Electrochemistry

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Electrochemistry
--the branch of chemistry that deals with the
relationships between electricity and chemical
reactions.
Rust
Batteries
Corrosion
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Oxidation-Reduction (Redox)
Reactions in which electrons are transferred
between substances
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Before we continue, we must review some things we
already learned about oxidation s and redox
reactions.
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Again, oxidation reduction reactions occur when
there is a transfer electrons from one species to
another in a reaction.

• If one reactant gains electrons another must lose
  electrons.
• Reduction is always accompanied by oxidation.

5
Determine Oxidation of red element in each of
the following
MnO2 4 KMnO4 7
Br2 0 HClO4 7
CaH2 -1 SO42- 6
6

• An atom that becomes more positively charged is
  oxidized.
• This is due to loss of e-.
• The gain of electrons by an atom is called
  reduction.

OIL -- RIG
Oxidation Involves Loss -- Reduction Involves
Gain
Leo the lion says Ger
Loss of electrons oxidation -- Gain of electrons
reduction
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Identify oxidation s of each of the following,
along with what is being oxidized and what is
being reduced.
2 Fe O2 ? 2 FeO
0
0
2
-2
oxidation
reduction
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26-2 0 26-2
0 CuSO4(aq) Zn(s) ?
ZnSO4(aq) Cu(s)
Oxidized
Reduced
What are the products?
What are the charges on each species?
What is oxidized and what is reduced?
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Redox Reactions

• oxidizing agent
• substance that causes oxidation by being reduced
• reducing agent
• substance that causes reduction by being oxidized.

26-2 0 26-2
0 CuSO4(aq) Zn(s) ?
ZnSO4(aq) Cu(s)
Reducing agent
oxidizing agent
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Direct Contact Redox Reaction
We will be determining whether or not reaction
between two substances is spontaneous or requires
energy in a bit.
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Voltaic Cell
Same reaction as previous slide, but external
wire is used to transfer electrons.
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loses mass
gains mass
reduction
oxidation
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• You must be able to remember
• Oxidation occurs at the anode, reduction occurs
  at the cathode. (Vowel goes with vowel, Consonant
  goes with consonant.)
• 2) Electrons flow from the anode to the cathode.
  (Since oxidation occurs at the anode, electrons
  must be lost at anode.)
• We consider the anode to be the negative
  electrode. (This is because it provides the
  negative charge potential its not an actual
  charge actual is 0.)
• Electrons arent actually coming directly from
  the anode. They are actually being pushed by the
  anode because of the higher potential that
  exists. (Think of a garden hose analogy.)

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http//www.chem.iastate.edu/group/Greenbowe/sectio
ns/projectfolder/flashfiles/redox/home.html
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• Draw a voltaic cell for the following reaction.
• In your drawing
• label the anode and the cathode
• indicate the sign of each electrode
• indicate direction of electron flow
• indicate where oxidation and reduction is
  occurring by drawing ions and arrows
• indicate which electrode is losing mass and which
  electrode is gaining mass.

Fe Sn2 ? Fe2 Sn
Homework 20.13-20.15 (Make drawings like above
for each.)
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Electromotor Force (EMF) (Cell Potential
Ecell Cell voltage) --a measure of the driving
force needed to push electrons through the
external circuit of a voltaic cell. --exists
because there is a difference in potential energy
between two electrodes in a voltaic cell. --the
difference in potential energy per electrical
charge is measured in units of volts. (positive
value for all reactions that proceed
spontaneously.)
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For now, we will focus on voltaic cells which
exist under standard conditions. 1 atm
25C 1 M Standard emf Standard cell
voltage Eºcell
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Ecell 1.10 V
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The total voltage of a cell depends on the
cathode and anode half-cells involved in the
reaction. We can calculate the standard
potential of a cell without actually doing any
experiments. The voltage that occurs when each
metal is reduced is tabulated in a data
table. (Given on the AP Exam)
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Table lists the voltages related to
reduction. Notice the signs.
Handout Table of Standard Reduction Potentials
STANDARD HYDROGEN ELECTRODE
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Ecell Ered(cathode) Eox(anode)
Because the values in the table are all listed as
reduction potentials and oxidation occurs at the
anode, not reduction, we must reverse the value
to represent reduction.
If we rearrange this equation, we can use the
values directly from the table.
Ecell Ered(cathode) - Ered(anode)
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Ecell Ered(cathode) - Ered(anode)
The cell potential equals the standard reduction
potential of the cathode reaction minus the
standard reduction potential of the anode
reaction.
If Ecell is , it is spontaneous. If it is
negative, energy must be supplied for the
reaction to occur.
Handout Table of Standard Reduction Potentials
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Cu2 (aq) 2e- ? Cu (s) E, V
0.340
Zn2 (aq) 2e- ? Zn (s) E, V -
0.763
Zinc is being oxidized in the reaction, but the
table lists the voltage associated with
reduction. (The negative sign fixes this for us.)
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Ecell Ered(cathode) - Ered(anode)
Ecell (0.340 V) - (- 0.763)
Ecell 1.103 V Spontaneous
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Where does reduction occur? Cathode or
Anode The table lists reduction potentials. The
more positive the voltage (higher on table), the
more likely it is for the element involved to be
reduced. Therefore, the cathode would be the
reaction listed higher in table. When given two
half reactions, if anode and cathode are not
known, we assign the half reaction with the more
positive potential as the cathode because it is
more likely to be reduced.
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Determine the standard cell potential (Emf) of
the following reaction in acidic
solution. Cr2O72- (aq) I- (aq) ? Cr3 (aq)
I2(s) (Must balance each half reaction first
keep as half reactions do not cancel electrons)
Cathode Cr2O72- 14H (aq) 6e- (aq) ? 2Cr3
(aq) 7H2O(s) Anode 6I- (aq) ? 3I2(s)
6 e-
Ecell Ered(cathode) - Ered(anode)
Ecell (1.33 V) - (0.54 V)
Ecell 0.79 V
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Ag e- ? Ag E 0.80 V Cu2 2e-
? Cu E 0.34 V Will Ag react with Cu2
spontaneously? Yes No Will Cu react with
Ag spontaneously? Yes No
Silver must be oxidized, which would make E
-0.46V
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Gibbs Free Energy Gibbs free energy is a quantity
related to the stability of a physical or
chemical system. The change in Gibbs free
energy (?G), which we will be calculating, allows
us to determine whether the overall process of a
reaction will be spontaneous. In the
thermodynamics chapter, Gibbs free energy is
dependent on the change in enthalpy (?H), the
temperature (T), and the change in randomness of
a system, or entropy (?S). ?G ?H T?S
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A negative value for ?G indicates a spontaneous
reaction. A positive value for ?G indicates that
energy is required for the reaction to occur. A
value of 0 indicates an equilibrium between two
systems.
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Relationship Between Emf and Gibbs Free Energy In
this chapter, ?G and Emf are related by the
following equation ?G - nFE n number of moles
of electrons transferred F (Faradays constant)
96,500 J V-1mol-1 Look at this equation on the
formula sheet. Because n and F are positive, a
Emf (spontaneous reaction) will produce a ?G
(spontaneous reaction).
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Calculate the standard free energy change for the
reaction below and determine spontaneity Zn (s)
2Ag (aq) ? Zn2 (aq) 2Ag (s) We must
figure out the values of n and E. ECell
Ecathode - Eanode Eºcell 0.80 V (-0.76 V)
1.56 V We already know this will be spontaneous
because of the positive value of E.
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In the reaction, two electrons are transferred
from Zn to 2Ag, which is evident by the half
reactions Zn (s) ? Zn2 (aq) 2e- (s) 2 Ag
(aq) 2e- ? 2 Ag (s) This tells us that n
2. We now substitute the values in our
equation ?G -nFE ?G -(2)(1.56 V)(96,500 J V-1
mol-1) ?G -3.01 x105 Joules/mol Usually given
in KJ/mol, so it equals -301 KJ
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Calculate the standard free energy change in
Kilojoules/mol for the reaction below Fe (s)
Cu2 (aq) ? Fe2 (aq) Cu (s) ?G -151 KJ
Read through sample exercise 20.10 on page 798 of
your textbook. Try the practice exercise.
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In all of the previous examples, we considered
standard conditions. What is the molarity of a
solution under standard conditions? 1.0
M. Consider the following cell, which we looked
at earlier.
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The reaction starts as a standard solution.
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Ag
Cu2
What happens to the concentration of Cu2 and Ag
as the reaction proceeds?
What do you think will happen to the voltage as
this occurs?
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The overall Emf (voltage) will decrease as the
reaction proceeds.(Just like in a battery)
Before we discuss the equation that allows us to
calculate this voltage change, we must discuss
the Reaction Quotient
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Reaction Quotient This equation allows us to
relate the concentrations of reactants and
products to the forward or reverse push of the
reaction.
The brackets indicate molarity. Only the ions
are considered because they contribute to the
concentration of the cells.
As previously learned, the lower case
superscripts indicate the coefficients from the
balanced equation.
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Determine the reaction quotient for the following
reaction
16H (aq) 2MnO4- (aq) 5C2O42- (aq) ?
2Mn2(aq) 8H2O(l) 10CO2(g)
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Determine the reaction quotient for the following
reaction
Cu (s) 4H (aq) 2NO3- (aq) ? Cu2
(aq) 2NO2 (g) 2H2O (l)
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Once we determine the reaction quotient, we can
determine the cell voltage using the Nernst
equation given to you on the AP Exam.
R gas constant (8.31 J mol-1 K-1)T
temperature in KelvinsQ thermodynamic reaction
quotientF Faraday's constant (96,500 J V-1
mol-1) n number of electrons transferred ln
natural logarithm (see LN on calculator)
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Consider the following voltaic cell.
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• What is the voltage of the cell under standard
  conditions?
• What is the voltage of the cell if after 10
  minutes the starting concentration of Ag is 0.80
  and 1.20 Cu2?

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Under standard conditions, we use the simple
equation
Ecell Ered(cathode) - Ered(anode)
Ecell (0.800 V) - (0.340 V)
Ecell 0.460 V
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Now that we know the standard cell voltage, we
must determine the reaction quotient of the
reaction.
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We must now use the Nernst Equation.
E .452 V Notice the voltage decreases as the
cell runs over time. (anode concentration
increases and cathode concentration decreases).
E .460
E .452
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1) What is the voltage of the cell under
standard conditions? 2) What is the voltage
of the cell if the starting concentration of Cu2
is 1.50 M and the concentration of Zn2 is 0.500
M.
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Under standard conditions, we use the simple
equation
Ecell Ered(cathode) - Ered(anode)
Ecell (0.340 V) - (-0.760 V)
Ecell 1.10 V
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Now that we know the standard cell voltage, we
must determine the reaction quotient of the
reaction.
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We must now use the Nernst Equation.
E 2.51 V Notice the voltage is higher if the
concentration of the anode is lower and
concentration of cathode is higher.
E 2.51 V
E 1.10 V
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Notice that the ion being reduced (cathode) is
the reactant As the reactant increases, the cell
voltage increases.
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Notice that on your formula sheet, there is a
second equation directly to the right of the
Nernst equation.
This is a simplified version of the first
equation using the base 10 logarithm instead of
the natural logarithm. We would get the same
values for both of the previous questions if we
used it. Because R is a constant, and because
the standard temperature didnt change, (25 C or
298 K), we can simplify the Nernst equation,
which gives the equation above. If only the
concentration changes, use this equation if you
find it easier.
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The base 10 logarithm (log) of any number is the
power which 10 must be raised to equal the
number. Ex log of 1000 3 because 103 equals
1000. The natural logarithm is the power to
which e (value of 2.71828) must be raised to
equal the number. E2.303 10 therefore, the
natural logarithm of 10 2.303. 2.303
(8.31)(298) / 96,500 .0592 which is given in
the simplified equation.
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So far, we have learned about Voltaic
Cells Driven by spontaneous chemical
reactions We will now discuss Electrolytic
Cells Requires electrical energy from external
source Non-spontaneous
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Electrolytic Cell --e- are provided by an
external source (dc power supply or
battery). --the power supply acts as an electron
pump, forcing e- onto cathode. --unlike in a
voltaic cell, the cathode acquires a negative
charge. --power supply also pulls electrons from
the anode making it positive.
Oxidation still occurs at anode
Reduction still occurs at cathode
-

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Electrolysis (electro refers to electricitylysis
refers to the breakup of something) --the
chemical breakdown of a substance due to
electricity Example electrolysis of sodium
chloride NaCl(l) ? Na(l) Cl2(g) Electroplating
and Purification of Metals There are two types
of electrodes in electrolytic cells
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Inactive or Inert Electrodes electrodes simply
provide a surface for oxid. and red. to
occur. (Example Electrolysis of water into H2
and O2.)
Active Electrodes electrodes actually participate
in the chemical processes. (Example
Electroplating of metals with other metals.)
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http//www.science.uwaterloo.ca/leroy/c123/Ch21/L
32-5.jpg
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Can we break down NaF (aq) into sodium metal and
chlorine gas by the process of electrolysis? Why
or why not?
aqueous ions water Must consider whether or
not the electrolysis of water is more likely to
occur over the electrolysis of the ions.
Possible reactions at the cathode Na(aq) e-
? Na (s) Ered -2.71 V 2H2O(l) 2e- ?
H2(g) 2OH-(aq) Ered -0.83 V
The more positive the Ered the more favorable
the reduction. The reduction of H2O ? H2 occurs
instead of Na ? Na(s).
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Possible reactions at the anode 2F-(aq) ? F2
(g) 2e- Ered 2.87 V 2H2O(l) ? O2(g)
4H(aq) Ered 1.23 V
Because oxidation is the reverse of
reduction The more negative the Ered the more
favorable the oxidation. The oxidation of H2O ?
H occurs instead of F- ? F2(g).
The actual salt, NaF, simply provides an
electrolyte which assists the movement of
electrons throughout the cell.
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Read through sample exercise on page 788 of
textbook. Try practice exercise.
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The battery charger forces the galvanic cell to
run backwards, turning it into an electrolytic
cell. Cu(s) Zn2 ? Cu2 Zn(s)
Reduction takes place at the Zn cathode (-)
electrode.Oxidation takes place at the Cu anode
() electrode. Cu2 2 e Cu(s)  Eo
0.34 voltsZn2 2 e   Zn(s)  Eo -0.76
volts
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Nickel Plating Steel
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Quantitative Aspect of Electrolysis
Na e- ? Na 1 mol e- needed to plate out 1
mol of Na metal
Cu2 2e- ? Cu 2 mol e- needed to plate out
2 mol of Cu metal
Al3 3e- ? Al 3 mol e- needed to plate out
3 mol of Al metal
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The charge on 1 mol of electrons is 96,500
Coulombs As stated earlier in the chapter, this
is called a Faraday. 1 F 96,500 Coulombs I
q/t (Given on Exam) I is current in amps, q is
charge in coulombs, t is time. 1 Coulombs amp x
sec
Common Type of Exam Question 10.0 amperes of
current is passed through an electrolytic cell
filled with molten lithium chloride, for 500 s.
How many grams of Lithium are collected at the
cathode?
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10.0 amperes of current is passed through an
electrolytic cell filled with molten lithium
chloride, for 500 s. How many grams of Lithium
are collected at the cathode?
Li e- ? Li Coulombs Amp x sec C 10.0 x
500 s 5000 C
1 mol e-
1 mol Li
6.491 g Li
5000 C
96,500 C
1 mol e-
1 mol Li
0.360 g Li
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Calculate the mass of aluminum produced in 1.00
hr by the electrolysis of molten AlCl3 if the
electrical current is 10.0 A.
Al3 3e- ? Al I q/t so, q I x t so
Coulombs Amp x sec C 10.0 x 3600 s 36000 C
1 mol e-
1 mol Al
27.0 g Al
36000 C
96,500 C
3 mol e-
1 mol Al
3.36 g Al
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2.65 g of silver are collected at the cathode in
an electrolytic cell filled with aqueous silver
nitrate. How long did it take to collect this
much silver if a constant current of 0.25 A was
applied?
Ag e- ? Ag Coulombs Amp x sec 1 mol e-
96,500 C
1 mol Ag
1 mol e-
96,500 C
1 Amp x s
2.65 g Ag
107.87 g Ag
1 mol Ag
1 mol e-
1 C
0.25 amp
9484 seconds or 158 minutes
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How long does 4.00 A have to be applied to an
electrolytic cell containing aqueous silver
nitrate to collect 9.5 g of silver at the cathode?
Ag e- ? Ag Coulombs Amp x sec 1 mol e-
96,500 C
1 mol Ag
1 mol e-
96,500 C
1 Amp x s
9.5 g Ag
107.87 g Ag
1 mol Ag
1 mol e-
1 C
4.00 amp
2124 seconds or 35.4 minutes