Using standard electrode
potentials to calculate
electrochemical cell voltages
Chapter 19 Pages 768-773
• An electrochemical (voltaic) cell consists
of an oxidation reaction and a reduction
reaction to produce a voltage for the cell.
Ex: Batteries
• Need to know: which electrode is the
cathode, which is the anode, and whether
the chemical reaction is spontaneous.
• Anode = negative electrode = where
oxidation occurs
• Cathode = positive electrode = where
reduction occurs
• Electrons travel from the anode to the
cathode, along a conductive wire = voltage
= electricity.
Anode
Cathode
When choosing your anode and
cathode follow this rule:
• Look at the reduction potential for the ions,
the cathode will be the more positive
reduction potential (p.11)
• Reason: higher reduction potential means
more easily reduced, therefore reduction
takes place there, therefore the cathode
Table
of standard reduction potentials
• Shows the relative ability for a substance
to be reduced (gain electrons). It
expresses this potential to gain electrons
by assigning a voltage for each reduction
half reaction in the table.
• Page 11 of our data booklet
Eonet= Eooxidation + Eoreduction
• Eonet = total net voltage of the cell (electric
potential energy of the cell)
• Must be greater than zero for the reaction
to be spontaneous
• If less than zero, than non-spontaneous
reaction and no electricity produced.
Eonet= Eooxidation + Eoreduction
• Eooxidation = oxidation half reaction taking
place in the cell
• Oxidation half reactions are the reverse
reaction from the table on p.11, so their
sign changes! If the reduction potential
of Ag+ = 0.80V, then the oxidation
potential of Ag(s) = - 0.80V
Eonet= Eooxidation + Eoreduction
• Eoreduction = reduction half reaction taking
place in the cell
• The larger the reduction potential (the
more positive), the greater the tendency
for the reaction to occur.
• The smaller the reduction potential (the
most negative), the least likely for a
reaction to occur.
Example Problem 1:
•
Design an electrochemical cell that uses
the reactions of the metals Zn and Ag,
and solutions of their ions, Zn+2 and Ag+.
a) Identify the anode and cathode
b) Write the oxidation and reduction half
reactions
c) Calculate the cell voltage
Solution:
Zn+2 + 2e-  Zn(s)
-0.76V
Ag+1 +1e-  Ag(s)
+0.80V
• Ag+ more easily reduced, therefore Ag(s) is the
cathode, so oxidation takes place at the anode
(Zn) and the oxidation potential half reaction is:
Zn(s)  Zn+2 + 2e+0.76V
c) Eonet = Eooxidation + Eoreduction
= (+0.76V) + ( +0.80V)
= +1.56V
a), b)
• http://www.chem.iastate.edu/group/Greenb
owe/sections/projectfolder/flashfiles/electr
oChem/volticCell.html
• Look at this electrochemical cell
demonstration. Do the same as our
example with Zinc and Silver
Exercise:
1. An electrochemical cell was designed using the
metals Mg and Al and solutions containing
Mg+2 and Al+3.
a) Identify the anode and cathode
b) Write the oxidation and reduction half reactions
c) Calculate the cell voltage
2. An electrochemical cell was designed using the
metals Zn and Pb and solutions containing the
positive ions of these metals, Zn+2 and Pb+2.
a) Identify the anode and cathode
b) Write the oxidation and reduction half reactions
c) Calculate the cell voltage