Transcript Electrochemistry

Electrochemistry
Why Do Chemicals Trade Electrons?
Galvanic (Voltaic) Cells
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Remember oxidation reduction reactions
Good, you can do half reactions no problem
Because you practiced them, over and over
Until they were easy
Lucky, because otherwise you would have
to learn them now in addition to all the rest
of the stuff in this chapter
Electrochem
• Oxidation involves loss of electrons (OIL)
– Increase of oxidation number
• Reduction involved gain of electrons (RIG)
– Reduction of oxidation number
• Oxidation agent is the one that’s reduced
• Reducing agent is the one that’s oxidized
• Potassium permanganate is added to an
acidified solution of iron (II) chloride.
• Write the net ionic reaction
MnO4- +H+ + Fe+2  Mn+2 + H2O + Fe+3
• Write the half reactions
8H+ + MnO4- + 5e-  Mn+2 + 4 H2O
5(Fe+2  Fe+3 + e-)
8H+ + MnO4- + 5Fe+2  Mn+2 + 4 H2O + 5Fe+3
Galvanic Cell
• A device that changes chemical energy to electrical
energy.
• The difference between a redox reaction and a cell
– Instead of mixing the chemicals in a flask and letting
the electrons move to a new substance
– Separate the chemicals
– Use moving electrons to do work
– Need to allow ions to flow!
• Or will only work for a moment
Voltaic Cells (Galvanic Cells)
• A voltaic cell is a chemical concept that led to
the manufacture of batteries.
• There are three major components to a voltaic
cell:
– Anode – oxidation reaction
– Cathode – reduction reaction
– Salt bridge – replaces ions to keep solutions
neutral
Salt bridge or porous disk
Allows ions to move
Oxidation occurs
Reduction Occurs
An OX
Red Cat
Cell Potential
• Between the anode and the cathode, electrons have a
difference in energy – the cathode is a lower energy state for
the electrons, so they will naturally “flow” there.
• This is a form of electrical potential energy.
– There is a “force” behind electrons moving
– Electromotive force (emf) or cell potential
• This potential can be measured with a potentiometer
• Like all chemical reactions there is a driving force.
– Which thermodynamic concept do you think would work here?
Zn(s) +
+2
Cu
=>
+2
Zn
+ Cu(s)
• What is the anode half reaction?
Zn  Zn+2 + 2 e-
• What is the cathode half reaction?
Cu+2 + 2e-  Cu
Cells
• The maximum amount of electrons that can
move (Current in Amps) is based on:
– Voltage – Volts - Pond on top of hill, water
flows downhill.
– Resistance – Ohms - Things in its way slow the
water down.
• Galvanic cell consists of two half reactions
– Trading electrons
– The force driving them is called
“voltage”
• Or potential
Water in a tank has a force
That pushes it out a pipe
Which pipe would have a higher
water pressure?
Analogous to a cell with
more electromotive force.
It has a higher voltage
Electromotive Force (EMF)
• The potential difference between anode and
cathode is called the electromotive force (the
force “pushing” electrons”), or emf.
• Ecell = emf
• EMF is measured in volts, where
– 1 Volt = 1 Joule per Coulomb
– 1 e- = 1.60 x 10-19 C
• To compute the value of the cell potential
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The maximum voltage
Need a standard for all half cells
Use hydrogen at [H+] = 1M, PH2 = 1 atm
All half cells are compared to hydrogen.
• Hydrogen half cell is assigned a value of zero.
Zero
Determining Relative Agent Strengths
• The strongest oxidizers (or oxidizing agents) are those
which have the most positive Eored values. (they are the
most easily reduced)
• The strongest reducing agents have the most negative
Eored values.
• F2 is the strongest oxidizing agent, while Li is the
strongest reducing agent.
Zn(s) +
+2
Cu
=>
+2
Zn
+ Cu(s)
• What is the anode half reaction? (An Ox)
Zn  Zn+2 + 2 e-
E° = +0.76
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What is the cathode half reaction? (Red Cat)
Cu Cu+2 + 2eE ° = -0.34
Eocell = Eored(cathode) – Eored(anode)
We subtract the anode’s Eored because oxidation happens
at the anode, and we are technically using the reverse
reaction.
• The cell potential is 0.76 + 0.34 = 1.10V
• What kind of reaction is this?
– We called this a single replacement reaction!
– Remember the activity series?
• Why are some substances more active then others?
Standard Reduction Potentials
• Calculate the standard emf for the voltaic cell based on
the reaction:
• Cr2O72- + 14 H+ + 6 I-  2 Cr3+ + 3 I2 + 7 H2O
– Eored (Cr2O72- + 14 H+ + 6 e-  Cr3+ + 7 H2O) = +1.33 V
– Eored (I2 + 2 e-  2 I-) = +0.54 V
• Note: even though there are coefficients, you do NOT
multiply emfs by them. EMFs are an intensive property
of matter and are not subject to the same rules as joules,
grams, etc.
Line Notation
Anode   Cathode
Mg(s)  Mg+2(aq)   Al+2(aq)  Al(s)
Mg + Al+2  Mg+2 + Al
For reactions producing/using gases or ions a
platinum electrode is used
Pt(s) ClO3-, ClO4-, H+   H+, MnO4-, Mn+2 Pt(s)
Two Cents Worth of Physics
• A galvanic cell consists of two half
reactions
• The “difference” in cell potentials is the
electromotive force E
• This force is related to the ability to do
work and ideally equal to the free energy
change
Equations
• emf = potential difference (V) = Work (J)
charge (C)
• Work is viewed from the point of view of the system!
• E = -w
q
Where E = electromotive force (voltage)
-w = work done (system lost energy)
q = charge (number of electrons x charge)
Free Energy
• Without the derivation!
∆G = -nF E
n = moles of electrons
F = 96,485 C/mol (the constant FARADAY)
E = emf = electromotive force
Under standard conditions
∆G° = -nF E ° (25 °C, 1 atm, [1M])
Units in Electrochem
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F = Faraday = 1 mole echarge on mole of electrons
F = 96,485 Coulombs / mol eCoulomb = C (unit of charge)
q = amount of charge in coulombs
E = electromotive force = EMF
E° = standard EMF
E = joule / C
Predicting Spontaneity
• ∆G for spontaneous reactions < 0 (negative)
• ∆G = -nF E
• Any process with a positive E
Will have a negative ∆G
Will be spontaneous, ∆G < 0, negative
K>1
Calculate ∆G° for the reaction
• Cu+2 + Fe → Fe+2 + Cu
• The half reactions are:
Cu+2 + 2e- → Cu
Fe
→ Fe+2 + 2 e-
E ° = 0.34 V
E ° = 0.44 V
• The E ° is 0.78 V so the delta G is
∆G ° = -nF E ° = -2 * 96500 * 0.78 = -15.1 kJ
What is the Voltage of a Cell?
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There are several different types of cells
Each has its uses
Some are batteries
Some are electrodes (pH meter)
Fuel cells use electrochemistry
– Instead of combustion
Concentration Cells
A galvanic cell can be made with the anode
and cathode with the same half reaction.
Which direction will
Electrons flow?
Porous disk
Ag
0.1M Ag+
Ag
1.0 M Ag+
Cathode
Nature will try to make
the concentration
On both sides
The same
Which is the cathode?
Which side plates silver metal?
Concentration Cells
• If the Eº of both the cells is the same, how
can a voltage be formed?
• Eº are at 1M concentrations
• So there is a potential difference between
the half cells
0.1M Ag+
1.0 M Ag+
The Nernst Equation
• The cell potential is dependent on
concentration
• Free energy is dependent on concentration
• Remember
G = Gº + RT lnQ
• Q and the concentration affects G
Nernst Again
G = -nFE
• Replacing the G’s with –nFE’s
E = E - RT lnQ
nF
At 25º C
E = E - 0.0591 log (Q)
n
2Al + 3Mn+2  2Al+3 + 3Mn
• [Mn+2] = 0.50M [Al+3] = 1.50M
– Remember heterogeneous equilibria
– and half reactions!
E = E - 0.0591 log (Q)
n
Al  Al+3 +3e-
Mn+2 +2e-  Mn n = 6 (3 x 2)
E = 0.48V – 0.0591 log (1.50)2
6
(0.50)3
=
0.48 – 0.01 = 0.47V
What is a Dead Battery?
• The cell will spontaneously discharge
• When the Q = K, the reaction is at equilibrium
• There is no longer a driving force
• The two cells have the same free energy
• E cell = G = 0
Lets Consider The Reaction
VO2+1 and Zn react.
2VO2+ + Zn  2VO+2 + Zn+2
Use half reaction method to get equation
2VO2+ +4H+ + Zn  2VO+2 +2H2O + Zn+2
What is the cell potential?
VO2+ +2H+ + e-  VO+2 +H2O
Zn  2 e- + Zn+2
E = 1.00V
E = 0.76V
2VO2+ +4H+ + Zn  2VO+2 +2H2O + Zn+2
Where T = 25 C, [VO2+] = 2.0M,
[H+] = 0.50M
[VO+2] = 1.0x 10-2M [Zn+2]= 1.0 x 10-1M
• Not at standard conditions, n = ? (number of electrons)
E = 1.76 - 0.0591 log ([VO+2]2[Zn+2]) = 1.76 + 0.13 = 1.89V
2
[VO2+]2 [H+]4
Ion Selective Electrodes
• pH electrodes are an example
– Sensitive to the H+ ion
– Others available
• Cd+2, Ca+2,Cu+2, K+1. Ag+1, Na+1
• Cl-1, Br-1, CN-1, F-1, NO3-1, S-2
• Works like a concentration cell
– Inside electrode is a half cell of known concentration
– Measures the potential difference
Equilibrium Constants
E = E - 0.0591 log (Q)
n
When Q = K, E = 0
E = +0.0591 logK
n
Solve for K.
K is typically large for a redox reaction.
(The useful ones anyway)
Electrolytic Cells
• An electrolytic cell consists of two electrodes
in a molten salt or a solution.
– Electrolysis is the decomposition of a molecule
using electricity.
• Example:
– 2 NaCl(l)  2 Na(l) + Cl2(g)
• Unlike a voltaic cell, electrolytic cells require
a voltage source (battery) to function.
Electroplating
• Electrolysis can be used for the industrial
practice of electroplating – that is, plating one
metal on another.
• Example: Copper plating of a quarter.
The Ampere
• An ampere is a coulomb per second
– It is the unit of electric current.
• 1 A = 1 C/s
• We can calculate the extent of electrolysis
given the electric current and time.
Moles of Electrons
• When you have total charge in coulombs, you
can use Faraday’s constant to convert to moles
of electrons.
• Faraday’s constant =
96,483 C/mol e• This will allow us to then use stoichiometry to
solve for the amount of substance
oxidized/reduced.
Practice Problems
• Calculate the number of grams of aluminum
produced in 1.00 hour by the electrolysis of
molten AlCl3 if the electrical current is 10.0 A
(amperes).
Electrical Work
• Gibbs’ free energy is the extra energy that can be used
to power other devices.
• ΔG = -nFEcell
• Therefore, the maximum amount of useful electrical
work (energy) from a voltaic cell is
– wmax = -nFEcell
• Or, if you are applying an external potential (like in an
electrolytic cell)…
– w = nFEext (positive because you are putting energy into the
system)
• q=nF
• wmax = qE
Electrical Work and Efficiency
• The unit of electrical work is the kilowatt-hour
(kWh).
• Based on the watt (1 J/s), a unit of electrical
power.
• 1 kWh = 3.6 x 106 J
• Efficiency = w x 100%
wmax
Practice
• Calculate the number of kilowatt-hours of
electricity required to produce 1.0 x 103 kg
of aluminum by electrolysis of Al3+ if the
applied voltage is 4.50 V.