been
measured and
tabulated.
19.1.3 Calculate cell potentials using standard
electrode potentials.
• For the oxidation in this cell,
 = −0.76 V
Ered
• For the reduction,
Ered
 = +0.34 V
19.1.3 Calculate cell potentials using standard
electrode potentials.
Ecell
 = Ered
 (cathode) − Ered
 (anode)
= +0.34 V − (−0.76 V)
= +1.10 V
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)
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.
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 ↔ Cu2+ + 2e- Eѳ = -0.34 V
2H+ + 2e- ↔ H2 Eѳ = 0.00 V
Eѳcell = -0.34 V
Answer= No
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.
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?
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:
H2O(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.
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-
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.
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: CuSO4(aq) → Cu2+(aq) + SO42-(aq)
• H2O(l) ↔ H+(aq) + OH-(aq)
• Inert (non-reactive) platinum or graphite
electrodes can be used.
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 Eѳ potentials.
(-) electrode: Cu2+(aq) + 2e- → Cu(s)
(+) electrode: 4OH-(aq) → 2H2O(l) + O2(g) + 4eBecause 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.
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-
IB HL Objective
• 19.2.2 Determine the relative amounts
of the products formed during
electrolysis.
• Determine: Find the only possible
answer.
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) + 2H2O(l) + 4eWe 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 H2: 1 mol O2
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?
1. The magnitude of current (increasing the flow of
electrons). The stronger the current, the faster
the reaction will take place.
2. Time: More time that current is allowed to pass,
more products will be formed.
3. Charge on the ions: Look at the half-equations,
and you can determine the mol ratios.
IB HL Objective
• 19.2.3 Describe the use of
electrolysis in electroplating.
• Describe: Give a detailed account.
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.
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.