Electrochemistry
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the study of the interchange of chemical and electrical energy
involves 2 types of redox reactions
o generation of electrical current from a spontaneous chemical reaction
o use of current to produce a chemical reaction (opposite of 1st one)
REVIEW OF REDOX PROBLEMS
OIL-RIG
Oxidation Is the Loss of electrons
Reductions Is the Gain of electrons
LEO the lion says GER
Lose Electrons in Oxidation
Gain Electrons in Reduction
Galvanic Cells (20.3)
Harness the energy of this reaction...
Galvanic cell
a device in which chemical energy is changed to electrical energy
spontaneous redox rxn producing current can be used to do work
Salt bridge/porous disk
tube connecting two solutions
needed to keep each compartment with a net charge of zero by
migration of ions not electrons
Anode
compartment where oxidation occurs (both begin with vowels)
An Ox--oxidation at the anode
Cathode
compartment where reduction occurs
Red Cat--reduction at the cathode
Current
flow of positive charge (opposite to the flow of electrons)
Cell potential (E) /Electromotive force (emf)
the driving force or pull of the electrons
--measured in volt (1 V = 1 Joule/Coulomb of charge)
Voltmeter
draws current through a known resistance – some waste from
frictional heating  measurement less than the maximum
cell potential -- digital voltmeters now used with
negligible current loss
Potentiometer
draws current without waste – variable
works against cell potential – opposite of voltmeter
Electrochemistry (Chapter 20)
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Example of a Galvanic Cell:
Standard Reduction Potentials
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all half reactions are given as reduction processes in standard tables – at 1 atm for gasses & 1M for solutes
– the degree sign indicates at standard state
the standard hydrogen potential is the reference potential against which all half reactions potentials are
assigned
when a half reaction is reversed, the sign of E is reversed
when a half reaction is multiplied by an integer, the E remains the same
a galvanic cell runs spontaneously in the direction that gives a positive value for E
the larger the E, the more spontaneous is the reaction
sometimes there are only ions involved in the redox reaction; in these cases, an inert conductor, such as Pt
must be used
Sample Problem 1- Follow example 20.5 on pg 766.
a. Consider a galvanic cell based on the reaction
Cu2+ + Zn  Zn2+ + Cu
The half-reactions are
Cu2+ + 2e-  Cu(s)
E = +0.34 V
Zn2+ + 2e- Zn(s)
E = - 0.76 V
Give the balanced cell reaction and calculate E for the cell. Label the galvanic cell above with the following: Anode, cathode,
direction of electron flow.
Electrochemistry (Chapter 20)
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b. A galvanic cell is based on the reaction MnO4- (aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn2+(aq) + H2O(l).
The half-reactions are
MnO4- + 5 e’ + 8 H+  Mn2+ + 4 H2O
E = 1.516V
ClO4- + 2 H+ + 2 e’  ClO3- + H2O
E = 1.19V
Give the balanced cell reaction and calculate the E for the cell.
Line Notation
- anode components on the left, cathode components on right
- separated by vertical parallel lines (indicating salt bridge or porous disk)
- phase boundary represented by a single vertical line
- anode material on far left, cathode material on far right
- catalysts are at the far ends separated by single vertical lines if only dissolved materials are present
Example:
Pt(s)  ClO3-(aq) , ClO4-(aq)  MnO4-(aq) , Mn2+(aq)  Pt(s)
Sample Problem 2 – Describe completely the galvanic cell based on the following half reactions under standard conditions:
Electrochemistry (Chapter 20)
Ag+ + e-  Ag
E = 0.80 V
(1)
Fe3+ + e-  Fe2+
E = 0.77 V
(2)
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Cell Potential, Electrical Work, and Free Energy
potential difference = emf (electromotive force) = voltage (V) = work (J) / charge (C)
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work depends on the push (thermodynamic driving force) behind the electrons
the driving force is the emf: potential difference between 2 points
w = -q E
w = work (J)
E =cell potential, emf
q = nF
q = quantity of charge transferred (C)
n = moles of electrons
F = Faraday’s constant = 96,485 C/mol of electrons
At standard conditions, Go = -nF E o
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
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the maximum cell potential is directly proportional to the change in free energy between the reactants and
products in the cell
this equation confirms that a galvanic cell runs in the direction that gives a positive cell potential
the E o of the cell correlates to a negative G, which means the reaction is spontaneous
In real spontaneous processes, some energy is wasted; work never is the maximum possible if any current is
flowing; lost to frictianl heating
efficiency = w / wmax x 100
Sampkle Problem 17.3: Calculate Go for the following reaction: Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)
Is the rxn spontaneous?
Smapel Problem 17.4: Preidct whether 1 M HNO3 will dissolve gold metal to form 1 M Au3+ solution.
Electrochemistry (Chapter 20)
4
The Dependence of Cell Potential on Concentration
If the cell concentration is not equal, use Le Chatelier’s principle to predict the change that will take place.
Ex: Cu(s) + 2 Ce4+(aq)  Cu2+(aq) + 2 Ce3+(aq)

as the concentration of Ce4+ increases, the reaction will proceed to the products’ side and increase the driving
force on the electrons
Sample Problem 17.5: For the cell reaction 2 Al(s) +3 Mn2+(aq)  2 Al3+(aq) + 3 Mn(aq) E o cell= 0.48 V
Predict whether the E o cell is larger or smaller than in the following cases.
a. [Al3+] = 2.0 M, [Mn2+] = 1.0M
b. [Al3+] = 1.0 M, [Mn2+] = 3.0M
Concentration Cells
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

Galvanic cells can be made where the concentration in each compartment is different but with the same ions
Nature tries to make both sides equal, so electrons will go to the side with the higher concentration
In these types of cells, the voltage is small – concentration is the only factor producing the cell potential
Ex: Figure 17.9
Sample Problem 17.6: Determine the direction of electron flow and designate the anode and cathode for the cell represented in Figure
17.10.
The Nerst Equation
Electrochemistry (Chapter 20)
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