Download File - Mr Weng`s IB Chemistry

IB CHEMISTRY
Topic 9 Redox processes
Higher level
9.1 Oxidation and reduction
OBJECTIVES
• Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss,
electron transfer or change in oxidation number.
• An oxidizing agent is reduced and a reducing agent is oxidized.
• Variable oxidation numbers exist for transition metals and for most main-group nonmetals.
• The activity series ranks metals according to the ease with which they undergo
oxidation.
• The Winkler Method can be used to measure biochemical oxygen demand (BOD), used
as a measure of the degree of pollution in a water sample.
• Deduction of the oxidation states of an atom in an ion or a compound.
• Deduction of the name of a transition metal compound from a given formula, applying
oxidation numbers represented by Roman numerals.
• Identification of the species oxidized and reduced and the oxidizing and reducing
agents, in redox reactions.
• Deduction of redox reactions using half-equations in acidic or neutral solutions.
• Deduction of the feasibility of a redox reaction from the activity series or reaction data.
• Solution of a range of redox titration problems.
• Application of the Winkler Method to calculate BOD.
1. REDOX CALCULATIONS
Oxidation
GAIN OF OXYGEN
2Mg + O2
——> 2MgO
magnesium has been oxidised as it has gained oxygen
REMOVAL (LOSS) OF HYDROGEN
C2H5OH
——> CH3CHO + H2
ethanol has been oxidised as it has ‘lost’ hydrogen
Reduction
GAIN OF HYDROGEN
C2H4 + H2
——> C2H6
ethene has been reduced as it has gained hydrogen
REMOVAL (LOSS) OF OXYGEN
CuO + H2 ——> Cu + H2O
copper(II) oxide has been reduced as it has ‘lost’ oxygen
However as chemistry became more sophisticated, it
was realised that another definition was required
Oxidation/Reduction in terms of electrons
Oxidation and reduction are not only defined as changes in O and H
...
OXIDATION
Removal (loss) of electrons ‘OIL’
species will get less negative or more positive
REDUCTION
Gain of electrons
‘RIG’
species will become more negative or less positive
REDOX
When reduction and oxidation take place
Oxidizing agent = substance that causes oxidation to occur.
The oxidizing agent is reduced.
Reducing agent = substance that causes reduction to occur
The reducing agent is oxidized.
Which one is the oxidising/reducing agent?
Mg + 2H+  Mg2+ + H2
Oxidised
Reducing agent
Reduced
Oxidising agent
Cr2O72- + 14H+ + 6Fe2+  2Cr3+ + 6Fe3+ + 7H2O
Reduced
Oxidising agent
Oxidised
Reducing agent
2MnO4¯ + 5C2O42- + 16H+  2Mn2+ + 10CO2 + 8H2O
Reduced
Oxidising agent
Oxidised
Reducing agent
2S2O32- + I2  S4O62- + 2I¯
Oxidised
Reducing agent
Reduced
Oxidising agent
REDOX agents
Oxidising agents
• Oxygen
• Chlorine and other halogens
• Iron (III) ion
• Hydrogen peroxide (H2O2)
• Manganate (VII) ion
(Permanganate ion MnO4-)
• Dichromate (VI) ion (Cr2O72-)
Reducing agents
• Hydrogen
• Carbon and carbon
monoxide
• Metals
• Iron (II) ion
• Iodide ion
• Thosulfate ion (S2O32-)
Oxidation states
Used to...
tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
construct half equations and balance redox equations
ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
atoms
Na in Na =
0
neutral already ... no need to add any electrons
cations
Na in Na+ =
+1
need to add 1 electron to make Na+ neutral
anions
Cl in Cl¯ =
-1
need to take 1 electron away to make Cl¯ neutral
Q.
What are the oxidation states of the elements in the following?
a) C (0)
b) Fe3+ (+3)
c) Fe2+ (+2)
d) O2- (-2)
e) He (0)
f) Al3+ (+3)
Rules for Determining Oxidation State
1.
Free elements are assigned an oxidation state of zero.
2.
The sum of the oxidation states of all that atoms in a species
must be equal to the net charge on the species.
3.
The alkali metals (Li, Na, K, Rb, and Cs) in compounds are
always assigned an oxidation state of +1.
4.
Fluorine in compounds is always assigned an oxidation state
of -1.
5.
The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and
also Zn and Cd in compounds are always assigned an
oxidation state of +2.
6.
Hydrogen in compounds is assigned an oxidation state of +1,
unless assigned to a metal then it is -1.
Oxygen in compounds is assigned an oxidation state of -2,
except for H2O2, where it is -1.
Halogen in compounds is assigned an oxidation state of -1.
7.
8.
Oxidation states in molecules
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
H in H2
=
0
both are the same and must add up to Zero
COMPOUNDS
C in CO2 =
O in CO2 =
+4
-2
1 x +4 and 2 x -2 = Zero
Explanation
• because CO2 is a neutral molecule, the sum of the oxidation states must be zero
• for this, one element must have a positive OS and the other must be negative
HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?
• the more electronegative species will have the negative value
• electronegativity increases across a period and decreases down a group
• O is further to the right than C in the periodic table so it has the negative value
Problems:
A.
The oxidation states of the elements other than O, H or F are
SO2
O = -2
2 x -2 = - 4
overall neutral
S = +4
NH3
H = +1
3 x +1 = +3
overall neutral
N=-3
NO2
O = -2
2 x -2 = - 4
overall neutral
N = +4
NH4+
H = +1
4 x +1 = +4
overall +1
N=-3
IF7
F = -1
7 x -1 = - 7
overall neutral
I = +7
Cl2O7
O = -2
7 x -2 = -14
overall neutral
Cl = +7
NO3¯
O = -2
3 x -2 = - 6
overall -1
N = +5
NO2¯
O = -2
2 x -2 = - 4
overall -1
N = +3
SO32-
O = -2
3 x -2 = - 6
overall -2
S = +4
S2O32-
O = -2
3 x -2 = - 6
overall -2
S = +2
S4O62-
O = -2
6 x -2 = -12
overall -2
S = +2½ ! (10/4)
MnO42-
O = -2
4 x -2 = - 8
overall -2
Mn = +6
What is odd about the value of the oxidation state of S in S4O62- ?
An oxidation state must be a whole number (+2½ is the average value)
(14/2)
(4/2)
Naming with oxidation states
THE ROLE OF OXIDATION STATE IN NAMING SPECIES
To avoid ambiguity, the oxidation state is often included in the name of a species
manganese(IV) oxide shows that
Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3
S is in the +6 oxidation state
dichromate(VI) for Cr2O72-
Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5
P is in the +5 oxidation state
phosphorus(III) chloride for PCl3
P is in the +3 oxidation state
Q.
Name the following...
PbO2
lead(IV) oxide
SnCl2
tin(II) chloride
SbCl3
antimony(III) chloride
TiCl4
titanium(IV) chloride
BrF5
bromine(V) fluoride
Determining redox with oxidation states
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S.
Species has been REDUCED
Q.
INCREASE in O.S.
Species has been OXIDISED
State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+
—>
Fe3+
O
+2 to +3
I2
F2
C2O42H2O2
H2O2
Cr2O72Cr2O72SO42-
—>
—>
—>
—>
—>
—>
—>
—>
I¯
F2O
CO2
O2
H2O
Cr3+
CrO42SO2
R
R
O
O
R
R
N
R
0 to -1
0 to -1
+3 to +4
-1 to 0
-1 to -2
+6 to +3
+6 to +6
+6 to +4
Balancing redox half equations
STEPS:
Atom
1. Balance the atoms of the oxidation species
Oxn no.
2. Write oxidation states
e-1
3. Balance with electrons
H+
4. Balance charge with H+
H2O
5. Balance H with H2O
Problem 1: Balance the half reaction of Iron(II) being oxidised to
iron(III).
Step 1
Step 2
Step 3
Fe2+
+2
Fe2+
——> Fe3+
+3
——> Fe3+
+
e¯
Problem 2:
Step 1
Step 2
Step 3
Step 4
Step 5
MnO4¯ being reduced to Mn2+ in acidic solution.
MnO4¯ ———> Mn2+
+7
+2
MnO4¯ + 5e¯ ———> Mn2+
MnO4¯ + 5e¯ + 8H+ ———> Mn2+
MnO4¯ + 5e¯ + 8H+ ———> Mn2+ +
4H2O
Problem 3:
Step 1
Step 2
Cr2O72- being reduced to Cr3+ in acidic solution.
Cr2O72- ———>
Cr2O72- ———>
2 Cr at +6
Cr3+
2Cr3+
2 Cr at +3
Step 3
Step 4
Cr2O72- + 6e¯
Cr2O72- + 6e¯ + 14H+
Step 5
Cr2O72- + 6e¯ + 14H+
——> 2Cr3+
——> 2Cr3+
——>
2Cr3+ +
7H2O
Combining half equations
A combination of two ionic half equations, one involving oxidation and the other
reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1
Step 2
Step 3
Step 4
Write out the two half equations
Multiply the equations so that the number of electrons in each is the same
Add the two equations and cancel out the electrons on either side
If necessary, cancel any other species which appear on both sides
Problem 1: The reaction between manganate(VII) and iron(II)
Step 1
Fe2+ ——>
MnO4¯ + 5e¯ + 8H+ ——>
Fe3+ + e¯
Mn2+ + 4H2O
Oxidation
Reduction
Step 2
5Fe2+
MnO4¯ + 5e¯ + 8H+
——>
——>
5Fe3+ + 5e¯
Mn2+ + 4H2O
multiplied by 5
multiplied by 1
Step 3
MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——>
Step 4
MnO4¯ + 8H+ + 5Fe2+ ——>
Mn2+ + 4H2O + 5Fe3+ + 5e¯
Mn2+ + 4H2O + 5Fe3+
2. TITRATION CALCULATIONS
Titration of Iron with manganate(VII)
5Fe2+ + MnO4- + 8H+  5Fe3+ + Mn2+ + 4H2O
purple
colourless
Manganate(VII) ion is the oxidizing agent
oxidizing iron ions.
Titration of with iodine-thiosulfate
reaction
Iodine is usually supplied in the from of KI.
Oxidation forces I- ions to I2 by oxidizing agents such as
KMnO4 or K2Cr2O7.
I-(aq) + oxidizing agent  I2(aq) + reduced product
Starch is now added to react with the I2 formed making
a blue-black complex:
This is titrated with thiosulfate, removing it from this
complex to make the solution clear.
IODOMETRY
Oxidation -1
state
Reaction IAnalyte
0

I2
Oxidizing agent
oxidized
reduced
IODIMETRY
Oxidation 0
state
Reaction I2
Analyte
-1

Reducing agent
Example: Vitamin C
I
reduced
oxidized
3. THE WINKLER METHOD
Biological Oxygen Demand (BOD)
Definition: BOD is the measure of the dissolved oxygen (in
ppm) needed to biologically decompose the organic
matter in water over 5 days, at 20°C.
• Fish require about 3ppm of dissolved oxygen (0.003gdm-3)
• Maximum oxygen solubility in water is 0.009gdm-3
BOD (ppm)
Quality of water
<1
Almost pure water
5
Doubtful purity
10
Unacceptable quality
100-400
100-10 000
Waste from untreated sewage
Waste water from meat-processing
plant
Measurement of BOD
Using the Winkler Method:
1. Water sample is saturated with O2 so known concentration is reached
(9ppm) by shaking with air.
2. The sample is incubated at a fixed temperature (20°C) for 5 days to
allow the organic matter present to decay by microbial action.
3. Excess Mn2+ (aq) is added with OH- to use up remaining O2 and
convert to MnO2 & H2O
2Mn2+(aq) + 4OH-(aq) + O2(g)  2MnO2(s) + 2H2O(l)
4. I- in H+(aq) is added to form I2 & water & reform Mn2+ (aq)
MnO2(s) + 2I-(aq) + 4H+(aq)  Mn2+(aq) + I2(aq) + 2H2O(l)
All the solid precipitate must be dissolved.
5. Finally the I2 is titrated with thiosulfate (S2O32-) to release the iodide
ions (a starch indicator could be used which will form a blue/black starch
I2 complex that turns clear as thiosulphate turns I2 to I-).
I2(aq) + 2S2O32-(aq)  S4O62-(aq) +2I-(aq)
6. Use reaction stoichiometry to determine the number of moles of O2 left
after 5 days. Following the red arrows, for every mole of S2O32- used, ¼
mole of O2 is used up. Then covert to mg O2 per dm3 to get DO in ppm.
7. Convert this to BOD (ppm)  9ppm less DO
8. BOD higher than 9ppm are determined by first diluting the original
sample before incubation and then multiplying by the dilution factor.
9.2 Electrochemical cells
OBJECTIVES
Voltaic (Galvanic) cells:
• Voltaic cells convert energy from spontaneous, exothermic chemical processes to
electrical energy.
• Oxidation occurs at the anode (negative electrode) and reduction occurs at the
cathode (positive electrode) in a voltaic cell.
Electrolytic cells:
• Electrolytic cells convert electrical energy to chemical energy, by bringing about
non-spontaneous processes.
• Oxidation occurs at the anode (positive electrode) and reduction occurs at the
cathode (negative electrode) in an electrolytic cell.
• Construction and annotation of both types of electrochemical cells.
• Explanation of how a redox reaction is used to produce electricity in a voltaic cell
and how current is conducted in an electrolytic cell.
• Distinction between electron and ion flow in both electrochemical cells.
• Performance of laboratory experiments involving a typical voltaic cell using two
metal/metal-ion half-cells.
• Deduction of the products of the electrolysis of a molten salt.
4. VOLTAIC CELLS
Comparison of Electrochemical Cells
Cations are
produced
at the cathode
Galvanic/voltaic
Electrolytic
Cations go to
the cathode
produces
electrical
current
anode (-)
cathode (+)
salt bridge
two
electrodes
conductive
medium
reduction at
cathode
oxidation at
anode
DG < 0
need
power
source
anode (+)
cathode (-)
DG > 0
Voltaic cell
5. ELECTROLYTIC CELLS
Electrolytic cell
Testing electrolytes for conductivity
Graphite and
platinum
electrodes are
called inert
electrodes as
they don’t take
part in the
reaction
Inert electrolysis - Predictions
1. If the metal is high in the reactivity series you
will get hydrogen
2. If the below is low in the reactivity series you
will get the metal
3. If the halide solution is concentrated you will
get the halogen (chlorine, bromine, iodine).
With other common negative ions you will
get oxygen.
Electrolysis of a molten salt (NaCl)
cathode half-cell (-) REDUCTION
Na+ + e-  Na
-
anode half-cell (+) OXIDATION
2Cl-  Cl2 + 2eoverall cell reaction
2Na+ + 2Cl-  2Na + Cl2
+
battery
Cl2 (g) escapes
e-
Na (l)
Cl-
Na+
Cl-
(-)
electrode halfcell
Na+ + e-  Na
NaCl (l)
Na+
Cl-
(+)
Na+
electrode halfcell
2Cl-  Cl2 + 2e-
OBJECTIVES
• A voltaic cell generates an electromotive force (EMF) resulting in the movement of electrons from
the anode (negative electrode) to the cathode (positive electrode) via the external circuit. The EMF is
termed the cell potential (E⁰).
• The standard hydrogen electrode (SHE) consists of an inert platinum electrode in contact with 1 mol
dm-3 hydrogen ion and hydrogen gas at 100 kPa and 298 K. The standard electrode potential (E⁰) is the
potential (voltage) of the reduction half-equation under standard conditions measured relative to the
SHE. Solute concentration is 1 mol dm-3 or 100 kPa for gases. E⁰ of the SHE is 0 V.
• When aqueous solutions are electrolysed, water can be oxidized to oxygen at the anode and
reduced to hydrogen at the cathode.
• ∆G⁰ = -nFE⁰. When E⁰ is positive, ∆G⁰ is negative indicative of a spontaneous process. When E⁰ is
negative, ∆G⁰ is positive indicative of a non-spontaneous process. When E⁰ is 0, then ∆G⁰ is 0.
• Current, duration of electrolysis and charge on the ion affect the amount of product formed at the
electrodes during electrolysis.
• Electroplating involves the electrolytic coating of an object with a metallic thin layer.
• Calculation of cell potentials using standard electrode potentials.
• Prediction of whether a reaction is spontaneous or not using E⁰ values.
• Determination of standard free-energy changes (∆G⁰) using standard electrode potentials.
• Explanation of the products formed during the electrolysis of aqueous solutions.
• Perform lab experiments that could include single replacement reactions in aqueous solutions.
• Determination of the relative amounts of products formed during electrolytic processes.
• Explanation of the process of electroplating.
Higher level
19.1 Electrochemical cells
Higher level
6. CELL POTENTIALS
A voltaic cell produces a potential difference
known as the electromotive force (EMF). The
cell potential or electrode potential (E) is
measured by comparing it to a standard which is
the standard hydrogen electrode (SHE).
Higher level
Electromotive force (EMF)
STANDARD ELECTRODE POTENTIALS (Eº) are the potential values determined by
comparison with the STANDARD HYDROGEN ELECTRODE which is the potential
created by 1 mole of hydrogen ions at 100kPa H2 at 298K, which has the Eº value
0.00V.
298K (25°C)
Solution of 1M H+(aq)
e.g. 1M HCl or 0.5M H2SO4
Hydrogen gas at
100kPa pressure
Platinum
electrode
Glass tube with holes to
allow gas to escape
The standard hydrogen electrode is assigned an E° value of 0.00V.
Equilibrium reached:
2H+(aq) + 2e- ⇌ H2(g)
Higher level
The standard hydrogen electrode (SHE)
SALT BRIDGE
HYDROGEN
(100kPa)
ZINC
ZINC SULPHATE (1M)
PLATINUM
ELECTRODE
HYDROCHLORIC
ACID (1M)
In the diagram the standard hydrogen electrode is shown coupled up to a zinc half cell.
The voltmeter reading gives the standard electrode potential of the zinc cell.
conditions
salt bridge
temperature
298K
solution conc.
1 Molar (1 mol dm-3) with respect to ions
gases
100kPa
filled with saturated potassium chloride solution
it enables the circuit to be completed
Higher level
Measurement of E⁰ values
Higher level
The electrochemical series
E° / V
F2(g) + 2e¯
2F¯(aq)
+2.87
Mn2+(aq) + 4H2O(l)
+1.52
2Cl¯(aq)
+1.36
2Cr3+(aq) + 7H2O(l)
+1.33
2Br¯(aq)
+1.07
Ag+(aq) + e¯
Ag(s)
+0.80
Fe3+(aq) + e¯
Fe2+(aq)
+0.77
H2O2(aq)
+0.68
2I¯(aq)
+0.54
Cu+(aq) + e¯
Cu(s)
+0.52
Cu2+(aq) + 2e¯
Cu(s)
+0.34
Cu2+(aq) + e¯
2H+(aq) + 2e¯
Cu+(aq)
H2(g)
+0.15
0.00
Sn2+(aq) + 2e¯
Fe2+(aq) + 2e¯
Sn(s)
Fe(s)
-0.14
-0.44
Zn2+ (aq) + 2e¯
Zn(s)
-0.76
MnO4¯(aq) + 8H+(aq) + 5e¯
Cl2(g) + 2e¯
Cr2O72-(aq) + I4H+(aq) + 6e¯
Br2(l) + 2e¯
O2(g) + 2H+(aq) + 2e¯
I2(s) + 2e¯
Layout
REACTION MORE
LIKELY TO WORK
SPECIES ON LEFT
ARE MORE POWERFUL
OXIDATION AGENTS
If species are arranged in order of their standard electrode potentials you
get a series that shows how good each substance is at gaining electrons.
All equations are written as reduction processes ... i.e. gaining electrons
A species with a higher E° value oxidises (reverses) one with a lower value
Higher level
The electrochemical series
E° / V
F2(g) + 2e¯
2F¯(aq)
+2.87
Mn2+(aq) + 4H2O(l)
+1.52
2Cl¯(aq)
+1.36
2Cr3+(aq) + 7H2O(l)
+1.33
2Br¯(aq)
+1.07
Ag+(aq) + e¯
Ag(s)
+0.80
Fe3+(aq) + e¯
Fe2+(aq)
+0.77
O2(g) + 2H+(aq) + 2e¯
H2O2(aq)
+0.68
2I¯(aq)
+0.54
Cu+(aq) + e¯
Cu(s)
+0.52
Cu2+(aq) + 2e¯
Cu(s)
+0.34
Cu2+(aq) + e¯
2H+(aq) + 2e¯
Cu+(aq)
H2(g)
+0.15
0.00
Sn2+(aq) + 2e¯
Fe2+(aq) + 2e¯
Sn(s)
Fe(s)
-0.14
-0.44
Zn2+ (aq) + 2e¯
Zn(s)
-0.76
MnO4¯(aq) + 8H+(aq) + 5e¯
Cl2(g) + 2e¯
Cr2O72-(aq) + I4H+(aq) + 6e¯
Br2(l) + 2e¯
I2(s) + 2e¯
AN EQUATION
WITH A
HIGHER E°
VALUE WILL
REVERSE AN
EQUATION
WITH A LOWER
VALUE
Application Chlorine is a more powerful oxidising agent - it has a higher E°
Chlorine will get its electrons by reversing the iodine equation
Cl2(g) + 2e¯ ——>
Overall equation is
2Cl¯(aq)
Cl2(g) +
and
2I¯(aq)
2I¯(aq) ——> I2(s) + 2e¯
——> I2(s)
+
2Cl¯(aq)
Will be
reduced,
gain
electrons
Will be
oxidized,
lose
electrons
• E° value can be used to predict the feasibility of
redox and cell reactions.
• In theory ANY REDOX REACTION WITH A
POSITIVE E° VALUE WILL WORK.
• In practice, it proceeds if the E° value of the reaction
is greater than + 0.40V.
• An equation with a more positive E° value reverse a
less positive one.
Higher level
Predicting with E⁰ values
Write out the equations
Cu2+(aq) + 2e¯
Cu(s)
; E° = +0.34V
Sn2+(aq) + 2e¯
Sn(s)
; E° = -0.14V
the half reaction with the more positive E° value is more likely to work
it gets the electrons by reversing the half reaction with the lower E° value
Cu2+(aq) ——> Cu(s)
therefore
Sn(s)
the overall reaction is
Cu2+(aq)
+ Sn(s)
——>
and
Sn2+(aq)
——> Sn2+(aq) + Cu(s)
the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V
If the value is positive the reaction will be spontaneous
Higher level
Problem 1: What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are
connected?
7. GIBBS
Higher level
ΔG⁰ = -nFE⁰
n = number of moles
F = Faraday constant, charge per mole, 96 500 C/mol
For non-standard conditions (Option C) we use the
Nernst Equation (in data booklet):
E = E⁰ – (RT/nF)lnQ
Higher level
Electrode potential (E⁰) and free
energy change (ΔG)
Sn4+(aq) + 2e- ⇌ Sn2+(aq)
Fe3+(aq) + e- ⇌ Fe2+(aq)
E⁰ = +0.015V
E⁰ = +0.77V
E⁰ = E⁰cathode + E⁰anode = 0.77 - 0.015 = + 0.62V
ΔG⁰ = -nFE⁰ = -(2mol e- x 96 500 C/mol x 0.62V)
= -119660 V C (J = VC)
= -1.2 x 102 kJ
Higher level
Problem 1: Calculate Gibbs free energy for the
following voltaic cell at standard conditions:
Higher level
8. PRODUCTS OF ELECTROLYSIS
1. If the metal is high in the reactivity series you
will get hydrogen
2. If the metal is low in the reactivity series you
will get the metal
3. If the halide solution is concentrated you will
get the halogen (chlorine, bromine, iodine).
With other common negative ions you will
get oxygen.
OH- > Cl- > SO42-
Higher level
Inert electrolysis – General predictions
Cathode
Formation of the metal or formation of
hydrogen
Anode
Formation of the non-metal, formation of
oxygen or oxidation of the electrode
Higher level
Products of electrolysis
1. Determine which products go the cathode and which
go to the anode
2. For the cathode determine which cation has the most
positive electrode potential. This will be reduced (gain
e-1). The higher the positive potential value the more
naturally a reaction will occur (atom + e-1  ion).
3. For the anode reverse the reactions and electrode
potential values before determining which is the most
positive and likely to occur (ion atom + e-1).
4. Include all setting out!!! (as seen in examples below)
Higher level
Using standard electrode potentials to
determine products of electrolysis
Higher level
Electrolysis of CuSO4(aq)
Ions present: Cu2+, SO42-, H2O(l)
From the data booklet (water must be on it’s own on one side of the equation,
polyatomic ions don’t react).
Cathode:
Anode:
Final equation:
Most positive
-1.23
Most positive
Higher level
Electrolysis of NaCl(aq)
Ions present: Na+, Cl-, H2O(l)
From the data booklet (water must be on it’s own on one side of the equation).
Cathode:
Anode:
Most positive
-1.23 Most positive BUT…
-1.36
BUT for concentrations of >25% Cl2 is preferred.
Final equation:
Water can be electrolyzed with weak sulphuric acid, (or weak NaOH).
Ions present: SO42-, H2O(l), H+
From the data booklet (water must be on it’s own on one side of the equation,
polyatomic ions don’t react).
Cathode:
Anode:
Final equation:
Most positive
-1.23
Most positive
Higher level
Electrolysis of water
Higher level
Factors affecting relative amounts of
products in electrolysis
Higher level
Electroplating
1. The item to be coated is placed on the
negative electrode so that metal cations will
be reduced.
2. The anode may be a metal block, often the
solution is cyanide.
3. The anode dissolves and releases cations into
the solution.
4. Temperature, current, constitution and
concentration of solution all controlled.
Higher level
Electroplating
Jewellery – cheaper to coat with gold than make
out of pure gold
Tin cans – coating with tin protects the steal
cans from rusting
Car parts and tools – coating with nickel and
chromium decreases friction, prevents rust,
and looks shiny
Galvanizing – coating with zinc prevents rusting.
Used on car bodies, pipes, wire, sheeting.
Higher level
Electroplating applications
Q = It
Q = charge (measured in C – coulombs), the
Faraday constant which is 96 500 C/mol
I = current (measured in A – amps)
t = time (measured in s – seconds)
Higher level
Quantitative electrolysis
Q = It = 1.50A x 3.25hr x 60min/hr x 60s/min = 17550C
96500C : 1mol
17500C : Xmol
∴ X = 0.1813 moles of electrons
Cu2+(aq) + 2e-  Cu(s)
2mol e- : 1 mol Cu
0.1813 : X
∴ X = 0.0910 moles of Cu
m(Cu) = nM = 0.0910mol x 63.55g/mol = 5.78g Cu
Higher level
Problem 1: Calculate the mass, in g, of copper
produced when a current of 1.50 A is passed through a
solution of aqueous copper(II) sulphate for 3.25 hours.