1. ELECTROCHEMISTRY

2. ELECTROCHEMISTRY HAVING COMPLETED THIS TOPIC I CAN:
describe the types of half cells
discuss the idea of cell potential
describe the use and make up of standard hydrogen electrode
describe how electrode potentials are measured
use the electrochemical series
combine half cells
use cell diagrams
explain the uses of E° values

3. TYPES OF HALF CELL METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS
These are systems involving oxidation or reduction
METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS
Reaction Cu2+(aq) e¯ Cu(s)
Electrode copper
Solution Cu2+(aq) (1mol L-1) - 1mol L-1 copper sulfate solution
Potential V
Half cell Cu2+/Cu
Reaction Zn2+(aq) e¯ Zn(s)
Electrode zinc
Solution Zn2+(aq) (1mol L-1) - 1mol L-1 zinc sulfate solution
Potential V
Half cell Zn2+/Zn

4. TYPES OF HALF CELL GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS
These are systems involving oxidation or reduction
GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS
Reaction 2H+(aq) e¯ H2(g)
Electrode platinum
Solution H+(aq) (1mol L-1) mol L-1 HCl or 0.5mol L-1 H2SO4
Gas hydrogen ( 1 atm pressure )
Potential 0.00V
Half Cell 2H+(aq), H2(g) / Pt (s)
Reaction Cl2(g) e¯ 2Cl¯(aq)
Solution Cl¯(aq) (1mol L-1) mol L-1 sodium chloride
Gas chorine ( 1 atm pressure )
Potential V
Half Cell Cl2(g), 2Cl-(aq) / Pt (s)

5. TYPES OF HALF CELL SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES
These are systems involving oxidation or reduction
SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES
Reaction Fe3+(aq) e¯ Fe2+(aq)
Electrode platinum
Solution Fe3+(aq) (1mol L-1) and Fe2+(aq) (1mol L-1)
Potential V
Half Cell Fe3+(aq),Fe2+(aq) / Pt (s)
SOLUTIONS OF OXIDISING AGENTS IN ACID SOLUTION
Reaction MnO4¯ + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l)
Solution MnO4¯(aq)(1mol L-1) and Mn2+(aq) (1mol L-1) and H+(aq)
Potential V
Half Cell MnO4-(aq),Mn2+(aq) / Pt (s)

6. CELL POTENTIAL Each electrode / electrolyte combination has its own
half-reaction which sets up a potential difference
The value is
affected by ... TEMPERATURE
PRESSURE OF ANY GASES
SOLUTION CONCENTRATION
Measurement it is impossible to measure the potential of a single electrode…
BUT you can measure the potential difference between two electrodes
it is measured relative to a reference cell under standard conditions
The reference used is the STANDARD HYDROGEN ELECTRODE.

8. THE STANDARD HYDROGEN ELECTRODE SOLUTION OF 1 mol L-1 H+(aq)
The reference used is the STANDARD HYDROGEN ELECTRODE.
HYDROGEN GAS AT 1 ATMOSPHERE PRESSURE
298K (25°C)
SOLUTION OF 1 mol L-1 H+(aq)
e.g. 1mol L-1 HCl or
0.5 mol L-1 H2SO4
PLATINUM ELECTRODE
The standard hydrogen electrode is assigned an E° value of 0.00V.

9. MEASUREMENT OF E° VALUES
SALT BRIDGE
HYDROGEN (1 ATM)
PLATINUM ELECTRODE
HYDROCHLORIC
ACID (1mol L-1)
ZINC
ZINC SULFATE (1mol L-1)
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 temperature 298K
solution conc. 1 mol L-1 with respect to ions
gases 1 atmosphere pressure
salt bridge filled with saturated potassium chloride solution
it enables the circuit to be completed by allowing the passage of ions from one half cell to the other. This avoids a build-up of charge.

11. THE ELECTROCHEMICAL SERIES
E° / V
F2(g) + 2e¯ 2F¯(aq)
MnO4¯(aq) + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l)
Cl2(g) + 2e¯ 2Cl¯(aq)
Cr2O72-(aq) + I4H+(aq) + 6e¯ 2Cr3+(aq) + 7H2O(l)
Br2(l) + 2e¯ 2Br¯(aq)
Ag+(aq) + e¯ Ag(s)
Fe3+(aq) + e¯ Fe2+(aq) O2(g) + 2H+(aq) + 2e¯ H2O2(aq)
I2(s) + 2e¯ 2I¯(aq)
Cu+(aq) + e¯ Cu(s)
Cu2+(aq) + 2e¯ Cu(s)
Cu2+(aq) + e¯ Cu+(aq)
2H+(aq) + 2e¯ H2(g)
Sn2+(aq) + 2e¯ Sn(s)
Fe2+(aq) + 2e¯ Fe(s)
Zn2+ (aq) + 2e¯ Zn(s)
Layout 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 oxidise (reverses) one with a lower value
REACTION MORE
LIKELY TO WORK
SPECIES ON LEFT
ARE MORE POWERFUL
OXIDISING AGENTS

12. THE ELECTROCHEMICAL SERIES
E° / V
F2(g) + 2e¯ 2F¯(aq)
MnO4¯(aq) + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l)
Cl2(g) + 2e¯ 2Cl¯(aq)
Cr2O72-(aq) + I4H+(aq) + 6e¯ 2Cr3+(aq) + 7H2O(l)
Br2(l) + 2e¯ 2Br¯(aq)
Ag+(aq) + e¯ Ag(s)
Fe3+(aq) + e¯ Fe2+(aq)
O2(g) + 2H+(aq) + 2e¯ H2O2(aq)
I2(s) + 2e¯ 2I¯(aq)
Cu+(aq) + e¯ Cu(s)
Cu2+(aq) + 2e¯ Cu(s)
Cu2+(aq) + e¯ Cu+(aq)
2H+(aq) + 2e¯ H2(g)
Sn2+(aq) + 2e¯ Sn(s)
Fe2+(aq) + 2e¯ Fe(s)
Zn2+ (aq) + 2e¯ Zn(s)
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¯ ——> 2Cl¯(aq) and I¯(aq) ——> I2(s) + 2e¯
Overall equation is Cl2(g) I¯(aq) ——> I2(s) Cl¯(aq)
AN EQUATION WITH A HIGHER E° VALUE WILL REVERSE AN EQUATION WITH A LOWER VALUE

13. ELECTROCHEMICAL CELLS
CELLS • electrochemical cells contain two electrodes
• each electrode / electrolyte combination has its own half-reaction
• the electrons produced by one half reaction are available for the other
• oxidation occurs at the anode
• reduction occurs at the cathode.
ANODE Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION
CATHODE Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION
The resulting cell has a potential difference (voltage) called the cell
potential which depends on the difference between the two potentials
It is affected by ... • temperature
• pressure of any gases
• solution concentrations

15. A TYPICAL COMBINATION OF HALF CELLS
1.10V
ZINC
ZINC SULFATE
(1 mol L-1)
COPPER
COPPER SULFATE (1mol L-1)
E° = V
E° = V
zinc is more reactive - it dissolves to give ions Zn(s) ——> Zn2+(aq) + 2e¯
the electrons produced go round the external circuit to the copper electrode
electrons are picked up by copper ions Cu2+(aq) + 2e¯ ——> Cu(s)
As a result, copper is deposited
overall reaction Zn(s) + Cu2+(aq) ——> Zn2+(aq) + Cu(s)

16. Zn Zn2+ Cu2+ Cu CELL DIAGRAMS
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OF
WITH A SOLUTION ARE JOINED VIA A COPPER IONS IN
OF ZINC IONS SALT BRIDGE CONTACT WITH COPPER
Zn Zn Cu2+ Cu

17. + _ Zn Zn2+ Cu2+ Cu CELL DIAGRAMS
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
• Draw as shown… the cell reaction goes from left to right
• the zinc metal dissolves Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION
• copper is deposited Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION
• oxidation takes place at the anode
• reduction at the cathode _
Zn Zn Cu2+ Cu +

18. cell voltage = +0.34V - (-0.76V) = +1.10V
CELL DIAGRAMS
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
• Draw as shown… the electrons go round the external circuit from left to right
• electrons are released when zinc turns into zinc ions
• the electrons produced go round the external circuit to the copper
• electrons are picked up by copper ions and copper is deposited
cell voltage = V - (-0.76V) = V _
Zn Zn Cu2+ Cu + V

19. USE OF Eo VALUES - WILL IT WORK?
E° values 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 V
An equation with a more positive E° value reverse a less positive one

20. USE OF Eo VALUES - WILL IT WORK?
An equation with a more positive E° value reverse a less positive one
What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected?
Write out the equations Cu2+(aq) + 2e¯ Cu(s) ; E° = V
Sn2+(aq) + 2e¯ Sn(s) ; E° = V
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
therefore Cu2+(aq) ——> Cu(s) and
Sn(s) ——> Sn2+(aq)
the overall reaction is Cu2+(aq) Sn(s) ——> Sn2+(aq) + Cu(s)
the cell voltage is the difference in E° values... (+0.34) - (-0.14) = V

21. USE OF Eo VALUES - WILL IT WORK?
An equation with a more positive E° value reverse a less positive one
Will this reaction be spontaneous? Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s)
Write out the appropriate equations Cu2+(aq) + 2e¯ Cu(s) ; E° = V
as reductions with their E° values Sn2+(aq) + 2e¯ Sn(s) ; E° = V
The reaction which takes place will involve the more positive one reversing the other
i.e. Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq)
The cell voltage will be the difference
in E° values and will be positive (+0.34) - (- 0.14) = V
If this is the equation you want then it will be spontaneous
If it is the opposite equation (going the other way) it will not be spontaneous

22. USE OF Eo VALUES - WILL IT WORK?
An equation with a more positive E° value reverse a less positive one
Will this reaction be spontaneous? Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s)
Split equation into two half equations Cu2+(aq) e¯ ——> Cu(s)
Sn(s) ——> Sn2+(aq) e¯
Find the electrode potentials Cu2+(aq) + 2e¯ Cu(s) ; E° = V
and the usual equations Sn2+(aq) + 2e¯ Sn(s) ; E° = V
Reverse one equation and its sign Sn(s) ——> Sn2+(aq) e¯ ; E° = V
Combine the two half equations Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s)
Add the two numerical values (+0.34V) + (+ 0.14V) = V
If the value is positive the reaction will be spontaneous

23. THE END