Chapter
21:Electrochemistry:
Chemical Change and Electrical Work
Homework:
Cells
Voltaic cells (galvanic
cells) are cells that spontaneously produce electricity through a chemical
reaction and are therefore batteries.
Electrolytic cells are not
spontaneous and must be driven by electricity.
Construction of Voltaic Cells
To make a voltaic cell, one
must define the chemical system to be exploited. We will consider the Daniel’s cell first, which uses a copper-zinc
chemical system. On page 813 in table
19.1 there is a list of several half reactions. Using the concept “The more
active element WILL replace the less active ion.”, we can write the reaction
that corresponds to the copper-zinc system.
The more active elements are at the top and the less active elements are
at the bottom. The two reactions as
they appear in the table are:
Zn2+(aq)
+ 2e- ---> Zn°(s) -0.76
Cu2+(aq)
+ 2e- ---> Cu°(s)
0.34
The more active element is
Zn° and it WILL replace the less active ion Cu2+. With this information, you see we must
reverse the first reaction so that when we add them the more active element can
interact with the less active ion giving:
Zn°(s) --->
Zn2+(aq) + 2e- 0.76 V
Oxidation half reaction (LEO)
Cu2+(aq)
+ 2e- ---> Cu°(s) 0.34
V Reduction half
reaction (GER)
------------------------------------------
Zn°(s) + Cu2+(aq)
---> Zn2+(aq)
+ Cu°(s) 1.10 V
Note that when we reversed
the first reaction we also changed the sign of the standard potential and that
the overall voltage is the sum of the half-reaction voltages. This will become
important later. Also, note that there
are no electrons in the balanced oxidation-reduction reaction.
To allow this reaction to do
work for us we separate the half reactions into two different beakers:
The zinc electrode and zinc
ion solution in one and the copper electrode and copper ion solution in the
other. To maintain electric neutrality,
a salt bridge must be introduced to allow ions to flow. When a circuit is
attached to the electrodes of the Daniel’s cell, electrons flow from the zinc
electrode to the copper electrode through the external circuit.
Notation for Voltaic Cells
A shorthand method for
representing the cell shows the oxidation species and reduction species as
follows:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Oxidation Reduction
(Anode) (Cathode)
The single lines represent a
phase boundary and the double line represents a salt bridge.
Electromotive Force (Potential Difference in Volts)
The driving force for a
voltaic cell is the electromotive force or voltage. The volt is defined as the amount of work per charge:
V = w/Q
Where V is volts, w is work
in joules and Q is the charge in coulombs.
The charge on one electron is 1.6x10-19C. The charge on a mole of electrons is
9.11x10-19C/e- x 6.022x1023 e-/mole e- = 96,500 C/mole e-.
96,500 C is called 1 farad
of charge.
Problem:
How much energy is released
when a Daniel’s cell having a potential difference of 1.10 V deposits 1.00 g of
copper on the copper electrode. Note: 1 volt = 1 Joule/1 coulomb
1 mol Cu 2 mol e- *6.022x1023 e- *9.11x10-19C
1.00 g Cu ´ ----------- ´ ---------- ´ ----------------- ´ --------------- ´1.10 V =
63.55 g Cu 1 mol Cu 1
mol e- 1 e-
1 mol Cu 2 Farads *96,500
C
1.00 g Cu ´ ------------ ´ ----------- ´ ----------- ´ 1.10 V =
63.55 g Cu 1 mol Cu 1
Farad
The maximum work that a
voltaic cell can do is given by:
DG = wmax = - nFEcell
where n is the number of
electrons transferred in the chemical equation, F is the faraday constant
96,500 C and Ecell is the voltage of the cell as calculated from
table 19.1 on page 813. Note for a
spontaneous cell, the Ecell must be positive in order for DG to be negative.
E°cell
= E°oxidation + E°reduction Note: The sign is different in the book.
Why?
Problem:
Calculate the E°cell
for the cell Li° | Li+|| Hg2+| Hg°.
Problem:
For the Li° | Li+|| Hg2+| Hg° system, what is a more powerful oxidizing
agent. What is the more powerful reducing agent?
Equilibrium Constants from EMF’s
Recall that the maximum work
for a system is called Gibbs Free Energy or simply free energy, DG°. Since
wmax = - nFE°cell ,
DG° = -nFE°cell = RT ln(Kc)
Problem:
Calculate the free energy
for the Daniel’s cell.
Problem:
Calculate the equilibrium
constant for the Daniel’s cell.
Dependence of emf on Concentration (The Nernst Equation)
The Nernst equation
describes how the Ecell varies with concentration. We start with the
equation that describes how the free energy varies with concentration:
DG = DG° + RT ln Q
Recall that Q is the
reaction quotient Q =[products]/[reactants] when not at equilibrium. Substituting for DG and DG°, -nFE and -nFE° we get:
-nFE = -nFE° + RT ln Q
dividing by -nF
E = E° - RT ln Q/(nF)
RT/F is a constant if we
assume 8.314 J/Kmol for R, 298 K for T, and 96,500 for F, and multiply by 2.303
to convert to base 10 logs the equation becomes:
E = E° - (.0592/n) log Q
which is the Nernst
equation.
Problem:
What is the voltage for a
standard Daniel’s cell (initially all concentrations 1M) after the cell has
used 99 % of its reactants.
Problem:
Calculate the equilibrium
concentrations for the standard Daniel’s cell, which has been allowed to die.
Electrolytic Cells (The Driven Cell)
Electrolytic cells are
driven by a battery or other source of emf. Usually two electrodes are placed
in a conductive medium and current is forced to pass through the medium. The medium may be aqueous or a molten
salt.
Electrolysis of Molten salts
Demo:
KClO3
electrolysis.
Aqueous Electrolysis
Demo:
Electrolysis of water
Demo:
Electroplating a quarter
with copper.
Problem:
A spoon and a gold electrode
are placed in an AuCl3 solution.
5.00 A of current is allowed to flow through the solution for 3
hours. Note: The ampere (A) is a unit
of electric current.
1 Ampere = 1
Coulomb/second or 1 A = 1 C/s
a) How many grams of gold are electroplated on the spoon?
b) If the spoon has an area of 30 cm2 how thick is the gold plating?