Transcript Ch 18 ppt.ppsx

ADVANCED PLACEMENT
CHEMISTRY
ELECTROCHEMISTRY
Galvanic cell- chemical energy
is changed into electrical energy
(also called a voltaic cell)
(spontaneous)
Electrolytic cell- electrical
energy is changed into chemical
energy (nonspontaneous)
Oxidation-reduction reaction
(redox)- involves the transfer of
one of more electrons. Oxidation
can not occur without
reduction!!!
Oxidation*loss of electrons
*increase in oxidation number
Reduction*gain of electrons
*decrease in oxidation number
OIL RIG
Oxidation is loss (of e-),
Reduction is gain (of e-)
LEO the lion goes GER
Lose electrons = oxidation,
Gain electrons = reduction
Oxidizing Agent- Substance
which is reduced and causes
another substance to be oxidized.
Reducing Agent- Substance
which is oxidized and causes
another substance to be reduced.
Half-reaction- Equation written
to show either the loss or the gain
of electrons (shows number of
electrons transferred)
Ex. Fe+2  Fe+3 + e−
Cu+2 + 2e−  Cu
Voltaic (Galvanic) Cell
The transfer of electrons can be used as an
energy source if it is harnessed. One way
that this may be done is by separating the
two half-reactions and requiring the
electron transfer to occur through a wire.
The electron flow (electricity) can be used
to do useful work.
The solutions must also be connected in
another way or the charge will become
unbalanced and the electrical flow will stop.
This connection is often by a salt bridge (tube
filled with conducting solution)or porous disk.
Ions flow from one compartment to the other
to keep the net charge zero.
Anode- The electrode at which
oxidation occurs. After a period
of time, the anode may appear to
become smaller as it falls into
solution.
Cathode- Electrode at which
reduction occurs. After a period
of time, it may appear larger, due
to ions from solution plating out
on it.
An Ox and a Red Cat
Oxidation occurs at the anode,
Reduction occurs at the cathode.
FAT CAT
Electrons travel
from anode to cathode
(from reducing agent
to oxidizing agent)
in a galvanic cell.
Animation
Cell potential (Ecell) or
electromotive force (emf):
*the voltage measured across the
electrodes of the half cells
*the driving force of the reaction
*cells with positive Ecell values are
always thermodynamically favored
(spontaneous)
*units are volts (1 joule of work per
coulomb of charge transferred)
1V = 1J/C
E°cell = E°ox + E°red
Standard cell potential is the sum of the
standard oxidation potential for the oxidation
half-reaction and the standard reduction
potential for the reduction half-reaction.
Standard conditions are 1 atm, 25°C,and 1 M
*at conditions other than standard,
Ecell = Eox + Ered
Standard potentials for half-reactions:
*usually written as reduction reactions (to
change to oxidation, reverse reaction and
change the sign of the potential)
*can't be determined directly
*are determined by the use of a standard
hydrogen electrode (SHE) which is
arbitrarily assigned a value of 0.00 V
*all other half reactions are expressed in
reference to the standard hydrogen electrode
*Substances with more positive reduction
potentials are more easily reduced. They act
as oxidizing agents (cause another substance
to be oxidized).
*The more positive the reduction potential,
the stronger the oxidizing agent.
*Substances with negative reduction
potentials (and thus positive oxidation
potentials) are more easily oxidized. They
act as reducing agents (cause another
substance to be reduced).
*The more negative the reduction potential,
the stronger the reducing agent.
http://www.jesuitnola.org/upload/
clark/Refs/red_pot.htm
Standard hydrogen electrode(SHE)- a platinum electrode in
contact with 1M H+ and bathed
by hydrogen gas at 1 atm
+
2H
+
2e
 H2
o
E
= 0.00 V
When combining half reactions
to obtain balanced redox
reactions and calculating cell
potential, you do not multiply the
cell potential by an integer even
if you multiply the half-reaction
by an integer when writing the
redox reaction.
If Ecell is positive the reaction is
thermodynamically favored
(spontaneous).
If Ecell is negative the reaction is not
thermodynamical favorable
(nonspontaneous).
Electrons will flow in the direction that
will make the Ecell positive.
Voltaic experiment
Ex. What will be the spontaneous reaction between
the following set of half-reactions? What is the value
of E°cell?
Cr3 +(aq) + 3e—  Cr(s)
E° = -0.74 V
MnO2(s) + 4H+(aq) + 2e—  Mn2 +(aq) + 2H2O(l) E° = +1.28 V
One rxn must be oxidation and the other must be reduction
(we must reverse one).
The sum of the two potentials must be positive.
2(Cr  Cr3+ + 3e−) E° = +0.74 V
3(MnO2 + 4H+ + 2e−  Mn2+ + 2H2O) E° = +1.28 V
2Cr + 3MnO2 + 12H+  2Cr3+ + 3Mn2+ + 6H2O
Eocell = 2.02 V
When diagramming an
electrochemical cell, the anode is
traditionally drawn on the left
and the cathode on the right.
(Don’t count on this on a test
question!)
The greater the positive potential of a cell,
the greater the spontaneity of the reaction.
Since G, K, and E° are all measurements
of the spontaneity of a reaction, we can
write equations to interrelate them.
G° = −nFE°
*G° is the standard free energy change (units are J)
*n is the number of moles of electrons transferred
*F is a Faraday = 96,500 Coulombs/mol
*E° is standard cell potential
If E° is positive, G is negative
To find K: G° = −RT lnK
Faraday -amount of electricity that
reduces one equivalent mass of a
substance at the cathode and oxidizes
one equivalent mass of a substance at the
anode (equivalent mass is the mass of a
species that will yield or consume one
mole of electron
*amount of electrical charge carried by 1
mol of electrons = 96,500 coulombs
If cell concentrations are not
standard (1 M) and 25°C, the cell
potential is affected. This effect can
be determined by using the Nernst
equation. This equation is derived
from the equation used for free
energy when pressures were not all 1
atm :
G= G° + RT lnQ
Nernst equation: E = E°-RT lnQ
nF
R = 8.314 J/Kmol
T = temp in Kelvin
Q = ratio of concentrations or pressures of products over
reactants each raised to the power of their coefficients
n = number of moles of electrons transferred
F = 96,500 Coulombs/mol
Another form of the Nernst equation is only valid at 25°C :
E= E°−(0.0592/n)log Q
The Nernst equation is not required on the AP Chem exam. You
should be able to determine whether changes in concentration increase
or decrease the cell potential through the use of LeChatelier’s principle.
The E value calculated is the
maximum potential at the very
beginning of the reaction. As the
reaction progresses, E decreases
until it reaches zero and the battery
is at equilibrium (dead).
Ex. A voltaic cell is prepared using zinc and copper strips. The
zinc strips are in a 0.40 M zinc nitrate solution and the copper
strips are in a 0.020 M copper(II) nitrate solution. At 25C, will
the cell have a voltage higher or lower than the standard cell
potential? Justify your answer.
Zn(s) + Cu2+ ⇄ Zn2+ + Cu(s)
Ecell = 1.10 V
Both the Zn2+ (a product) and the Cu2+ (reactant)
concentrations are less than 1 M. However, the
[Cu2+] is significantly lower than the [Zn2+],
causing the reaction to shift toward the Cu2+
(reactants). This makes the reaction less
thermodynamically favorable and the voltage will
decrease.
.
Equilibrium constants can also be
calculated using cell potentials. At
equilibrium E = 0 and Q = K.
E° = (0.0592/n) log K
Ex. Determine whether the following reaction is
spontaneous and calculate its equilibrium
constant at 25°C.
Sn(s) + Ni2+  Sn2+ + Ni(s)
The half-reactions would be as follows:
Sn(s) Sn+2 + 2 e− E° = 0.14 V
Ni+2 + 2e− Ni(s) E° = −0.23 V
E°cell = 0.14 + (−0.23) = −0.09
nonspontaneous
E° = (0.0592/n) log K
−0.09= (0.0592/2) log K
K = 9 × 10−4
Electrolytic cellsWhen the desired chemical reaction is
nonspontaneous (E°cell is negative) the
reaction can proceed by the addition of
electrons from an outside source. This
results in the conversion of electrical
energy into chemical energy. Examples
are electrolysis of water, NaCl solution,
silver plating
Faraday's Law- The amount of
substance being oxidized or
reduced at each electrode during
electrolysis is directly
proportional to the amount of
electricity that passes through the
cell.
Coulomb = amp × s : amount of charge that
passes a given point when 1 amp of electrical
current flows for 1 second.
Ampere (amp) : flow rate of 1 coulomb/s
(i is used as symbol for current)
Joule (J): amount of energy absorbed or
evolved when 1 coulomb of electrical charge
moves through a potential difference of 1 V
(1J = 1 volt-coul and 1F = 1 J/V mol)
Conversion Factors
Amps ×time(s) = Coulombs
96,500 Coulombs = 1 F
1 F = 1 mole of electrons
I = q/t
I = current in amps
q = charge in Coulombs
t= time in seconds
Dimensional Analysis is very helpful for solving electrolysis problems.
current(amps) × time 
coulombs  Faradays
mol reactant  g reactant
or
g reactant  mol reactant 
Faradayscoulombs
current(amps) × time
Ex. How many grams of Cr can be plated out
by passing 2.05 amps through acidic CrO3 for
1.0 × 104 s?
Cr must go from an ox # of +6 to zero, therefore 6 electrons
must be gained.
Cr+6 + 6e−  Cr
10000s 2.05C 1 mol e− 1 mol Cr 52.0 g Cr = 1.84 g Cr
1s
96,500 C 6 mol e− 1 mol Cr
Chrome plating
Ex. How many hours would it take to
produce 25.0 g of Cr from a solution of
CrCl3 by a current of 2.75 A?
Cr+3 + 3e−  Cr
25.0g Cr 1 mol Cr 3 mol e− 96500 C 1 s
1 hr =14.0 hr
52.0g Cr 1 mol Cr 1 mol e− 2.75 C 3600s
When a molten salt is electrolyzed, the
cation is always reduced (and formed at
the cathode) while the anion is oxidized
(and formed at the anode).
The electrolysis of molten NaCl:
NaCl(l) Na(l) + ½ Cl2(g)
anode: Cl-  ½ Cl2 + e−
cathode: Na+ + e−  Na
When electrolysis is carried out in
aqueous solutions, there is always the
possibility that water will be oxidized
and/or reduced instead of the solute. In
order to predict the products produced
when aqueous solutions are electrolyzed,
you must determine the potentials of
each possibility.
Ex. The electrolysis of aqueous NaBr
Possible anode reactions (oxidation):
2Br−(aq) Br2(g) + 2e−
E° = −1.09 V
2H2O(l)  O2(g) + 4H+(aq) + 4e− E° = −1.23V
The first reaction occurs because it is less
negative (more spontaneous)
Possible cathode reactions (reduction):
Na+(aq) + e−  Na(s)
2H2O(l) + 2e−  H2(g) + 2OH−(aq)
E° = −2.71V
E° = −0.83V
The second reaction occurs because it is less
negative (more spontaneous). The overall reaction is:
2H2O + 2Br−  H2 + 2OH− + Br2
Electrolysis of KI
This method of determining whether water is
oxidized or reduced does not always work. In
the electrolysis of salt water (brine), we would
predict that water would be both oxidized and
reduced. In fact, the chloride ions are
oxidized instead of the water. This is due to
some fairly complex factors. This
phenomenon is called overpotential or
overvoltage.
The electrolysis of aqueous
CuSO4 :
Anode: H2O  ½ O2 + 2H+ + 2e−
+2
−
Cathode: Cu + 2e  Cu
The electrolysis of water:
Anode: 2H2O(l)  O2(g) + 4H+(aq) + 4e−
E° = −1.23 V
Cathode: 2H2O(l) + 2e−  H2(g)+ 2OH−(aq)
E ° = −0.83 V
E ° cell = −2.06 V
Overall:
6H2O  2H2 + O2 + 4H++ 4OH−
2H2O  2H2 + O2