Unit 6: Solubility and Redox Reactions
Section 5: Rate Laws - Cell Potential/Voltage
* Cell Potentials
* Reduction Potentials
* Voltage And WorkCell Potentials
In a galvanic cell, electrons are pulled away from the anode by the power of
the oxidizing agent. This force by which the electrons are pulled, is called
cell potental (Ecell) or electromotive force (emf.) The unit of emf is the
volt which is equal to one joule per coulomb.Reduction Potentials
In each redox reaction (galvanic cell), there is an oxidizer and reducer.
Each redox reaction can be broken up into half reactions. The half reactions
describe what is happening to each element. For instance:Cu+2 + Zn -----> Zn+2 + Cu
can be broken up into:
Zn -----> Zn +2 + 2 e-
Cu+2 + 2 e- ------> CuUsing a table of reduction potentials, voltage of the cell can be predicted.
What this means is that you can look up the voltage fro the Cu+2 + 2 e------> Cu
and the voltage for the Zn+2 + 2 e- -----> Zn. Then, you reverse
the reaction of Zn+2 + 2 e- -----> Zn to create the final reaction, so you
reverse the voltage for the reaction from the reduction potential table and
add the two together to get the voltage for the cell. For example:the voltage of Cu+2 + 2 e- -----> Cu is .34
and the voltage of Zn+2 + 2 e- -----> Zn is -.76
but because you reverse the reaction, you also must make the -.76 a .76.
Finally, add the two together. The final voltage of the cell is 1.1 volts.
Voltage and Work
Another important formula is:
Voltage = work (w) in joules / charge (q) in coulombs
but because we are looking at the flow of electrons in the system the work
(j) must be negative so, the formula that you would use is:Voltage = - w / q
Charge is in the unit of coulombs. A faraday is the charge of 1 mole of
electrons being transfered. The faraday is equal to 96,485 coulombs.
Therefore, mathematically:
q = 96485 f(Note: Teachers love to tell you the voltage and work and ask you for an
answer in faradays. Make sure you read the question and use both formulas if
necessary.)