Intro to Electrochemistry
David Fabian
04/28/2014
Electrochemistry basics
Helpful pneumonic devices:
OIL RIG Oxidation is losing electrons, reduction is gaining electrons
AnOx RedCat
Anode oxidation, cathode reduction
Electrons flow from anode to cathode What are the half cell reactions?
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/electr
o.php#voltaic
What are the half cell reactions?
Zn2+ +2e‐
Zn Eo red = ‐ 0.76 vs. SHE
Cu2+ + 2e‐
Cu Eo red = 0.34 vs. SHE
Positive potential = favorable
The reduction of Cu2+ is favorable  this will be the reduction reaction
Overall Cell potential?
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/electr
o.php#voltaic
Overall Cell potential
Eocell = EoCu – EoZn
Eocell = 0.34 – (‐ 0.76)
Eocell = 1.10 V
Anode and Cathode
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/electr
o.php#voltaic
Line notation
Anode
Cathode
Single line = phase boundary
Double line = two phase boundaries (one at each end of the salt bridge)
0.10 M Cu2+
1.00 M Zn2+
http://chemed.chem.purdue.edu/genchem/topicreview/bp/
ch20/electro.php#voltaic
Mass of Zn plated on Cu electrode
The following reaction is observed at the cathode (copper): (1) ZnSO4(aq) + 2e‐  Zn(s) + SO42‐(aq)
A current probe is used to measure the current (I) passing through the system as a function of time (t in sec). The electrical charge (q) can then be calculated using the following equation (Faraday’s Law):
(2) q = I t
If the electrical charge is known, the mass of zinc plated on the copper electrode can be calculated from the following equation:
(3) m(Zn) = q M / n F
M = atomic mass of Zn: 65.39 g/mol
n = number of electrons in the half reaction
F = Faraday’s constant: 96 485 C/mol
,
Calculate Avogadro’s Number, NA
To calculate Avogadro’s number, NA, use the experimentally determined values of the mass of Zn(s) plated onto the copper strip and the total amount of electrical charge passed, qT,
(4)
NA  F / e  (qT * M ) /(n * m( Zn ) * qe)
where qe is the charge of one electron