Transcript Unit 15 Electrochemistry

Chapter 20 Problem Set: p. 890-898
3, 11, 13, 17, 23, 27, 31, 35, 43, 45, 47,
53, 61, 63, 69, 72, 75, 79, 85, 87, 95
The energy released in a spontaneous
redox reaction that is used to perform
electrical work is harnessed in Voltaic Cells
Also called Galvanic Cells
Electrons transfer through an external
pathway rather than directly between
reactants
Electrodes: metals used in the circuit
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
Anode: electrode at which oxidation occurs
Cathode: electrode at which reduction occurs
Salt Bridge: a connection between two half-cells
that allows for the flow of ions
Composed of a strong electrolyte
Must be a solution of a compound that will
not produce a precipitate with any other ion
in solution
Salt anions migrate toward the anode;
cations migrate toward the cathode
**Electrons flow from the anode to the cathode**
Keep your vowels together and your consonants
together
Anode = oxidation
Cathode = reduction
An ox and Red cats
Anode is oxidation, Cathode is reduction
Electrons flow from A to C (in alphabetical
order)
Used to determine which metal will be the
cathode/anode for the reaction to be spontaneous
Cell EMF: Electromotive Force
Caused by a difference in potential energy
between the different electrodes.
Allows electrons to be pushed
Denoted Ecell , measured in volts, V
Also called cell potential
Cell potential is
positive for spontaneous reactions
negative for nonspontaneous reactions
Depends on
1. Reactions Occurring (Metals Used)
2. Concentrations of Solutions (Molarity)
3. Temperature (normally 25oC)
Eocell cell potential at standard conditions
The more positive the Eored value for a half
reaction, the greater the tendency for the
reactant of the half reaction to be reduced
and, therefore, to oxidize another species
The half reaction with the smallest reduction
potential is most easily reversed as an
oxidation
The Eored table acts as an activity series for
which substances act as oxidizers and
reducers
Eocell based on standard reduction potentials
Potential associated with each electrode is the ability for
reduction to occur at the electrode.
Eored = standard reduction potential for metal (based
on Standard Hydrogen Electrode, SHE – use
Appendix E p. 1128!)
Eocell = Eored (cathode) – Eored (anode)
If the stoichiometry of the reaction changes DO NOT
multiply the value by the the cell potential for that metal
The more positive the Eored for a metal, the greater the
ability for reduction
Thus the metal with a larger Eored will be the better
cathode in a voltaic cell
Determine voltage produced in spontaneous
reaction.
Determine voltage produced in spontaneous reaction.
Eored (Cd2+ /Cd ) = -0.403 V
Eored (Sn2+ /Sn ) = -0.136 V
Cathode: Sn2+ + 2e- Sn
Anode: Cd Cd2+ + 2eEocell = Eocathode – Eoanode
= -0.136 V – (-0.403 V)
= 0.267 V
Na Na+ + 1ePbO2 Pb2+
+ 2.71 V
+ 1.50 V
(Na Na+ + 1e-)x2 *DO NOT MULTIPLY Eored by 2*
4H+ + 2e- + PbO2 Pb2+ + 2H2O
4H+ + PbO2 + 2Na 2Na+ + Pb2+ + 2H2O
Overall potential for the battery = 2.71 + 1.50 = 4.3V
Reduction Potentials can be used to calculate
energies of reactions when using redox processes
Remember:
E=+
Spontaneous
E=Nonspontaneous
E can be converted into G for Gibbs Free Energy or
other thermochemical quantities
∆Go = -nFEo
n= number of electrons transferred
F = Faraday’s constant = 96500 C/mol or J/Kmol
Eo = Calculated from Eored values
Start with Half-Reactions
Fe Fe2+ + 2eCu Cu2+ + 2eFlip the copper reaction so it matches the overall equation
Fe Fe2+ + 2e0.44 V
Cu2+ + 2e- Cu
0.34 V
Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq)
0.78V
Solve for ∆Go = -nFEo
∆Go = -(2)(96500)(0.78)
= -150517 J/mol = -151 kJ/mol
What is the Keq for this reaction? ∆Go = -RTlnKeq
As voltage is produced, concentrations vary.
Reactants are consumed and products are
produced.
When E = 0, the cell is dead, the concentrations
cease to change due to equilibrium being reached.
For different concentrations, the Nernst equation
is used:
To calculate voltage, use the Nernst Equation
Will be
zero
Cell EMF and Equilibrium
∆G = 0, ∆E = 0 at Equilibrium
Using Nernst
Solving for the
equilibrium
constant
Start with the half reactions
2(2H+ + 1e- + VO21+ VO2+ + H2O )
Zn Zn2+ +2e4H+ + 2VO21+ + Zn Zn2+ + 2VO2+ + 2H2O
Write the equilibrium expression
1.00 V
0.76 V
1.76 V
4 x 10-5
=1.76 – (-0.13)
E = 1.89 V
The voltage went up!
Use LeChatelier’s Principle to explain why
What will the voltmeter read for
the following electrochemical
cell?
Anode Reaction – Oxidation dilute solution becomes more
concentrated.
Ag(s) → Ag+ (0.1 M) + e- +0.8 V
Cathode Reaction – Reductionconcentrated solution becomes
more dilute.
Ag+ (1.0 M) + e- → Ag (s) -0.8 V
Net Cell Reaction
Ag+ (1.0 M) → Ag+ (0.1 M) 0.0V
V
You DON’T need your calculator for this one!!!
E = 0.0592 V
Self-contained electrochemical power source
consisting of 1 or more galvanic cells.
Voltages are additive when batteries are
connected due to a continuation of the flow of
electrons
A voltaic cell that converts chemical energy into
electrical energy
Not self-contained
Produces current by combustion and electronexcitation
A spontaneous redox reaction in which a metal is
attacked by some substance in the environment and
converted to an unwanted compound
Is often prevented, or reversed, by use of a sacrificial
anode (a more active metal)
The use of electricity to cause nonspontaneous
reactions to occur by driving the reaction in the
opposite direction (basis of rechargeable batteries
Take place in electrolytic cells
Cathode is connected to – terminal to accept e-,
anode is attached to + terminal to donate e(opposite of voltage)
Ampere = units to measure current in
Coulombs/second
Measures the rate of electron flow
Faraday’s Constant, F = 96485 Coulomb/mol
Ampere = units to measure current in
Coulombs/second
Measures the rate of electron flow
Faraday’s Constant, F = 96485 Coulomb/mol
Use a t-table!
Time = 158823 sec = 2647 min = 44 hours = 1.84 days
3 e + Cr3+ Cr
Use a t-table!
Mass of Cr = 2.36 x 102 g
Thin coating is plated onto an existing metal to
help protect it
Metal to be plated is attached to the cathode.
Metal coating is produced by the metal
attached to anode and a “like” solution