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Electrochemistry  review and
quiz
Chemical reactions involving transfer of electrons are called oxidation
and reduction reactions or redox reactions. When properly
set up, these reactions generate power in a
Battery, or galvanic cell.
A galvanic cell consists of at least two half cells, each of which
consists of an electrode and a electrolyte solution.
A redox reaction can be divided into two
Half Reactions,
an oxidation reaction and a reduction half reaction.
Each half reaction can be set up as a
Half Cell
and putting two half cells together makes a simple battery.
A galvanic cell can be considered as a battery, but batteries
include packages of galvanic cells in series to supply higher voltages
than a single galvanic cell.
Oxidation of a species (atom, ion, or molecule) is a loss of electron(s),
and reduction of a species a gain of electron(s). Some conventions
are used to define the
Oxidation States,
and oxidation of an atom causes a increase in its oxidation state.
Conversely, reduction causes a decrease in its oxidation state.
Balance Redox Equations
is a complicated task, but the use of half
reactions gives a very good strategy for balancing redox reactions.
Energy is the driving force for chemical reactions.
Energies of oxidation and reduction reactions are related to the
electrochemical potentials (E),
Cell EMF,
of the galvanic cells. The standard reduction
potentials for half cells are values that make sensible comparisons,
because a set of conventions have been followed. The energy (called
Gibb's free energy G) is the electric work of the reaction. Therefore,
DG^{o} =
n F DE^{o}.
The difference of Gibb's free energy (DG^{o})
between products and reactants equals to the charge n F times the
potential difference DE^{o}.
Since n electrons are transferred in the equation, n F
(F is the Faraday constant of 96485 C) is the charge involved
for the equation in terms of moles.
The equilibrium constant K is related to DG^{o},
DG^{o} =  R T ln K.
For the reaction,
a A + b B = c C + d D
The
Nernst equation
is a natural result,
R T [C]^{c}[D]^{d}
DE = DE^{o}   ln 
n F [A]^{a}[B]^{b}
Notations Used
In the summary given above, we have used many energy and potential
related notations. For completeness, the notations are given
as follows:
DG  difference in Gibb's free energy
DG^{o}  difference in Gibb's free energy at standard condition
E^{o}  standard reduction cell potential
E^{*}  standard oxidation cell potential
emf or EMF  Electromotive force
DE^{o}  standard potential of a battery or galvanic cell
DE  Potential of a galvanic cell not necessary at standard conditions
F = 96485 C, Faraday constant
Q  charge, unit coulomb (C)
I = Q / t, current Ampere
R = 8.3145 J/mole.K, gas constant
T  temperature in K
ln(10) R T / F = 0.0592 at 298 K
Electrochemistry skills expected in
Quiz

Explain the meaning of terms used in electrochemistry and properly use
the cell notations.
For example, galvanic cells or voltaic cells are devices
that use spontaneous chemical reactions to produce electric
currents. Other terms are reduction potentials, oxidation potentials,
oxidation state, redox reactions, cathode, anode, cell potentials,
Nernst equations, etc.
The definition for anode and cathode apply to both galvanic
cells and electrolytic cells.
Oxidation takes place on anodes, for example,
reaction: Zn > Zn^{2+} + 2 e E^{*} = 0.762
notation: Zn  Zn^{2+} E^{*} =  E^{o}
Note that we have defined and used E^{*} to represent the
oxidation potential which has the same absolute value as the reduction
potential E^{o}, but a different sign. This notation
is not used in most text books.
Reduction takes place on cathodes, for example,
reaction: Hg_{2}^{2+} + 2 e > 2 Hg E^{0} = 0.796
notation: Hg^{2+}  Hg
The details regarding oxidation and reduction half reactions have been
given on the page of
HalfCell reactions.

Calculate DE^{o} and
DE for galvanic cells
according to the conventions used for cell notations.
The standard cell potential DE^{o}
is the sum of the reduction potential of the cathode and the oxidation
potential of the anode,
DE^{o} = E^{o} + E^{*}.
The data for calculating DE^{o}
may be given using cell notations or in the forms of equation.
If a cell notation is given, you must be able to search for
proper reduction and oxidtation couples for the cell and then
evaluate E^{o}'s from a table of standard reduction
potentials. Proper use of such tables is a basic skill, and some
usages have been given in the modules
Battery or Galvanic Cells, and
Electromotive Force.,

If the cells are not at standard conditions, you will be required to
calculate the Gibb's free energy, DG,
or the cell potential DE.
The standard Gibb's free energy, DG^{o},
is the negative of the maximum available electrical work.
The electrical work W_{e} is the product
of the charge and the potential of the cell.
W_{e} = q E
DG =  q E
=  n F E
where, q is the charge in C and E is the potential in V.
Note also that q = n F, and F = 96485 C is the
Faraday constant, whereas n is the number of moles of electrons in
the reaction equation.
The change of Gibbs free energy at standard condition is derived in the
following way:
DG^{o} =  W_{e}
=  q E^{o}
=  n F E^{o};
This equation is for a galvanic cell under the standard conditions.
The notations E^{o} and DG^{o}
are used for cells under standard conditions.
When the cells are not at standard conditions,
DE and DG^{o}
are used for these quantities.
Note also that DG^{o} is the
theoretical amount of energy per reaction equation as written.
The amount of energy for a given system depends on the quantities of
the reactants and their conditions (concentration or pressure etc.).

The Nernst equation allows us to evaluate the cell potentials when the
condition of the cell is not at standard conditions. The details have
been given on the page of
Nernst equation.

For the calculation of Gibb's energies and quantities generated in
Electrolysis
and electroplating, skills in solving stoichiometric
problems are also required.
Confidence Building Questions
 How many minutes will be required for a 1.5 A current to electroplate
1.97 g of gold (at. mass, 197.0)?
Hint...
1 mol Au 1 mol e 96485 C 1 s 1 min
1.97 g Au      = ? min
197 g Au 1 mol Au 1 mol e 1.5 C 60 s
Answer 10.72
Consider...
The reaction is
Au(CN)^{} + e = Au + 2 CN^{}
In exams, the reaction
Al^{3+} + 3 e = Al
is usually used.
Pay attention to the number of electrons involved in the reaction.
 For the Daniell cell,
Zn  Zn^{2+}  Cu^{2+}  Cu,
Calculate the dE^{0}.
Look up data from a table of standard reduction potential.
You need to have the skill.
That is why no data is given.
Hint...
Did you get these data?
Zn > Zn^{2+} + 2 e E^{*} = 0.762
Zn  Zn^{2+} Note: E^{*} = E^{0}
Cu > Cu^{2+} + 2 e E^{*} = 0.339
Cu  Cu^{2+} Note: E^{*} = E^{0}
dE = E^{0} + E^{*}
Note sign for convention? Calculate dE^{0}
Answer 1.10 V
Consider...
The reaction is Zn + Cu^{2+} = Zn^{2+} + Cu
 For the Daniell cell,
Zn  Zn^{2+}  Cu^{2+}  Cu,
dE^{0} = 1.10 V. Calculate the K_{c} for the reaction,
Cu^{2+} + Zn = Zn^{2+} + Cu
Hint...
At equilibrium, dE = 0;
dE^{0} = ^{0.0592}/_{2}logK_{c}
K_{c} = antilog 37.2 = ?
Answer 2e37
Consider...
You may be asked to calculate
[Zn^{2+}] / [Cu^{2+}] = ?
or
[Cu^{2+}] / [Zn^{2+}] = ?
The large K_{c} value means that the reaction is almost
quantitative. The Zn metal almost causes all Cu^{2+} ions to
deposit as copper metal.
Cu^{2+} + Zn = Zn^{2+} + Cu
 For the Daniell cell,
Zn  Zn^{2+} (0.010 M)  Cu^{2+} (1.0 M)  Cu,
Calculate dE
Hint...
Assume you've done the previous problems.
dE = dE^{0}  ^{0.0592}/_{2}log 0.01/1.0
= ?
Answer 1.16
Consider...
The voltage is higher than the standard potential of 1.10 because
Zn^{2+} concentration is less than 1.0 M in this case.
The Nerns equation enables us to give a quantitative value
when the conditions change.
 For the Daniell cell,
Zn  Zn^{2+}  Cu^{2+}  Cu,
dE^{0} = 1.10 V. Calculate dG^{0} in J
Hint...
dG^{0} =  n F dE^{0}
=  2 * 96485 * 1.10 = ?
Answer 212267 J
Consider...
More often, you'll see 212 kJ rather than 212000 J
 For the Daniell cell,
Zn  Zn^{2+} (0.010 M)  Cu^{2+} (1.0 M)  Cu,
Calculate dG in J
Hint...
From dE = dE^{0}  0.0592/2 log K = 1.26 V
dG =  n F dE =  2 * 96485 * 1.16 = ?
Answer 223845 J
Consider...
dG^{0} =  212 kJ; dG for the said condition is 224 kJ.
Concentration makes a difference.

What is the dE^{0} for the reaction
Pb(s) + Co^{2+} = Pb^{2+} + Co(s)?
Hint...
The purpose of this problem is to ask you to search the table
for the proper reduction potentials.
If you a dE^{0} = 0.151 V, you are probably right.
The negative potential indicates that the reaction should be reversed.

What is the equilibrium constant for the reaction
Pb(s) + Co^{2+} = Pb^{2+} + Co(s)?
Hint...
Use the Nernst equation to derive the equilibrium constant,
If you get a value of K = 7.6x10^{6},
you have acquired the skill.
We have used mostly the Zn/Cu cell in these questions. A similar
set of questions can be set up using any two couples of reduction
potentials.
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