1. Electrochemistry
An electrochemical cell produces electricity using a chemical reaction.
It consists of two half-cells connected via an external wire with a salt bridge connecting the solutions
An electrolytic cell uses an external electricity source to produce a chemical reaction. This is usually called electrolysis.

2. Historically Historically oxidation involved reaction with O2.
i.e., Rusting
4 Fe(s) + 3O2 (g) Fe2O3 (s)
Another example
Zn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s)
In this reaction:
Zn(s) g Zn2+(aq) Oxidation
Cu2+(aq) g Cu(s) Reduction
In a redox reaction, one process can’t occur without the other. Oxidation and Reduction reactions must simultaneously occur.

3. Redox Between Zn and Cu
If Zn(s) and Cu2+(aq) is in the same solution, then the electron is a transferred directly between the Zn and Cu.
No useful work is obtained. However if the reactants are separated and the electrons shuttle through an external path...

4. Electrochemical Cells
Voltaic / Galvanic Cell Apparatus which produce electricity
Electrolytic Cell Apparatus which consumes electricity
Consider: Zn Cu
Initially there is a flow of e-
After a very short time the process stops
Electron transport stops because of charge build up on both sides
Build up of positive charge
Build up of negative charge
The charge separation will lead to process where it cost too much energy to transfer electron.

5. Completing the Circuit
Electron transfer can occur if the circuit is closed
Parts:
Two conductors
Electrolyte solution
Salt Bridge / Porous membrane
Process that must happen if e- is to flow.
A. e- transport through external circuit
B. In the cell, ions must migrate
C. Circuit must be closed using a salt bridge (no charge build up)
Anode (-)
Green
Negative electrode generates electron
Oxidation Occurs
Cathode (+)
Red
Positive electrode accepts electron
Reduction Occurs A C B
Anode/Anion (-)
Cathode/Cation(+)

6. Voltaic Cell Electron transfer can occur if the circuit is closed
Parts:
Two conductors
Electrolyte solution
Salt Bridge / Porous membrane
3 process must happen if e- is to flow.
A. e- transport through external circuit
B. In the cell, ions a must migrate
C. Circuit must be closed (no charge build up)
Anode (-)
Black
Negative electrode generates electron
Oxidation Occur
Cathode (+)
Red
Positive electrode accepts electron
Reduction Occur
Anode/Anion (-)
Cathode/Cation(+)

7. Completing the Circuit: Salt Bridge
In order for electrons to move through an external wire, charge must not build up at any cell. This is done by the salt bridge in which ions migrate to different compartments neutralize any charge build up.

8. Sign Convention of Voltaic Cell
@ Anode: Negative Terminal (anions).
Source of electron then repels electrons. Oxidation occurs.
Zn(s) g Zn+2(aq) + 2e- : Electron source
@ Cathode: Positive Terminal (cation)
Attracts electron and then consumes electron. Reduction occurs.
Electron target: 2e- + Cu+2(aq) g Cu(s)
Overall:
Zn(s) + Cu+2(aq) g Zn+2(aq) + Cu(s) E° = 1.10 V
Note when the reaction is reverse: Cu(s) + Zn+2(aq) g Cu+2(aq) + Zn(s)
Sign of E ° is also reversed E° = V
Oxidation: Zn(s) g Zn+2(aq) E° = 0.76 V
Reduction: Cu+2(aq) g Cu(s) E° = 0.34 V
1.10 V = E°CELL
or E°CELL = E°red (Red-cathode) - E°red (Oxid-anode)

9. Another Voltaic Cell Zn(s) + 2H+ (aq) g Zn+2(aq) + H2 (g) E° = 0.76 V
@ Anode: Negative Terminal (anions): Zn(s) g Zn+2(aq) + 2e- :
Source of electron then repels electrons. Oxidation occurs.
@ Cathode: Positive Terminal (cation): 2e H+(aq) g H2 (g)
Attracts electron and then consumes electron. Reduction occurs.
Overall: Zn(s) + 2H+ (aq) g Zn2+ (aq) + H2 (g)

10. Another Voltaic Cell Zn(s) + 2H+ (aq) g Zn+2(aq) + H2 (g) E° = 0.76 V
@ Anode: Negative Terminal (anions): Zn(s) g Zn+2(aq) + 2e- :
Source of electron then repels electrons. Oxidation occurs.
@ Cathode: Positive Terminal (cation): 2e H+(aq) g H2 (g)
Attracts electron and then consumes electron. Reduction occurs.
Overall: Zn(s) + 2H+ (aq) g Zn2+ (aq) + H2 (g)

11. Line Notation Convention
basically: metal | ionic solution || ionic solution | metal
This is a convenient way of representing cells
1. Anode g Cathode
[oxidation (-) ] [reduction (+)]
2. | represents a phase boundary
3. || represents the salt bridge
4. Concentration of component
Zn(s) ZnSO4 (aq,1.0M) CuSO4 (aq,1.0M) Cu(s) 1 4 3 2

12. Line Notation Examples
Consider : Zn(s) + Cu+2(aq) g Zn+2(aq) + Cu(s
Anode: Zn g Zn e-
Cathode: Cu e- g Cu
Shorthand “Line” notation
Zn (s) Zn+2 (aq)(1.0M) Cu+2(aq) (1.0M) Cu(s)
2nd Example : Zn(s) + 2H+ (aq) g Zn+2(aq) + H2(g)
Cathode: 2H e- g H2 (g)
Zn (s) Zn+2 (aq)(1.0M) H+(aq) (1.0M), H2(g, 1atm) Pt(s)

13. EMF - ElectroMotive Force
Potential energy of electron is higher at the anode. This is the driving force for the reaction (e- transfer) e
Anode (-)
D P.E. = V = J
e - C
e- flow toward cathode
(+) Cathode
Larger the gap, the greater the potential (Voltage)

14. Charge (C) = Current (A) x Time (s)
ElectroMotive Force
For electrochemical cells
Emf – electromotive force
Potential energy difference between the two electrodes
Units of emf: V - Volts :
1V - 1 Joule / Coulomb
1 Joule of work per coulomb of charge transferred.
For electrolysis
When a current of one amp flow for 1 second this equals one coulomb of charge passed.
Charge (C) = Current (A) x Time (s)
96,400 coulombs = 1 Faraday of charge
1 Faraday is equal to 1 mole of electrons

15. Standard Electrode Potentials
Cell Potential is written as a reduction equation.
M+ + e- g M E° = std red. potential
Most spontaneous <Reduction occurs> Oxidizing Agent
Written as reduction
Most non-spontaneous Spontaneous in the reverse direction. <Oxidation occurs> Reducing Agent

16. Zoom of Std. Electrode Potentials
Cell Potential is written as a reduction equation.
M+ + e- g M E° =
F2 (g) e- g 2 F - (aq) V
Ce e- g Ce3+ (aq) V
2H e- g H2 (g) V
Li+(aq) + e- g Li(s) V
Most spontaneous Reduction Oxidizing Agent
Written as reduction
Most non-spontaneous Spontaneous in the reverse direction. Oxidation Reducing Agent
All reaction written as reduction reaction. But in electrochemistry, there can’t be just a reduction reaction. It must be coupled with an oxidation reaction.

17. Eº(cell) = Eº(red) – Eº(ox)
E°Cell Evaluation
E°Cell Function of the reaction
g Oxidation Process (Anode reaction)
g Reduction Process (Cathode reaction)
E°Cell = E°red (cathode) - E°red (anode)
Simple to remember:
Eº(cell) = Eº(red) – Eº(ox)
E = E redox
The potential difference of the cell is equal to the electrode potential of the reduced component minus the electrode potential of the oxidised component.

18. Standard Electrode Potential
How is E°red (Cathode) and E°red (Anode) determine.
E° (EMF) - State Function; there is no absolute scale
Absolute E° value can’t be measured experimentally
The method of establishing a scale is to measure the difference in potential between two half-cells.
Consider:
Zn g Zn e E°=?
Can’t determine because the reaction must be coupled
Question? - How can a scale of reduction potential be determined ?
Answer - Use a half reaction as reference and assign it a potential of zero.
Measure all other half cells in comparison to this reaction.

19. Side-Bar: Relative Scale
Consider a baby whose weight is to be determine but will not remain still on top of a scale. How can the parents determine the babies weight?
Carry the child in arms and weight both child and parent then subtract the weight of the parent from the total to yield the baby weight.

20. Standard Reference Half Cell
Selected half reaction is:
H+ / H2 (g) couple half reaction: 2H+ (aq, 1.0M) + 2e- g H2 (g,1atm)
by definition c E° = 0.0 V, the reverse is also 0.0 V
H+/H2 couple - Standard Hydrogen Electrode (SHE)
To determine E° for a another half reaction, the reaction of interest needs to be connected to this SHE. The potential measured is then assigned to the half-reaction under investigation.
Example:
E°Cell = V = E°red (Cat) - E°red (Anode)
0.0 V (?)
E°red (Anode) = V
\Zn+2/Zn E° = V reduction reaction

21. Determining Other Half-Cell Potentials
Now consider the reaction:
Zn(s)|Zn+2 (1.0 M)||Cu+2(1.0 M)|Cu(s)
E°Cell = V
E°Cell = E°red (Cat) - E°red (Anode)
recall, E° Zn+2/Zn = V
Therefore,
E°Cell = E°Cu+2/Cu - E° Zn+2/Zn
1.10 V = (?) ( V)
E°Cu+2/Cu = V

22. Example: Half-Cell Potential
For the reaction: Tl Cr2+  Tl Cr3+ E°Cell = 1.19 V
i) Write both half reaction and balance
ii) Calculate the E°Cell Tl+3  Tl+
iii) Sketch the voltaic cell and line notation
i) Tl e-  Tl+
(Cr2+  2Cr3+ + 2e- ) x E° = 0.41 V
ii) E°Cell = 1.19 V = E°red (Cat) - E°red (Anode)
1.19 V= E°red (Cat) V
for Tl e-  Tl+ :
1.19 V = E°red (Cat) = V Pt
Cr2+ g Cr3+
Tl3+ g Tl+
1.19 V

23. Voltaic Vs. Electrolytic Cells
Voltaic Cell
Energy is released from spontaneous redox reaction
Electrolytic Cell
Energy is absorbed to drive nonspontaneous redox reaction
General characteristics of voltaic and electrolytic cells. A voltaic cell generates energy from a spontaneous reaction (DG<0), whereas an electrolytic cell requires energy to drive a nonspontaneous reaction (DG>0). In both types of cell, two external circuits provides the means or electrons to flow. Oxidation takes place all the anode, and reduction takes place at the cathode, but the relative electrode changes are opposite in the two cells.
System does work on load (surroundings)
Surrounding (power supply) do work on system (cell)
Anode (Oxidation)
Oxidation Reaction
X g X+ + e-
Oxidation Reaction
A- g A + e-
Reduction Reaction
e- + Y+ g Y
Reduction Reaction
e- + B+ g B
Overall (Cell) Reaction
A- + B+ g A + B, DG> 0
Overall (Cell) Reaction
X + Y+ g X+ + Y, DG = 0