Chapter 18
1




Electrochemistry - is the study of the relationships
between chemical reactions and electricity.
Oxidation Number – The charge on an ion for a
monoatomic ion. Otherwise, it is the hypothetical
charge based on a set of rules. (See Slides 6 - 7.)
If a substance has lost electrons, its oxidation
number has become more positive and thus
oxidized
If a substance has gained electrons, its oxidation
number is more negative and thus has been reduced.
2
LEO:
GER
 Lose Electrons
Oxidation
LEO says GER
GER:
 Gain Electrons
Reduction
3

Redox – Shorthand name for an oxidationreduction reaction. Oxidation states are really
just used as ‘electron bookkeepping’ and don’t
represent an actual charge (unless you are
talking about monoatomic ions). It does
represent, though, a movement of electrons
from one atom(s) to another. Electrons are
particles of matter, so the number that leave
one substance must the same number that go to
another substance.
4
1) Oxidation number of an element is zero.
2) For monoatomic ion, the oxidation number is
equal to its charge. (ex: Na+ is +1, F- is -1)
3) Oxygen usually has a -2 oxidation number.
(except in peroxides where it’s -1)
4) Hydrogen’s oxidation number is +1 when its
attached to a nonmetal (e.g. HCl) and -1 when
attached to a metal (e.g. CaH2) as a hydride.
5
5) Fluorine is always -1 in compounds.
6) The other halogens are mostly -1 in binary
compounds. When they combine w/oxygen,
they can be positive (ClO4-).
7) The sum of the oxidation numbers in a neutral
compound is zero.
8) The sum of the oxidation numbers in a
polyatomic ion is equal to its charge.
 Example from #6 & #8: oxygen is always -2
and there are 4 of them, Cl has to be +7 in order
for the ClO4- ion to be -1.
6
COMPOUND

P2O5

NaH

Cr2O72-

SnBr4

K2O2
OXIDATION NUMBER OF
RED BOLDFACE ELEMENT
7

Determine the oxidation number of every
element in the following reactions.
1.
Sn(s) + 2H+(aq)  Sn2+ (aq) + H2(g)
2.
2MnO4-(aq) + 5HSO3-(aq) +H+
2Mn2+(aq) + 5SO42-(aq) + 3H2O
8
Oxidation-Reduction Reactions
• Zn added to HCl yields the spontaneous reaction
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g).
• The oxidation number of Zn has increased from 0
to 2+.
• The oxidation number of H has reduced from 1+
to 0.
• Zn is oxidized to Zn2+ while H+ is reduced to H2.
9
Oxidation-Reduction Reactions
10
Balancing Oxidation-Reduction
Reactions
• Law of conservation of mass: the amount (moles) of
each element present at the beginning of the reaction
must be present at the end.
• Conservation of charge: electrons are not lost in a redox
chemical reaction, just moved around
Half Reactions
• Half-reactions are a convenient way of separating
oxidation and reduction reactions.
11
Balancing Oxidation-Reduction
Reactions
Half Reaction Method
• The skeleton reaction is:
MnO4-(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)
• We want to make this such that it is easy to balance the
atoms and the charges.
12
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
So, how do we balance the previous reaction?
1. Write down the two half reactions.
2. Balance each half reaction:
a. First with elements other than H and O.
b. Then balance O by adding water.
c. Then balance H by adding H+.
d. Balancing the charge by using electrons
13
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
3. Multiply each half reaction to make the number of
electrons equal.
4. Add the reactions and simplify.
5. Check!
14
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
1. The two incomplete half reactions are
MnO4-(aq)  Mn2+(aq)
Fe2+(aq)  Fe3+(aq)
2. Adding water (to balance O’s) and H+ yields:
8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O
• There is a charge of 7+ on the left and 2+ on the right.
Therefore, 5 electrons need to be added to the left:
5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O
15
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
• In the iron ion reaction, there is a 2+ charge on the left
and a 3+ charge on the right, so we need to add an
electron:
Fe2+(aq)  Fe3+(aq) + 1e• Now there is the same overall +2 charge on both sides.
16
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
3. To balance the 5 electrons for permanganate and 1
electron for iron, we need to have 5 electrons move from
1 species to the other. Multiplying the iron equation by
5 gives:
5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O
5Fe2+(aq)  5Fe3+(aq) + 5e-
17
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
3.
5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O
5Fe2+(aq)  5Fe3+(aq) + 5e• Notice now that the number of electrons is the same for
both equations, appearing on opposite sides.
• In reduction half-reaction, e- are reactants
• In oxidation half-reaction, e- are products
18
Balancing Oxidation-Reduction
Reactions
Balancing Equations by the Method of Half
Reactions
4. Adding gives:
8H+(aq) + MnO4-(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l)
+ 5Fe3+(aq)
5. Which is balanced!
Notice that atoms and charges are both balanced. There is
a +17 charge on left (8-1+5(2)), and +17 charge on right
(2+5(3)).
19
Balancing Oxidation-Reduction
Half-Reactions Practice
Fe2+(aq) + Al(s)  Fe(s) + Al3+(aq)
20
Balancing Oxidation-Reduction
Half-Reactions Practice
Mn2+(aq) + NaBiO3(s)  Bi3+(aq) + Na+(aq) + MnO4-(aq)
1
Balancing Oxidation-Reduction
Reactions
Balancing Equations for Reactions Occurring in
Basic Solution
• We can use OH- and H2O rather than H+ and H2O,
respectively.
• However, it’s easier to balance in acid, and then
‘neutralize’ the acid with OH-. But if you add OH- to one
side, you have to add it to the other side.
• In other words, balance in acidic solution and then
convert it to a basic solution.
1
Balancing Oxidation-Reduction
Half-Reactions Example
Balance the following reaction in a basic solution
Cr(OH)3(s) + ClO-  CrO42- + Cl2(g)
1. H2O + Cr(OH)3  CrO42- + 5H+ + 3e2. 2e- + 4H+ + 2ClO-  Cl2(g) + 2H2O
3. Multiply equation 1 by 2 and equation 2 by 3
2H2O + 2Cr(OH)3  2CrO42- + 10H+ + 6e6e-+ 12H++ 6ClO-  3Cl2(g) + 6H2O
Add to get:
2H+ + 2Cr(OH)3 + 6ClO-  2CrO42- + 3Cl2 + 4H2O
1
Balancing Oxidation-Reduction
Half-Reactions Example
2H+ + 2Cr(OH)3 + 6ClO-  2CrO42- + 3Cl2 + 4H2O
+2OH+2OH“Neutralize” by adding OH- to both sides.
2H2O +2Cr(OH)3 +6ClO-2CrO42- +3Cl2+ 4H2O+2OHCross out common substances on both sides.
2Cr(OH)3 +6ClO-2CrO42- +3Cl2+ 2H2O+2OH-
1
Ch. 18.2 - Galvanic Cells
• The energy released in a spontaneous redox reaction is
used to perform electrical work.
• Galvanic cells or voltaic cells are devices in which
electron transfer occurs via an external circuit.
• Galvanic cells are spontaneous.
• If a strip of Zn is placed in a solution of CuSO4, Cu is
deposited on the Zn and the Zn dissolves by forming
Zn2+.
1
Galvanic Cells
• Zn is spontaneously oxidized to Zn2+ by Cu2+.
• The Cu2+ is spontaneously reduced to Cu0 by Zn.
• The entire process is spontaneous.
1
Spontaneity and Potential Energy
 Redox
reactions occurring in voltaic cell are
spontaneous
 Why do electrons flow spontaneously from
one electrode to other?

Flow spontaneously due to difference in potential
energy between anode and cathode
27
Voltaic Cell: Cathode Reaction
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Voltaic Cell: Anode Reaction
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Galvanic Cells
• Galvanic cells consist of
–
–
–
Anode: Zn(s)  Zn2+(aq) + 2e2- (Oxidation)
Cathode: Cu2+(aq) + 2e-  Cu(s) (Reduction)
Salt bridge (used to complete the electrical circuit): cations
move from anode to cathode, anions move from cathode to
anode.
• The two solid metals are the electrodes (cathode and
anode).
1
Galvanic Cells
• The flow of electrons from anode to cathode is
spontaneous.
• Electrons flow from anode to cathode because the
cathode has a lower electrical potential energy than the
anode.
• Potential difference: difference in electrical potential.
Measured in volts.
• One volt is the potential difference required to impart
one joule of energy to a charge of one coulomb:
32
Electron Flow and Potential Energy

Anode


higher potential
energy
Cathode

lower potential
energy
33
Height is an analogy for voltage
High Voltage
Low Voltage
34
Galvanic Cells
• As oxidation occurs, Zn is converted to Zn2+ and 2e-.
The electrons flow towards the cathode where they are
used in the reduction reaction.
• We expect the Zn electrode to lose mass and the Cu
electrode to gain mass.
• “Rules” of galvanic cells:
1. At the anode electrons are products. (Oxidation)
2. At the cathode electrons are reactants. (Reduction)
3. Electrons cannot swim.
1
Galvanic Cells
• Electrons flow from the anode to the cathode.
• Therefore, the anode is negative and the cathode is
positive. (Think anion is negative, cation is positive.)
• Electrons cannot flow through the solution, they have to
be transported through an external wire. (Rule 3.)
1
Galvanic Cells - Line Notation
Used to describe electrochemical cells.
Anode components are listed on the left.
Cathode components are listed on the right.
Separated by double vertical lines which indicated salt
bridge or porous disk.
The concentration of aqueous solutions should be
specified in the notation when known.
Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s)
 Mg → Mg2+ + 2e–
(anode)
 Al3+ + 3e– → Al
(cathode)






1
1
Galvanic Cells
• Anions and cations move through a porous barrier or salt
bridge.
• Cations move into the cathodic compartment to balance
the excess negatively charged ions since the Cu2+ ions
are being reduced. (Cathode: Cu2+ + 2e-  Cu)
• Anions move into the anodic compartment to neutralize
the excess Zn2+ ions formed by the oxidation of Zn
metal.
1
Galvanic Cells
A Molecular View of Electrode Processes
During the spontaneous redox reaction between Zn(s) and
Cu2+(aq):
• Zn(s) is oxidized to Zn2+(aq) and Cu2+(aq) is reduced to
Cu(s).
• On the atomic level, a Cu2+(aq) ion comes into contact
with a Cu(s) atom on the surface of the electrode that has
extra electrons.
• Electrons form from Zn(s), forming Zn2+(aq) , travel
through the circuit to the Cu electrode where Cu2+(aq)
gain those electrons and form more Cu(s).
1
1
Ch. 18.3 – Standard Reduction Potentials
Standard Reduction (Half-Cell) Potentials
• Convenient tabulation of electrochemical data.
• Standard reduction potentials, Ered are measured relative
to the standard hydrogen electrode (SHE).
1
Standard Reduction Potentials
Standard Reduction (Half-Cell) Potentials
1
Standard Reduction Potentials
Standard Reduction (Half-Cell) Potentials
• The SHE is the cathode. It consists of a Pt electrode in a
tube placed in 1 M H+ solution. H2(g) is bubbled through
the tube.
• For the SHE, we assign
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
• Ered of SHE is zero by definition.
• The emf of a cell can be calculated from standard
reduction potentials:
E cell  E red cathode   E red anode 
1
1
Standard Reduction Potentials
Standard Reduction (Half-Cell) Potentials
• Consider Zn(s)  Zn2+(aq) + 2e-. We measure Ecell
relative to the SHE (cathode):
Ecell = Ered(cathode) - Ered(anode)
0.76 V = 0 V - Ered(anode).
• Therefore, Ered(anode) = -0.76 V.
• Standard reduction potentials must be written as
reduction reactions:
Zn2+(aq) + 2e-  Zn(s), Ered = -0.76 V.
1
Standard Reduction Potentials
•
•
•
•
•
Standard Reduction (Half-Cell) Potentials
Since Ered = -0.76 V we conclude that the reduction of
Zn2+ in the presence of the SHE is not spontaneous.
The oxidation of Zn with the SHE is spontaneous.
Changing the stoichiometric coefficient does not affect
Ered. (Measures potential energy/unit charge (Volt =
J/C) Intensive property.)
Therefore,
2Zn2+(aq) + 4e-  2Zn(s), Ered = -0.76 V.
Reactions with Ered > 0 are spontaneous reductions
relative to the SHE.
1
Standard Reduction Potentials
Standard Reduction (Half-Cell) Potentials
• Reactions with Ered < 0 are spontaneous oxidations
relative to the SHE.
• The larger the difference between Ered values, the larger
Ecell.
• In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is
more positive than Ered(anode).
• Recall
E cell  E red cathode   E red anode 
1
1
Standard Reduction Potentials
•
•
•
•
Ease of Oxidation and Reduction
The more positive Ered the more easily reduced the
reactant on the left.
The more negative Ered the more difficult the reactant is
reduced. Conversely, the more easily the product is
oxidized.
A species higher and to the left of the table of standard
reduction potentials will spontaneously oxidize a species
that is lower and to the right (product) in the table.
(Called the Northwest-Southeast Rule)
That is, F2 will oxidize H2 or Li; Ni2+ will oxidize Al(s).
1
1
Increasing ease of oxidation
Increasing ease of reduction
Most easily reduced
Most easily
oxidized
Standard Reduction Potentials
Standard Reduction (Half-Cell) Potentials
Rank the halogens in
order of ease of being
reduced.
Which of the halogens
is capable of oxidizing
Ag to Ag+.
Which of these metals is
easiest to oxidize? Fe,
Ag, Zn, Mg, Au



1
Standard Reduction
Potentials
Cell EMF
Calculate the Eocell for the following
reaction:
2Al(s) + 3I2(s)  2Al3+ + 6I-(aq)
 1st: Determine anode & cathode
 2nd: Write the reduction half-reactions.
2Al3+ + 6e-  2Al(s) Eored = -1.66 V
3I2(s) + 6e-  6I-(aq) Eored = +0.54 V
 3rd: Calculate Eocell =Eocathode – Eoanode
 Eocell = 0.54 – (-1.66) = +2.20 V

1
Cell Practice

A voltaic cell is based on a Co2+/Co half
cell and a AgCl/Ag half cell.


What is the reaction at the anode?
What is the standard cell potential?
Co2+ + 2e-  Co(s) Eored = -0.277
 AgCl + 1e-  Ag(s) + Cl-(aq) Eored = 0.22 V
 Anode will be oxidation, more neg
reduction potential (Co(s)Co2+ + 2e-)
 Eocell = 0.22 – (-0.277) = 0.497 V

1
Cell Practice
The Galvanic cell below has a cell potential of 1.19 V.
Tl3+(aq) + 2Cr2+(aq)  Tl+(aq) + 2Cr3+(aq)




What are the half reactions?
Determine the Ered for Tl3+ to Tl+.
Write the shorthand notation for the cell.
Tl3+ + 2e-  Tl+(aq) Eored = ??
2Cr3+ + 2e-  2Cr2+(aq) Eored = -0.41 V
The cell potential is positive, so see what was oxidized
(Cr2+)
Eocell = Eocathode – Eoanode then
Eocathode = Eocell+Eoanode
=1.19 + (-0.41) = +0.78
Cr2+(aq)|Cr3+(aq)||Tl3+(aq)|Tl+(aq)







1
Ch. 18.4 - Cell Potential, Electrical
Work and Free Energy
•
•
•
•
1J
1V 
1C
Electromotive force (emf) is the force required to push
electrons through the external circuit.
Cell potential: Ecell is the emf of a cell.
For 1M solutions at 25 C (standard conditions), the
standard emf (standard cell potential) is called Ecell.
Electrical work is viewed from the point of view of the
system.
1
WORK
 Work is never the maximum possible if any current
is flowing.
 In any real, spontaneous process some energy is
always wasted – the actual work realized is always
less than the calculated maximum.
1
Maximum Cell Potential

Directly related to the free energy difference
between the reactants and the products in the
cell.

ΔG° = –nFE°
 F = 96,485 C/mol e–
1
Spontaneity of a Cell Reaction
EMF and Free-Energy Change
• The Free Energy of a Cell (ΔG) is related to it’s EMF:
G  nFE
• G is the change in free-energy, n is the number of moles
of electrons transferred, F is Faraday’s constant, and E is
the emf of the cell.
• We define
1F  96,500 C/mol  96,500 J/V·mol
• Since n and F are positive, if G > 0 then E < 0.
• And if E>0, then ΔG <0 and cell is spontaneous.
59
Spontaneity of a Cell Reaction
• In a galvanic (voltaic) cell (spontaneous) Ered(cathode) is
more positive than Ered(anode) since
E cell  E red cathode   E red anode 
• More generally, for any electrochemical process
Ecell  Ered  reduction process   Ered  oxidation process 
• A positive E indicates a spontaneous process (galvanic
cell).
• A negative E indicates a nonspontaneous process.
60
Spontaneity of a Cell Reaction

What is n for the following reaction?
 What is the ∆G0 for the reaction?
Cu(s) + Ba2+(aq)  Cu2+(aq) + Ba(s)
n=2
 E0 = -2.90V (E0cathode) – 0.337(E0anode) = -3.24 V
 ∆G0 = -nFE0

=-(2 mol)(96,500J/V mol)(-3.24V)

=625 kJ
=> Reaction is not spontaneous.
1
Ch. 18.5 - Dependence of Cell
Potential on Concentration
The Nernst Equation
• A voltaic cell is functional until E = 0 at which point
equilibrium has been reached.
• The point at which E = 0 is determined by the
concentrations of the species involved in the redox
reaction.
• The Nernst equation relates emf to concentration using
G  G  RT ln Q
and noting that
 nFE  nFE   RT ln Q
1
Concentration and Cell Potential
The Nernst Equation
• This rearranges to give the Nernst equation:
RT
E  E 
ln Q
nF
• The Nernst equation can be simplified by collecting all
the constants together using a temperature of 298 K:
0.0592
E  E 
log Q
n
• (Note that change from natural logarithm to base-10 log.)
• Remember that n is number of moles of electrons.
1
Concentration and Cell Potential
Example
• Recall earlier we had a voltaic cell reaction of
2Al(s) + 3I2(s)  2Al3+ + 6I-(aq) E0cell = +2.20 V
• What is the Ecell when [Al3+] is 4.0x10-3M and [I-] =
0.010 M?
• Ecell = E0cell – 0.0592/n log [Al3+]2[I-]6
•
= 2.20 V – 0.0592/6 log [4.0x10-3]2[0.010]6
•
= 2.20 V – 0.0099*log 1.6x10-5(1x10-12)
•
= 2.20 V – 0.0099*log 1.6x10-17
•
= 2.20 V – 0.0099*(-16.796)
• 1 = 2.20 V + 0.17 = 2.37 V
Concentration and Cell Potential
Concentration Cells
• We can use the Nernst equation to generate a cell that has
an emf based solely on difference in concentration.
• One compartment will consist of a concentrated solution,
while the other has a dilute solution.
• Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq).
• The cell tends to equalize the concentrations of Ni2+(aq)
in each compartment.
• The concentrated solution has to reduce the amount of
Ni2+(aq) (to Ni(s)), so must be the cathode.
1
Concentration and Cell Potential
Concentration Cells
1
Concentration and Cell Potential
Cell Potential and Chemical Equilibrium
• A system is at equilibrium when G = 0.
• From the Nernst equation, at equilibrium and 298 K (E =
0 V and Q = Keq):
0.0592
0  E 
log K eq
n
nE 
log K eq 
0.0592
1
Concentration
Spontaneity of and
Redox
Cell
Reactions
Potential
Calculate the Keq for the following reaction
Fe(s) + Ni2+  Fe2+(aq) + Ni(s)
 Fe2+ + 2e-  Fe(s) E0= -0.440 V
 Ni2+ + 2e-  Ni(s) E0 = -0.280 V
 E0cell = -0.280 – (-0.440) = +0.16V
 Log Keq = nE0/0.0592

= 2(0.16)/0.0592

= 5.405

Keq = 2.54x105

1
Ch. 18.8 - Electrolysis
Electrolysis of Aqueous Solutions
• Nonspontaneous reactions require an external current in
order to force the reaction to proceed.
• Electrolysis reactions are nonspontaneous.
• In voltaic and electrolytic cells:
–
–
–
1
reduction occurs at the cathode, and
oxidation occurs at the anode.
However, in electrolytic cells, electrons are forced to flow
from the anode to cathode. (Electrons are forced to flow from
low potential to high potential. So it’s a measure of the work
you put into the system.)
Electrolysis
Electrolysis of Aqueous Solutions
1
1
Electrolysis
•
•
•
•
Electrolysis of Aqueous Solutions
Example, decomposition of molten NaCl.
Cathode: 2Na+(l) + 2e-  2Na(l)
Anode: 2Cl-(l)  Cl2(g) + 2e-.
Industrially, electrolysis is used to produce metals like Al.
1
Electrolysis
Electroplating
• Active electrodes: electrodes that take part in electrolysis.
• Example: electrolytic plating.
1
1
Electrolysis
Electroplating
• Consider an active Ni electrode and another metallic
electrode placed in an aqueous solution of NiSO4:
• Anode: Ni(s)  Ni2+(aq) + 2e• Cathode: Ni2+(aq) + 2e-  Ni(s).
• Ni plates on the inert electrode.
• Electroplating is important in protecting objects from
corrosion.
1
Electrolysis
Quantitative Aspects of Electrolysis
• We want to know how much material we obtain with
electrolysis.
• Consider the reduction of Cu2+ to Cu.
–
–
–
–
1
Cu2+(aq) + 2e-  Cu(s).
2 mol of electrons will plate 1 mol of Cu.
The charge of 1 mol of electrons is 96,500 C (1 F).
Since q = It, the amount of Cu can be calculated from the
current (I) and time (t) taken to plate. (C = Ampere*second)
Electrolysis
In an electrolysis cell, Ni(s) is deposited on
the cathode in a solution of Ni2+ ions. If the
current is 0.150 A for 12.2 min, how much Ni
is plated?
 1st: Calculate how many Coulombs of e- are
needed.
Q=0.150 A * 12.2 min * 60 sec/min
=109.8 Coulombs
2nd: Calculate the number of moles of electrons
(Faradays) that is by…
109.8 C = 0.0011 mol e- (F)
96500C/mol

1
Electrolysis
3rd: Determine the stoichiometric ratio of
electrons to Ni2+
 Ni2++ 2e-  Ni(s)
 0.0011 mol e-(1 mol Ni2+)(58.7g Ni/mol Ni)
2 mol e =0.0334 g or 33.4 mg of Ni

1
Electrolysis
Metallic magnesium can be made by the
electrolysis of molten MgCl2. How many
minutes are needed to form 10.00 g of Mg
from molten MgCl2 using a 3.50 A current?
 1st: Calculate how many moles of Mg you
need to make.
10.00g Mg*1mol/24.3 g = 0.4115 mol Mg
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Electrolysis
2nd: Determine stoichiometric ratio of e- to Mg
Mg2+ + 2e-  Mg(s)
0.4115 mol Mg* 2 mol e-/mol Mg
=0.8230 mol e3rd: Determine the charge on 0.8230 mol e0.8230 mol e- * 96,500 C/mol =
79423 C
4th: Calculate the time using t=q/I or q * 1/I
t = 79423C * 1 sec/3.5C * 1min/60sec
= 378.2 min
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Electrolysis Practice
An unknown metal (M) is electrolyzed. It took
52.8 sec for a current of 2.00 amp to plate
0.0719 g of the metal from a solution containing
M(NO3)3.
What is the metal?
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AP Exam Practice
 2011
#3(d-g) (Fuel Cell question)
 2010 #6(c-g)
 2009B #6
 2008 #3
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