Transcript Electrochemistry

Chapter 17
Electrochemistry
Electrochemistry
• The branch of chemistry that links chemical
reactions to the production or consumption of
electrical energy.
• In chemistry, electrical energy is stored in
electrons.
– In other words, electrochemistry is based upon
the principles oxidation-reduction reactions.
Redox Revisited
• Earlier this year, we examined Redox reactions
and how to write acid-base half reactions
• To review principles of redox reactions:
– A redox reaction is the sum of two half reactions, the
reduction and oxidation reactions.
– Reduction and oxidation reactions happen
simultaneously, so the number of electrons gained
and lost must match exactly
– Oxidation=loss of electrons
– Reduction=gain of electrons.
Redox revisited
• Lets look at a common reaction of Zinc metal
immersed in a copper sulfate solution.
The entire reaction is represented as a single
replacement reaction where the blue CuSO4
solution becomes clear as the Zinc replaces the
copper:
Zn + CuSO4 → Cu + ZnSO4
Redox Revisited
Zn + CuSO4 → Cu + ZnSO4
The Redox half reactions are then represented
as:
Zn(s) → Zn+2(aq) + 2eCu+2(aq) + 2e- → Cu(s)
Zn(s) + Cu+2(aq) + 2e- → Cu(s) + Zn+2(aq) + 2eSo:
Zn(s) + Cu+2(aq) → Cu(s) + Zn+2(aq)
Electrochemical (Galvanic) cells
• An apparatus that
converts chemical
energy into electrical
work, or vice versa
– Contains two
compartments
– Bridge that allows flow
of energy (electrons)
• “salt bridge”
• Usually a piece of tubing
filled with an electrolyte
Electrochemical (Galvanic) cells
• Anode-compartment in which the oxidation
half reaction takes place
• Cathode-compartment which the reduction
half reaction takes place.
– We represent the reactions that take place using
cell diagrams
• Cell diagrams are symbols that show how the
components of an electrochemical cell are connected.
Electrochemical (Galvanic) cells
Salt bridge replaced with porous disk to allow ion flow and
minimum mixing of solutions. Oxidizing agent “pulls”
electrons through wire from reducing agent. Let there be
light.
Homework
• Pg 879-880 # 13-19.
Cell Potential
• The “pull” on the electrons is called the cell
potential (E °cell), or “electromotive force”
(emf), is measured in volts.
– Volt: 1 joule of work per coulomb of charge
transferred. 1 J/C
– coulomb: defined as the charge transported by a
constant current of one ampere in one second:
Standard Reduction Potentials
• E °--standard reduction potentials in Volts.
–E
° =
cell
E
°
cathode-
E
°
anode
• Pay close attention to sign of E for certain reactions
• If the reaction is in reverse, change the sign for the
reduction potential.
• See Table 17.1 on page 843 in your book.
• All reduction potentials are given with all solutes at
1M and all gases at 1 atm pressure.
• E °cell is always positive
Standard Reduction Potentials
Zn + CuSO4 → Cu + ZnSO4
The Redox half reactions are then represented
as:
-E °
Zn(s) → Zn+2(aq) + 2eanode=.76
Cu+2(aq) + 2e- → Cu(s)
E °cathode=.34
Zn(s) + Cu+2(aq) → Cu(s) + Zn+2(aq) E °cell = 1.1V
• This cell produces 1.1 volts of electrical energy
(work).
Standard Reduction Potentials
• Another example, consider the galvanic cell
based on the reaction:
Al+3(aq) + Mg(s) → Al(s) + Mg2+(aq)
Give the redox half reactions, make sure to balance
the reactions (equal # of electrons), and calculate
E °cell (E ° for half reactions on table 17.1)
Standard Reduction Potentials
Al+3(aq) + Mg(s) → Al(s) + Mg2+(aq)
The Redox half reactions are then represented
as:
-E °
3(Mg(s) → Mg+2(aq) + 2e)anode =2.37V
2(Al+3(aq) + 3e- → Al(s))
E °cathode=-1.66V
2Al+3(aq) + 3Mg(s) → 2Al(s) + 3Mg+2(aq) E °cell = .71V
• This cell produces .71 volts of electrical energy
(work).
Standard Reduction Potentials
• Sometimes there are multiple possibilities for redox
potentials, in this case, pay close attention to what
the equation they give states the cell is based on.
• Example: a galvanic cell is based on the reaction:
– MnO4-(aq) + H+(aq) + ClO3-(aq) → ClO4-(aq) + Mn+2(aq) + H2O(l)
• There are multiple oxidation reactions for MnO4-(aq) , so you must
consult table 17.1 and match the reactants and products.
Representing Cells with Line Notations
• Consider key components of this galvanic cell:
Copper solid electrode
Zinc solid electrode
Zinc ions in solution
Spectator ions
Spectator ions
Salt bridge
Copper ions in solution
Steps to Representing Cells with Line
Notations
• Rule #1…list everything
• Separate the cathode/anode with the double line
notation (II) that represents the salt bridge
• Separate substances in different states of matter
in the same compartment with a single line (I),
separate substances in the same state in the
same compartment with a comma.
• Dispense spectator ions (usually water)
• When there is no solid electrode listed, assume
there is Platinum (Pt) in both.
Representing Cells with Line Notations
• Consider key components of this galvanic cell:
Copper solid electrode
Zinc solid electrode
Zinc ions in solution
Spectator ions
Spectator ions
Salt bridge
Copper ions in solution
homework
• Pg 880, #’s 25-35 odd
Determining Spontaneity in Galvanic
cells
• 1. Any reduction reaction is spontaneous
when paired with the reverse of any reaction
below it on Table 17.1.
• 2.If the Cell potential calculated is negative,
the reaction is not spontaneous (yes, I know I
told you that the cell potential is always
positive…in a cell that works).
Cell Potential, Electrical Work, And
Free Energy (∆G)
• Potential Difference (V)= work (J)/charge©
– When a cell produces a current (V), the cell
potential is positive, and the current can be used
to do work.
w  qE
– Since the work does not stay in the system (the
cell), the sign for work is negative. So:
w
E 
q
w  qE
q=charge in coulombs transferred from anode to
cathode
Cell Potential, Electrical Work, And
Free Energy (∆G)
w  qE
Work measured in Joules
Cell potential
difference in V or J/C
Coulombs based on
Faraday Constant
The Faraday Constant: the charge on one mole of electrons is
96,485 Coulombs of charge . When 1.33 moles is transferred:
q  nF = 1.33 mol e- x 96,485 C/mol eThis equation calculates amount of work done. HOWEVER, since some work and
energy is always lost to the surroundings as heat, there is a way to calculate Maximum
work
Cell Potential, Electrical Work, And
Free Energy (∆G)
• All galvanic cells have a maximum potential
that they never reach because of energy lost
as heat. To calculate maximum work, use the
maximum potential in your equation. And
then the Maximum work equals ∆G.
w max  G  qE max  nFE max
Dependence of a Cell on
Concentration
• Simply put…if the product concentration is
raised above 1.0M, E °cell will be less than
what is listed 17.1
• If the reactant concentration is above 1.0M,
E °cell will be greater than listed in table 17.1
Dependence of a Cell on
Concentration
• The dependence of cell potential on
concentration relies directly on the
dependence of free energy on concentration.
Remember that G  G  RT ln(Q)
• And since Gcell  qE max  nFEcell
• And G    nFE 
• Then:
RT
E  E 
nF
ln(Q)
“Nerst Equation”
Nerst Equation
At 25°C:
E=E°-
.0591
log(𝑄)
𝑛
n= moles of electrons
homework
• Pg 881, #’s 39-59 odd.