Chapter 06 : Electrochemistry

 Chapter 06 : Electrochemistry 

• Electrochemistry is the study of redox reactions that occur in electrochemical cells.
Galvanic cells are where a redox reaction is used to generate an electric current
Electrolytic cells are where an electric current is used to drive a redox reaction.
A half-cell is a reactant placed in an electrolyte solution. When two half-cells are connected by a conducting wire and a salt bridge, an electrochemical cell is created.


• In electrochemical cells,
Oxidation occurs at the anode
Reduction occurs at the cathode.
The electrical potential can be calculated by identifying the oxidation and reduction half reactions, and using a table of standard reduction potentials.


• Example: A Zn-Cu battery:
The half reactions are:
Cu2+(aq) + 2e- → Cu(s)
Zn(s) → Zn2+(aq) + 2e-
At the anode, the Zn is oxidized, and at the cathode, the Cu is reduced.
The standard reduction potentials are:
Cu2+(aq) + 2e- → Cu(s) ; Eo = 0.339 V
Zn2+(aq) + 2e- → Zn(s) ; Eo = - 0.76 V
Since the Zn is being oxidized, the reduction potential for Zn must be reversed, so +0.76 V
The standard potential of the cell is 0.339 + 0.76 = 1.10 V


• The charge on an electron is 1.602x10-19 C (coulombs).
• The charge on one mole electrons is called a Faraday (F), which is: ?
F = (6.022045x1023 / mol) x (1.602x10-19 C) = 96485 C/mol
• One ampere is a measure of an amount of electrical charge going through an electric circuit in one second: 1 C/sec
• The free energy of an electrochemical cell is given by: ΔG = nFEo


• Sample Question: In an electroplating experiment, 10 amps are applied to a Cu2+ solution for 96.4 seconds. How many moles of Cu(s) would be deposited?
10 amps = 10 C/s
10 C/s x 96.5 s = 965 C
964 C / 96385 C/mol = 0.01 mol electrons
There are 2 electrons required to reduce Cu2+ to Cu(s)
Therefore, 0.01 mol/2 or 0.005 mol, or 5 mmol of Cu would be deposited.